I heard a wise person say: The purpose of a conversation or a discussion is not to try and convince that you are correct and they are wrong or vice versa, but to allow them to hear your point of view and you their point of view.
That is much needed in this time of comment mudslinging... thank you. It even goes a step further IMO: it's not even trying to convince one another anymore, it's about speaking the loudest to drown out the others.
A long time ago when I was in high school I was on a debating team. There we patiently waited while the other side made their case, we then responded, then the other side got a chance to rebut those statements and then we did the same. That has been lost somehow and now they don't even listen to what the other side has to say and as you indicated continue to interrupt and drown out the other side.
IN SHORT, escape velocity is the velocity that, at some distance from the centre of a star or planet, is sufficient to escape the gravitational pull of that star or planet to reach that famous point-of-no-return. Given a certain escape velocity, during that trip to the point-of-no-return no additional energy is needed to get there. N.B. Kinetic energy = 1/2 . m . v(esc)
@@freddythamesblack8479 I have exposed Einstein. I have outsmarted Einstein. I have surpassed Newton and Einstein. BALANCE AND completeness go hand in hand. The following is consistent with WHAT IS E=MC2 AND F=ma. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is gravity. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/AS) inertia/INERTIAL RESISTANCE, AS ELECTROMAGNETISM/energy is gravity; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. This is consistent with what is E=MC2 AND F=ma. Now, ON BALANCE, carefully consider what is the fully illuminated AND setting/WHITE MOON. E=MC2 IS F=MA. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution. ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE). Great !!!! WHAT IS GRAVITY is an INTERACTION that cannot be shielded or blocked ON BALANCE !!! GREAT !!! BALANCE AND completeness go hand in hand. TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE) !!! WHAT IS E=MC2 is F=ma. CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky (ON BALANCE). Consider TIME AND time dilation ON BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution. By Frank Martin DiMeglio The gravity of WHAT IS THE EARTH/ground is BALANCED WITH and by WHAT IS E=MC2 (AND TIME). Indeed, TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. Great. By Frank Martin DiMeglio TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE. WHAT IS E=MC2 is taken directly from F=ma. The rotation of WHAT IS THE MOON matches the revolution. The stars AND PLANETS are POINTS in the night sky ON BALANCE. Consider TIME AND time dilation ON BALANCE. ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE). By Frank Martin DiMeglio WHAT IS E=MC2 is taken directly from F=ma, as the rotation of WHAT IS THE MOON matches the revolution; as TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). The stars AND PLANETS are POINTS in the night sky ON BALANCE. Great. It is proven. WHAT IS E=MC2 is taken directly from F=ma, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky. Consider TIME (AND time dilation) ON BALANCE. Consider WHAT IS THE EYE ON BALANCE. Great. Consider what is the fully illuminated (AND setting/WHITE) MOON ON BALANCE. WHAT IS E=MC2 is taken directly from F=ma. Great. TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). By Frank Martin DiMeglio WHAT IS E=MC2 is taken directly from F=ma, AS the rotation of WHAT IS THE MOON matches the revolution; AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE); AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS WHAT IS GRAVITY is an INTERACTION that cannot be shielded or blocked ON BALANCE; AS c squared CLEARLY represents a dimension of SPACE ON BALANCE. Great. INDEED, consider what is the fully illuminated (AND setting/WHITE) MOON ON BALANCE; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE); AS WHAT IS E=MC2 is taken directly from F=ma. GREAT !!! By Frank Martin DiMeglio
escape velocity has NOTHING to do with REALITY as we know it.. it is SOLELY for a HYPOTHETICAL object.. THROWN out to space... not SELF-PROPELLED rockets and machines. "Escape Velocity' for a rocket, is whatever speed it has when LIFTING OFF the Earth!!! or CENTIMETERS per Second!!!! in other words, it doesnt need to go fast... it just needs to KEEP GOING..... at ANY VELOCITY!!!!!
Throw a rock up in the air and it will always come back down. Imagine a rocket to be a rock. They will always come back down unless they move fast enough to either go into orbit or get away from the Earth. The question is what is that velocity? (answer: the orbital velocity or the escape velocity)
im seriously confused ..to me that analogy is terrible a rock does not have a propulsion system ..my question is ..theoretically what would happen to a rocket with alot (infinite even) of fuel travelling away from earth at under escape velocity..... ? the mind naturally thinks it would have to continue onwards and escape earth... if not why ? doesnt gravity lessen with distance from earth
Shouldn't gravity be strongest at the surface and weaker the further you go away from the surface? If very little energy is needed to throw a rock up into the air and velocity is irrelevant to the rock temporarily leaving the surface, why is velocity even a question for escaping earth? Shouldn't a rocket be able to just fly up, experience less gravitational pull the further it goes with available fuel being the only factor necessary to reach the vacuum of space?
yes, i agree with other comments... “escape velocity” sounds absurd, according to physics we see every day... if the rocket is fast enough to get off ground = fast enough to escape gravity.. ON the surface where the pull is supposed to be stronger!! but when it goes 300 miles up.. it requires 40,000 km/h to “escape” that WEAKER gravity??? why??? and at such speeds,, why dont the rockets burn up, when they burn at 20,000 km/h re-entry!!!
@@adamfirst3772 Couple of thing : 1) rockets dont burn up because they are well outside the thickest atmosphere by the time they reach those speeds. On the way back in, you are falling back down and are at those speeds when you reach thick atmosphere. 2) Your intuition is correct. Escape velocity is not some magical speed something must reach to escape the Earth, but rather the hypothetical instantaneous velocity some projectile that has no means to thrust itself would need to be given to escape. The escape velocity of any celestial body is from the surface. For Earth, it is about 11.2 km/s. This does not mean rockets have to reach 11.2 km/s to escape Earth. It simply means that this is the instantaneous velocity something would need to be given from the surface of the Earth and ignoring drag and assuming it is given no further propulsion, it will escape. As your intuition says, the actual escape velocity decreases as your distance from Earth increases due to the gravity getting weaker, so there is nothing in physics preventing something with propulsion (like a rocket) escaping at any speed. For example, if you had the fuel, you could escape the Esrth never going faster than 1 mph. Youd need a lot of fuel because that would be a pretty inefficient rocket, you would eventually be far enough away that the escape velocity was less than 1 mph at which point you could turn the engines off and would have escaped never having gone faster than 1 mph. The theoretical escape velocity quoted simply does not account or allow for powered flight.
Willoughby Krenzteinburg thanks for your informative and detailed response... however, it raises more questions.. probably due to my ignirance more than my curiosity... 1.. in the falcon heavy launch.. it shows the rocket reaching mach3 within the same altitude as the titanium alloy SR71 flight ceiling... where it is said to ELONGATE by half a foot!! due to heat caused by friction.. im assuming the same atmosphere would do wonders with steel and alluminum rockets sealed with RUBBER o-rings which are claimed to be “critical” components, as in the SS Challenger... 2. thanks for verifying my assumptions regarding “escape velocity” and its irrellavance to self-propelled rockets... however, i still dont get the 11.2 km/s figure you mentioned... for example, a fly, which takes off from the ground, surely is not receiving an “instantaneous velocity” of 11.2 km/s???! as i stated earlier.. wouldnt ANY velocity, that gets an object off the ground, be a sufficient to “escape gravity,”?. at what point does that HUGE number factor in?? am i missing something?.
It seems you're still interested in replying to comments, so I want to ask you this as I've been unable to find a good answer to it. How does escape velocity work, in terms of non negligible forces, when gravity acts with an essentially infinite radius? If you were to escape a planet in a bare universe at any X velocity, shouldn't the force of gravity always win?
not to nitpick, though I'm sure you love to disprove those who do, but kinetic energy can never be infinite. However, given an infinite playing field, potential energy can be. So how come kinetic energy has the ability to overpower the potential energy?
The gravitational force does indeed reach out to "infinity", but it becomes weaker with increasing distance ( 1/R^2) and the force becomes so weak with increasing distance that the force cannot slow down a rocket enough, that leaves the Earth with sufficient velocity, to stop it completely. That sufficient velocity is called the escape velocity. Another way to look at it is this: The faster a rocket goes the farther it can reach before gravity pulls it back (you can do that with tossing a ball upwards). Eventually if the rocket goes fast enough it will never come to a complete stop.
@@almscurium Lol, crazy. This was 3 years ago. It was really interesting for me! A lot of the advanced concepts are only taught to math and engineering majors... The physics I did was not calculus based, so... It can be good and bad depending on the attitude you go into it with I think.
Is is possible to escape the earth's gravity/orbital at a lower speed so long as you apply a continuious force to oppose the earth's gravition pull but don't increase or at least don't dramatically increase your speed? Much like a balloon filled with hydrogen or helium but instead of working of buoyancy it operates under its own proprelent force like a rocket or techonolgy that is yet to be discovered? Will you eventually pull away from earth's gravity at a speed below escape volcity or will this simply keep you at a cirtian orbital distance/radius until you accelatrate faster to either send you into a higher orbit or then to or past escape volcity to break free from the earth's orbit. It seems very clear, not accounting for air friction that once escape volocity is achieved no further energy is required to escape the earth's gravitional pull. Although it seems very confusing as to when a continuous force is applied to send you away from earth but not achieve anywhere near escape volocity what exactly would happen. If you could maybe do a video explaining this that would be helpful to alot regular people like me who are confused about it
Yes, you don't technically need to ever achieve escape velocity to escape the earth's gravity. At least not surface escape velocity, which is 11.2km/s. But this decreases with distance from earth, so at the altitude where the moon orbits, it's already only 1.4km/s (www.wolframalpha.com/input/?i=sqrt%282*mass+of+earth*G%2Fradius+of+moon%27s+orbit%29). So it is sufficient to just continuously push with higher acceleration than whatever the current g is, until you get to about the moon's orbit, achieve velocity 1.4km/s and you have left the earth's gravitational pull. But of course, you will have spent much more fuel with this approach, so it's not really practical.
I'm no physicist, in fact a lot of these equations are over my head!; but i just want to talk from an observation stand point. When i see a rocket launching into space i know that its not travelling 25,000 mph yet its still beating earth's gravity (even with all the weight its carrying!!). Just to understand Escape velocity, is it an equivalent of the force needed from a rocket to generate such speed? and also does the EV need to be maintained till it escapes the earth's gravity or just initially so that there is enough kenetic energy to reach orbit?
Zim Zimmer, that observation is probably correct when the rocket is getting 'up to speed' initially upon launch. But you have to consider that the velocity will increase to the point where it really does need to travel that fast to leave earths orbit! I find it hard to believe too, but when you look at a spacecraft moving in space, you don't really have a reference point to see just how fast it's really travelling! Kind regards.
But surely, theoretically, if a rocket launches and maintains a speed of 100mph it would eventually leave earth's gravitational pull so it would never have to reach that escape velocity?
I don't understand escape velocity. The further from the centre of the Earth, the less the Earth's gravitational force therefore the escape velocity should be constantly changing. As your distance gets further and further from Earth, you're required less and less velocity to escape because Earth's gravitational force keeps getting weaker and weaker as you get further and further therefore how can we define 1 specific excape velocity? I've watched 6 escape velocity videos now and this seemingly simple question/confusion/misunderstanding of mine is yet to be answered...
The escape velocity decreases as you move away from Earth, so your intuition is correct. The escape velocity of a celestial body if no distance from that celestial body is explicitly stated is simply from the surface of that celestial body. The 11.2 km/s escape velocity of the Earth is from the surface. The escape velocity of the Earth from 10,000 km away from Earth is less than that. In fact, you should be able to imagine that if you launched a projectile straight up, it would immediately start slowing down due to gravity of Earth pulling it back down. If it were launched from the surface at exactly escape velocity - or about 11.2 km/s, then obviously it would start slowing down, but never reach zero and start falling back down. At any point along the way, you would note that the velocity is decreasing all the time, and ignoring all other factors including other gravitational sources and drag, you could say that the projectile is at all times traveling at the exact escape velocity of Earth from that location relative to Earth. Obviously, this must necessarily be the case since the projectile will indeed escape, and it doesn't travel at a constant 11.2 km/s as it is escaping. It is always slowing down. So, when it is traveling 5.5 km/s, the escape velocity from that distance from Earth is exactly 5.5 km/s. When it is traveling 1 cm/s, the escape velocity from that distance from Earth is exactly 1 cm/s, and so on. There is a common misconception that the escape velocity of 11.2 km/s is some magical speed a rocket has to reach in order to escape the Earth, and that's not what escape velocity is. Escape velocity is just a theoretical velocity at which a projectile will escape the Earth if given this velocity instantaneously and no other forces will act on it (other than gravity with respect to the body to which the escape velocity is attributed). In other words, if you wanted to fire a bullet so that it escaped the Earth, and you fired it from the surface (also ignoring things like drag), then you'd have to fire that bullet at 11.2 km/s. Because the bullet will never receive any other force. However, if you wanted to put that bullet in a rocket and send that rocket so far away that the escape velocity is only 5 km/s, then you could shoot that bullet at 5 km/s from that distance and it would still escape. Rockets are under constant power (or can be), so it is possible to escape the Earth while never reaching anywhere close to 11.2 km/s - because you can just ride that rocket at 5 km/s constantly until you reach a point where the escape velocity has decreased to 5 km/s and turn the engines off. This would be highly inefficient, but doable. In fact, in theory - you could ride a rocket that travels no faster than 1 m/s and just ride that rocket so high that you reach a point where the escape velocity is under 1 m/s. You will have escaped Earth having never traveled any faster than 1 m/s.
Dear Michel van Biezen, just one doubt: If, hypothetical we build a indestructible sphere (with total mass of just 100 mg, I don't know made of carbyne or still don't known material) of 1m^3 "fill" with interestelar vaccum (10^-17 Pa), how higher these sphere can go out with boyound forces? Like the dream theorically of vaccum airships or evacuated airships of NASA. If we build these sphere at space at same interestelar vaccum would be easier, like fill a ballon in air than the botton of ocean, these sphere could be use to lift things from earth to space (assume even the rope is made of carbyne)? How weight it could support?
But, theorically, these special ballon wich I mentioned, how higher could be reach? Nasa say that much higher than Helium balon, but not specify how much. And, if we make a "ballon" outer space, could be used to anchor things to move up?
The height depends on the size of the balloon and on the size of the payload. You could not reach space since there is no atmosphere to push the balloon any higher.
A Quick and (Somewhat) Dirty Way to Calculate Escape Velocity - I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius of each of these bodies, ASSUMING THE ACCELERATION DUE TO GRAVITY REMAINED CONSTANT DURING THE FALL. Escape velocity is the minimum velocity needed to escape a gravitational field. For example, escape velocity on earth, Vesc = 40,270 km/h (given). Using the simple formula V2 = (2ar)^0.5, where V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of earth (initial velocity = 0), and a = 9.80665 m/s^2 r = 6,378,000 m V2 = (2*9.80665*6378000)^0.5 = 11,184.53 m/s = 40,264.29 km/h This value of 40,264 km/h compares very closely to the escape velocity of 40,270 km/h I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body. Moon: Escape velocity = Vesc = 8,533.6 km/h a = 1.62 m/s^2 r = 1,737,150 m V2 = (2*1.62*1737150)^0.5 = 2,372.418 m/s = 8,540.7 km/h Again, a very close fit. Mars: Escape velocity = Vesc = 18,108 km/h a = 3.72761 m/s^2 r = 3,389,500 m V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s = 18,097 km/h Another very close approximation. I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth). I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known. Sources: www.livescience.com/50312-how-long-to-fall-through-earth.html keisan.casio.com/exec/system/1360310353 www.wolframalpha.com/input/?i=escape+velocity+Mars
@@Alexandru.Popescu Nothing. You are confused. Orbital velocity _of a particular body_ is related to the escape velocity _of that same body._ The orbital velocity of the Earth itself around the sun is not related to the escape velocity of Earth. However, the orbital velocity of the Earth around the sun is related the SUN'S escape velocity from that same distance.
Most equations and derivations in physics at the lower division level, ignore air resistance. For applications in space flight, air resistance must be taken into account.
Can someone explain why he substituted radVM/R as 7900? Is that orbital velocity at earth's surface? Because I'm confused why that's a known number. Why would an object ever being orbiting earth at its surface?
Yes I understand that, but isn't that only the orbital velocity for that SPECIFIC radius? As in, 7900m/sec is only the orbital velocity at the surface of earth? I'm confused how that's possible since nothing orbits earth at its surface
+MutatedGamer Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
So escape velocity is the initial speed one needs to have from the surface of an object to escape the gravitational pull of that object (assuming you are in a vacuum and only factor in gravity). However if one could fly an aircraft in a vacuum at Mach1 you could still reach the moon from Earth it would just take an insane amount of energy because of gravitational drag?
We are assuming the when the initial velocity (escape velocity) is reached, the engines are turned off. (There isn't enough fuel to keep the engines going the entire time)
1) You do not have to escape the Earth to reach the moon. The moon clearly hasn't escaped the Earth (since it is in orbit around Earth), ergo reaching the moon wouldn't require escaping Earth. 2) Your intuition is correct. You would have to go a lot farther away than just the moon, but there is nothing preventing a rocket or anything that has some sort of fuel supply that can propel itself from escaping the Earth at any speed. As long as you still have fuel. You still have to reach "escape velocity", but escape velocity is not a constant. The velocities quoted for various celestial bodies are from their surface. In reality, there is an escape velocity from any arbitrary distance from any body, and that escape velocity decreases with distance, so if you got on a rocket and went 1 mph away from Earth, you would eventually reach a point where the escape velocity is under 1 mph, and you can turn off the engines as you are exceeding escape velocity, and never went faster than 1 mph.
This is kind of misleading. Escape velocity still applies to rockets and the like. I mean, rockets must still reach escape velocity to escape the Earth. The issue is that escape velocity doesn't allow for powered flight. The calculated escape velocity simply assumes all that velocity is given instantaneously, and then the rest is just a ballistic trajectory. Rockets can burn their engines and get themselves into a position where the escape velocity is lower. They still need to reach that escape velocity though. For example, a rocket could escape the Earth at 1 mph. Eventually, it would be far enough away that the escape velocity is under 1 mph, and it could then cut off the engines. It still had to reach escape velocity, so it very much applies to rockets.
I understand the way that you found velocity using kinetic and potential energy, but thinking about this problem from a kinematics perspective has me confused. From my logic, in order to escape the pull of gravity, your force applied would have to be greater than the force of gravity right? So because force is equal to mass times acceleration, as long as your acceleration is greater that the acceleration of gravity g you should escape the pull of gravity, which i know is wrong. Why is it instead an escape velocity and not an escape acceleration? I'm trying to find the flaw in my logic of thinking because i know its wrong.
Your way of thinking about it is correct. If you continue to apply a force greater than your weight (which would diminish as you move farther away from Earth), you will eventually get away from the gravitational force of the Earth. But you would have to apply that force for a very long time and distance. Better to speed up the the escape velocity and then turn off the engines (or you will run out of fuel).
The notion of escape velocity assumes that some theoretical projectile is given some instantaneous velocity and no other forces (other than gravity) are ever exerted on it. That's escape velocity. From the surface of the Earth (and ignoring drag from the air), the velocity you would need to give some object after which you leave that object at the sole mercy of gravity alone is 11.2 km/s. This does not in any way mean that something MUST reach 11.2 km/s to escape the Earth. Obviously, we don't launch our rockets with instantaneous velocities; we accelerate them up to speed. There is nothing preventing you from propelling yourself out of Earth's gravity under constant power - and doing so at speeds much lower than escape velocity. A "rule" of escape velocity is that once given that initial velocity, no further propulsion is allowed.
But the denominator of the orbital velocity is R which equals the radius of Earth and the height, which means that orbital velocity is not constant and the escape velocity is not equal root two times orbital velocity, am I right?
No. The orbital velocity depends on the height of the orbit, that is correct. The escape velocity will always be sqrt(2) times the orbital velocity from the location of the orbit.
Michel van Biezen Does/can(without intentionally attempting to) helium reach escape velocity? I'm sure by the time you've collected the information you need you'll be able to formulate questions that I wouldn't have dreamed of asking. So just remember to right down the questions that come to mind during this endeavour because I’m sure we'd love to hear it all, that is if you are willing and up to the challenge...
Well based on pure calculations I have a formula to find the variation in 'g' with depth For a body Above the surface of the earth at height h, g is given by- 9.81/ {(Radius of earth + h) ÷ Radius of earth}^2 For a body inside the earth at depth d, g is given by- 9.81 x (Radius of earth - d) ÷ Radius of the earth 9.81 is the value of 'g' at the surface of the earth
@@MichelvanBiezen Can you provide me the link, also how do I calculate the time period of a body in space which stops rotating in elliptical path and is falling into sun
All physics problems at the undergraduate level ignore air resistance. (which is typically fine as it offers a good approximation and it makes the problem a lot easier). The plan is to cover higher level physics after we complete many of the topics we are still working on.
This makes no sense to me. If I was able to fly straight up for any amount of time eventually I will "escape" the earth's gravitational pull no matter how fast or slow I go as long as I keep putting distance between me the surface. Am I wrong here?
If you continue to provide fuel to the engines in order to keep pushing the rocket further out. Spaceships don't have enough fuel to do that. They just carry enough fuel to give the spaceship enough speed to escape Earth. They then turn the engines off and float through space. (That is how the Apollo missions made it to the Moon).
Catalyst - your logic is sound. Escape velocity is merely a theoretical threshold. It is a property of Earth's gravity. If you had the means, you could indeed propel yourself out of Earth's gravitational well at any speed. The quoted "escape velocity" of Earth is merely the instantaneous velocity a projectile would need if launched from the _surface_ of Earth - to never fall back down. Since Earth's gravity decreases as you get farther away, this escape velocity decreases as well. About the average peak velocity for the Apollo missions was about 34,200 ft/s (23,318 mph or 10.4 km/s). This is the fastest they every went on the way to the moon, and while they aren't on the surface when they reached this speed, they were only a little over 100 miles away, so technically the escape velocity from that altitude is only 11.0 km/s. Apollo missions never reached escape velocity; they didn't need to. A mission to the moon does not require escaping Earth gravity. Even while standing on the surface of the moon, you are still very much under the influence of Earth's gravity; the moon's influence is just stronger. The moon is still well within Earth's gravitational well, so a visit to the moon does not require escaping Earth. Back to the original point, if you had the fuel, you could fire up a rocket and go 10 m/s and never go any faster. You would eventually reach a distance where the escape velocity was under 10 m/s at which point you could turn off the engine. This would be a horribly inefficient rocket, and it couldn't actually exist, but in principal, your logic is correct.
The vast majority of the viewers have expressed that they like this format. My students also prefere the small lectures where each one covers a specific topic or problem. So we stuck with this approach.
It is not a matter of what speed it needs to start at, but what speed you can build up to. To reach orbital speed you must reach a speed of 18,000 miles per hour (29,000 km/hour). Theoretically (ignoring wind resistance), you can reach space if you had that initial velocity.
If "space" is 100 km up, you ignore air resistance and assume the projectile starts with the instantaneous speed it needs : v = √(2gh) = √(2 * 9.8 m/s² * 100,000 m) = √1,960,000 = 1,400 m/s (3,132 mph)
thanks for the video.. i have a few questions that pop to mind if you could help me with them... ...why couldnt the rocket keep at the same speed as TAKE-OFF?? obviously its enough to overcome gravity FROM THE SURFACE... which, im assuming, stronger than a few 100 miles up?? no?. if the rocket keeps propeling upwards, gravity shouldnt be getting stronger, and pulling back down?! no? ....if escape velocity is around 40,000 kph.... and re-entry is 20,000 kph.. why do things burn up coming down, but not burn when going up... AT TWICE THE SPEED.. IN THE SAME ATMOSPHERE??? thanks in advance..
Re-entry is about 40,000 km/h. By the time the rocket accelerates to higher velocities, it has reached the upper regions of the atmosphere where it is really thin and doesn't cause extreme temperatures like during re-entry.
Think of it in terms of escape energy. Sure, you could escape earth's gravity going two miles per hour constantly, that would require less force per period of time over a longer time interval. Integrating would yield the same escape energy. That's how I think about it anyway
I have a question about escape velocity. No matter how fast you go you have a limited amount of kinetic energy and even though gravity weakens as you go farther away from Earth the gravitational force never goes to zero which means after a long enough time wouldn't anything traveling away from Earth eventually come back unless it was moving at an infinite speed?
That is the concept of the escape speed. If you are traveling at a speed greater than the escape speed, the rocket will continue to slow down forever, but will never come to a complete stop, as it will continue to move away forever, although at a slower and slower speed.
If the speed is equal to the escape speed, yes. if the speed is much greater than the escape speed, then the final velocity will be much greater than zero
Why are you able to say theres no kinetic nor potential energy in the final state. Whered all the energy go? Surely there’s still velocity at the final state right? KE?
By definition, the escape velocity is the velocity necessary to break free from the Earth's gravitational force, in such a way that when you reach an infinite distance the velocity will be zero. Therefore, there will be NO KE and NO PE.
Because small g is actually decreasing the further away you are from the center of gravity. Small g = GM/(r^2). Big G is the gravitational constant, big M is the mass of the source of gravity (Earth, in our case), and r is the distance between the object and the center of gravity. As you can see, as the distance r tends toward infinity, small g gets smaller and smaller and approaches zero, which makes PE also approaches zero.
Amandeep Kumar because both move away from the center of the planet, it does not matter if it was 90 degrees or 10 degrees from the horizon. both bodies will be moving away from the center.
Yes it does affect the velocity - the higher you go, the lower the escape velocity from that height. Mt. Everest won't make much of a difference though. You don't need to reach escape velocity to escape the Earth - this is a misconception. You only need to reach escape velocity if you will never give yourself any further propulsion. If you have a way to power yourself (like with a rocket), you can escape the Earth at any speed you wish. Theoretically, you could build a ladder to infinity and climb away from Earth at 1 mph and you could successfully escape the earth. The idea of "escape velocity" is that instantaneous velocity needed that - if you were to never apply any other propulsive force to it every again - would still escape.
@@willoughbykrenzteinburg first and foremost, I'm an idiot, and I've been drinking. I have a hypothesis that if you could construct an aircraft that could provide the initial thrust to put a spacecraft into the upper atmosphere, then eject said spacecraft (in upper earth atmo) that can propel itself using some type of ion drive that works both in atmo and vacuum, would it be possible that such a craft to exit earths gravity? Again, I've been drinking and I don't know what my brain is up to at the moment lol
That would be equal to the escape velocity of the Sun. Ve = sqrt(2) * sqrt(GM/R) where M is the mass of the Sun and R is the radius of the Sun (in meters).
@@MichelvanBiezen That's not right. If something fell into the sun from an infinite distance away starting at zero velocity, it would impact the sun at the sun's surface escape velocity, but from just 150,000,000 km away, it would be less than the sun's escape velocity. Just logically, you could deduce that something that fell from a distance farther from Earth would hit the sun harder than the Earth would by simply pointing out that by the time that object reached the Earth-distance from the sun, it would already have some non-zero velocity.
If the distance from the Earth to the rocket at infinity is infinite shouldn't we also have infinite potential energy? Why does it goes to 0? Like if potencial energy = m g h shouldn't that be an insanely large number at the very least?
As "h" increases, "g" decreases. As h approaches infinity, g approaches zero. It doesn't matter how big h is. If it's all multiplied by zero, the answer is zero.
I haven't made those types of videos. (That gives us a good idea to make some like that). It would be similar to finding the electric field due to a ring of charge. Take a look at: Physics - E&M: Electric Field (8 of 16) Ring of Charge
Ray Kay Ray, You are correct. The force of gravity will always slow down an "escaping" object no matter how far away. However sine the force of gravity is proportional to the square of the distance, the farther the object the smaller the force of gravity. When the force is very small, the acceleration is very small, and if it is too small compared to its velocity, the object will never stop.
+Ray Kay Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
So this formula is abstract ? We cant apply this in real cases, because we are in a solar sistem, not just 2 bodys interacting with eachother, we have the moon, the sun... right ?
+Tom Siddle I made a mistake in my previous comment. The initial (gravitational) potential energy is actually (GMm)/Re... where Re is the distance of the object from the centre of the earth.However, the "g" (9.8m/s^2) variable in the simplified potential energy equation (mgh) is actually the number you get when you multiply GM/Re^2. Pretty cool!! I'm not the too knowledgable on the subject but I hope that helps!!
A weather balloon filled with helium can reach space, as demonstrated in a famous youtube video of a legoman going into space, so what is the relevance of escape velocity, when you can clearly escape at a lower speed?
oh, seems like this was already answered below. Escape velocity is the speed needed if there is no additional energy added to the system. Anyhow, we should be taking helium balloons into space and then turning on the rockets later. We would save a ton of fuel.
If a balloon gets you to space your velocity would be very small. To get away from the pull of the Earth, you would still need to reach a velocity of 25,000 miles per hour.
Actually, a weather balloon does not reach space (which starts at about 65 miles (100 km). At that altitude the air is too thin to provide the buoyancy force needed to push a balloon that high.
1) The helium balloon didn't reach space 2) Getting to space is not "escaping the Earth". The MOON is 240,000 miles away, and it hasn't escaped the Earth. The Apollo astronauts that went to the moon did not escape the Earth. 3) The only reason rockets go to space is to avoid most of the drag air causes. Otherwise, they'd be able to orbit anywhere higher than any obstacles on Earth (higher than the mountains). Rockets need all that fuel NOT so they can get high - but so they can get fast. They need to reach upwards of 17,000 miles per hour LATERALLY across the surface to get into an orbit. Launching from a higher altitude does not add any horizontal speed, so they'd need just about the same amount of fuel. Certainly any savings in fuel would be more than offset by the cost of the gargantuan mechanism it would take to reliably and safely lift those rockets up to that altitude in the first place.
The guy that jumped out of the redbull stratos, Felix Baumburger or something, was nowhere near the edge of space, even though it looked that way. He was only about a quarter the way.
Why can you not travel at a low speed (all the way up) to escape earth's gravity? Say if you had a ship that could ascend at a rate of 50 feet a second or something, and do it all the way till earth's gravity was zero. Why is that not possible??
It is theoretically possible, but it would require so much fuel that the rocket would be so heavy you would barely get it off the ground. It is much more efficient to build up a lot of speed quickly and that will require much less fuel.
It is possible, and in fact - you wouldn't have to do this all the way until Earth's gravity was zero. Though, it's important to point out that there is no point where Earth's gravity is zero - since it technically reaches to infinity, BUT obviously there would be a point where the gravity of some other body was stronger than Earth's. Escape velocity is simply the instantaneous velocity something needs to never fall back down from a specific body. This varies depending on how far away from the body you already are. From the surface of Earth, it is about 11.2 km/s and this completely ignores drag. This is the instantaneous velocity something would need to escape Earth without further propulsion. There is nothing preventing something from simply traveling much slower than this under constant power. You could climb a ladder out of Earth's gravity well theoretically. As for your example, you wouldn't have to travel 50 ft/s forever. If you had some mechanism that constantly powered you and maintained this 50 ft/s, you would eventually reach a point where this 50 ft/s exceeded the escape velocity. For the Earth, this would be about 2.1 billion miles (distance somewhere between Uranus and Neptune's distance to the sun). Once you got this far away, you could shut your engine down and you'd never return. As has been pointed out though, this would take an incredible amount of fuel. You're better off just building up as much velocity as quickly as possible.
why we can't escape from earth even if we do not move with escape velocity. we can get away from earth easily since gravitational pull will be decreasing. so why we need escape velocity. please answer
The escape velocity is the theoretical velocity needed to hurl an (inanimate) object away from a larger object (e.g. a rock from the surface of the earth) out to where gravity is no longer acting. To fire a rock from the surface to space without any additional on-board propulsion you need to achieve Mach 33 at sea level. It is therefore much easier to send a rocket up with its own on-board propulsion to reach an altitude where its orbital velocity has significantly increased and the escape velocity has diminished to where it can move to a place where it can remain without needing further propulsion to stop it tumbling back to Earth (i.e. "escaped" Earth gravity).
in order to make the faulty logic work... the real question would be.. why do we need escape velocity?? escape velocity is the force needed to GET OFF GROUND.. for the MASS of the given object.. and CONTINUE to apply that same propulsion until far enough from the surface... one would think that force would DECREASE the further the rocket got from the center of the surface/planet.. NOT INCREASE TO 25,000 miles/ hr!!!!
Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
" If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity." You're comparing velocity (v) to the force of gravity (F = ma, where a is the rate of change of v)
first and foremost, I'm an idiot, and I've been drinking. I have a hypothesis that if you could construct an aircraft that could provide the initial thrust to put a spacecraft into the upper atmosphere, then eject said spacecraft (in upper earth atmo) that can propel itself using some type of ion drive that works both in atmo and vacuum, would it be possible that such a craft to exit earths gravity? Again, I've been drinking and I don't know what my brain is up to at the moment lol
i guess what im asking is does this formula only affect relative vertical EV or can it be reduced by following the curvature of the earth using centripetal force?
so is the earth flat ? just kidding.i am troubled by all of the flat earth people . we really need some flights over the south pole ,so these people can understand .i think it is the only way they would believe .although some claim that flight would just be flying in a circle. i point out the north star and none have replied yet . i feel bad for them.they have been lied to and trick so much that even the shape of the planet is hard for them to believe .what is a simple way to really set them straight in a nice way?
escape velocity has NOTHING to do with REALITY as we know it.. it is SOLELY for a HYPOTHETICAL object.. THROWN out to space... not SELF-PROPELLED rockets and machines. "Escape Velocity' for a rocket, is whatever speed it has when LIFTING OFF the Earth!!! or CENTIMETERS per Second!!!! in other words, it doesnt need to go fast... it just needs to KEEP GOING..... at ANY VELOCITY!!!!!
The explanation about escape-speed that the initial speed can take an object to as a far distance as infinity away from a gravitational field or influence of the Earth is totally nonsensical It is so obviously absurd to say so simply because the Earth's gravitational force can never reach anywhere farther beyond this solar system. By saying that an object from the Earth has to get such an initial speed that it can continue traveling to infinity to escape the Earth's gravitational force , you guys are automatically proving yourselves to be so ignorant as idiots that you guys cannot even know that the Earth's gravitational force can influence only objects within a certain limited space within the solar system , but not even the big lion share of the solar system, and the only form of existence in the solar system which has influence on all other forms of existence within the solar system is the Sun Each of the forms of existence within the solar system alone can never ever have any influence on any other form of existence outside the solar system within this galaxy. Only the whole solar system with the interreactions among the Sun and all other forms of existence , and the interactions among all other forms of existence in a continuous chain of interreactions can have a certain influence on the nearby systems within this galaxy, and the whole galaxy system can have interractions with nearby galaxies which are still within the limit of distance from this galaxy It seems that all you guys have been doing is repeating after, like a monkey or parrot , whatever were said so long ago by scientists of the previous centuries when human knowledge of science is far much poorer than it is today , instead of activating your idle minds to see the flaws and defective holes of such obsolete notions of the old worlds
WTF? I mean, thank you for teaching something about 'pure physics' that wasn't clear to me yet. I mean that. But still, WTF?????????? I am pretty sure that you don't have to travel to infinity away from the earth before another gravity well (i.e. another planet, a star, a black hole .... a tooth brush another traveler dropped.....) *Sigh .... but hey .. in the mean time maybe you will find out if there are truly no white crows.
I heard a wise person say:
The purpose of a conversation or a discussion is not to try and convince that you are correct and they are wrong or vice versa, but to allow them to hear your point of view and you their point of view.
Why did it become - GmM/R^2 at 1:36
Harish55 Equation for force due to gravity
Harish55 G being a constant
That is much needed in this time of comment mudslinging... thank you.
It even goes a step further IMO: it's not even trying to convince one another anymore, it's about speaking the loudest to drown out the others.
A long time ago when I was in high school I was on a debating team. There we patiently waited while the other side made their case, we then responded, then the other side got a chance to rebut those statements and then we did the same. That has been lost somehow and now they don't even listen to what the other side has to say and as you indicated continue to interrupt and drown out the other side.
IN SHORT, escape velocity is the velocity that, at some distance from the centre of a star or planet, is sufficient to escape the gravitational pull of that star or planet to reach that famous point-of-no-return. Given a certain escape velocity, during that trip to the point-of-no-return no additional energy is needed to get there. N.B. Kinetic energy = 1/2 . m . v(esc)
Fascinating and interesting.
@@freddythamesblack8479 I have exposed Einstein. I have outsmarted Einstein. I have surpassed Newton and Einstein.
BALANCE AND completeness go hand in hand. The following is consistent with WHAT IS E=MC2 AND F=ma. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is gravity. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/AS) inertia/INERTIAL RESISTANCE, AS ELECTROMAGNETISM/energy is gravity; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. This is consistent with what is E=MC2 AND F=ma. Now, ON BALANCE, carefully consider what is the fully illuminated AND setting/WHITE MOON. E=MC2 IS F=MA. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution. ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE). Great !!!! WHAT IS GRAVITY is an INTERACTION that cannot be shielded or blocked ON BALANCE !!! GREAT !!! BALANCE AND completeness go hand in hand.
TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE) !!! WHAT IS E=MC2 is F=ma. CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky (ON BALANCE). Consider TIME AND time dilation ON BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE. Accordingly, ON BALANCE, the rotation of WHAT IS THE MOON matches the revolution.
By Frank Martin DiMeglio
The gravity of WHAT IS THE EARTH/ground is BALANCED WITH and by WHAT IS E=MC2 (AND TIME). Indeed, TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. Great.
By Frank Martin DiMeglio
TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE. WHAT IS E=MC2 is taken directly from F=ma. The rotation of WHAT IS THE MOON matches the revolution. The stars AND PLANETS are POINTS in the night sky ON BALANCE. Consider TIME AND time dilation ON BALANCE. ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE).
By Frank Martin DiMeglio
WHAT IS E=MC2 is taken directly from F=ma, as the rotation of WHAT IS THE MOON matches the revolution; as TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE. c squared CLEARLY represents a dimension of SPACE ON BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE). The stars AND PLANETS are POINTS in the night sky ON BALANCE. Great. It is proven.
WHAT IS E=MC2 is taken directly from F=ma, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE).
CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky. Consider TIME (AND time dilation) ON BALANCE.
Consider WHAT IS THE EYE ON BALANCE. Great. Consider what is the fully illuminated (AND setting/WHITE) MOON ON BALANCE. WHAT IS E=MC2 is taken directly from F=ma. Great. TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE, AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE).
By Frank Martin DiMeglio
WHAT IS E=MC2 is taken directly from F=ma, AS the rotation of WHAT IS THE MOON matches the revolution; AS ELECTROMAGNETISM/energy is CLEARLY (AND NECESSARILY) proven to be gravity (ON/IN BALANCE); AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS WHAT IS GRAVITY is an INTERACTION that cannot be shielded or blocked ON BALANCE; AS c squared CLEARLY represents a dimension of SPACE ON BALANCE. Great. INDEED, consider what is the fully illuminated (AND setting/WHITE) MOON ON BALANCE; AS TIME is NECESSARILY possible/potential AND actual ON/IN BALANCE; AS ELECTROMAGNETISM/energy is CLEARLY AND NECESSARILY proven to be gravity (ON/IN BALANCE); AS WHAT IS E=MC2 is taken directly from F=ma. GREAT !!!
By Frank Martin DiMeglio
1:36 when you show up to a party with your homeboy then spend the whole party flirting with your crush
lmao
escape velocity has NOTHING to do with REALITY as we know it..
it is SOLELY for a HYPOTHETICAL object.. THROWN out to space... not SELF-PROPELLED rockets and machines.
"Escape Velocity' for a rocket, is whatever speed it has when LIFTING OFF the Earth!!! or CENTIMETERS per Second!!!!
in other words, it doesnt need to go fast... it just needs to KEEP GOING..... at ANY VELOCITY!!!!!
@@adamfirst3772 what the hell are you on about, this is a jokey haha comment sir
James Eastgate
i put that under your comment by mistake... my bad.
@@adamfirst3772 all good vro have a good day
Throw a rock up in the air and it will always come back down.
Imagine a rocket to be a rock. They will always come back down unless they move fast enough to either go into orbit or get away from the Earth. The question is what is that velocity? (answer: the orbital velocity or the escape velocity)
im seriously confused ..to me that analogy is terrible a rock does not have a propulsion system ..my question is ..theoretically what would happen to a rocket with alot (infinite even) of fuel travelling away from earth at under escape velocity..... ? the mind naturally thinks it would have to continue onwards and escape earth... if not why ? doesnt gravity lessen with distance from earth
Shouldn't gravity be strongest at the surface and weaker the further you go away from the surface? If very little energy is needed to throw a rock up into the air and velocity is irrelevant to the rock temporarily leaving the surface, why is velocity even a question for escaping earth? Shouldn't a rocket be able to just fly up, experience less gravitational pull the further it goes with available fuel being the only factor necessary to reach the vacuum of space?
yes, i agree with other comments... “escape velocity” sounds absurd, according to physics we see every day...
if the rocket is fast enough to get off ground = fast enough to escape gravity.. ON the surface where the pull is supposed to be stronger!!
but when it goes 300 miles up.. it requires 40,000 km/h to “escape” that WEAKER gravity??? why???
and at such speeds,, why dont the rockets burn up, when they burn at 20,000 km/h re-entry!!!
@@adamfirst3772 Couple of thing :
1) rockets dont burn up because they are well outside the thickest atmosphere by the time they reach those speeds. On the way back in, you are falling back down and are at those speeds when you reach thick atmosphere.
2) Your intuition is correct. Escape velocity is not some magical speed something must reach to escape the Earth, but rather the hypothetical instantaneous velocity some projectile that has no means to thrust itself would need to be given to escape. The escape velocity of any celestial body is from the surface. For Earth, it is about 11.2 km/s. This does not mean rockets have to reach 11.2 km/s to escape Earth. It simply means that this is the instantaneous velocity something would need to be given from the surface of the Earth and ignoring drag and assuming it is given no further propulsion, it will escape. As your intuition says, the actual escape velocity decreases as your distance from Earth increases due to the gravity getting weaker, so there is nothing in physics preventing something with propulsion (like a rocket) escaping at any speed. For example, if you had the fuel, you could escape the Esrth never going faster than 1 mph. Youd need a lot of fuel because that would be a pretty inefficient rocket, you would eventually be far enough away that the escape velocity was less than 1 mph at which point you could turn the engines off and would have escaped never having gone faster than 1 mph. The theoretical escape velocity quoted simply does not account or allow for powered flight.
Willoughby Krenzteinburg
thanks for your informative and detailed response...
however, it raises more questions.. probably due to my ignirance more than my curiosity...
1.. in the falcon heavy launch.. it shows the rocket reaching mach3 within the same altitude as the titanium alloy SR71 flight ceiling... where it is said to ELONGATE by half a foot!! due to heat caused by friction.. im assuming the same atmosphere would do wonders with steel and alluminum rockets sealed with RUBBER o-rings which are claimed to be “critical” components, as in the SS Challenger...
2. thanks for verifying my assumptions regarding “escape velocity” and its irrellavance to self-propelled rockets...
however, i still dont get the 11.2 km/s figure you mentioned... for example, a fly, which takes off from the ground, surely is not receiving an “instantaneous velocity” of 11.2 km/s???!
as i stated earlier.. wouldnt ANY velocity, that gets an object off the ground, be a sufficient to “escape gravity,”?. at what point does that HUGE number factor in?? am i missing something?.
Deserve more views......Thank you very much for posting out the 20 videos about this topic.Really cleared my mind.Thanks.
A really well explained and short video. Loved every bit. Thank you, Sir.
Thankyou sir
Love from India ❤️
Thank you and welcome to the channel!
@@MichelvanBiezen
I love the way you teach
I love science ❤️
We appreciate your comment. We are glad that the videos are inspiring you to love science.
It seems you're still interested in replying to comments, so I want to ask you this as I've been unable to find a good answer to it. How does escape velocity work, in terms of non negligible forces, when gravity acts with an essentially infinite radius? If you were to escape a planet in a bare universe at any X velocity, shouldn't the force of gravity always win?
not to nitpick, though I'm sure you love to disprove those who do, but kinetic energy can never be infinite. However, given an infinite playing field, potential energy can be. So how come kinetic energy has the ability to overpower the potential energy?
The gravitational force does indeed reach out to "infinity", but it becomes weaker with increasing distance ( 1/R^2) and the force becomes so weak with increasing distance that the force cannot slow down a rocket enough, that leaves the Earth with sufficient velocity, to stop it completely. That sufficient velocity is called the escape velocity. Another way to look at it is this: The faster a rocket goes the farther it can reach before gravity pulls it back (you can do that with tossing a ball upwards). Eventually if the rocket goes fast enough it will never come to a complete stop.
Love physics. Can not wait until I go to college to learn quantum physics/mechanics :-). To bad my ninth grade class does not have physics :(
+Tristen Ess So nice to hear! :)
Inteon Nothing How was physics in college😐😐?
@@yaredarcher3774 I want to know too
@@almscurium Lol, crazy. This was 3 years ago. It was really interesting for me! A lot of the advanced concepts are only taught to math and engineering majors... The physics I did was not calculus based, so... It can be good and bad depending on the attitude you go into it with I think.
Is is possible to escape the earth's gravity/orbital at a lower speed so long as you apply a continuious force to oppose the earth's gravition pull but don't increase or at least don't dramatically increase your speed?
Much like a balloon filled with hydrogen or helium but instead of working of buoyancy it operates under its own proprelent force like a rocket or techonolgy that is yet to be discovered?
Will you eventually pull away from earth's gravity at a speed below escape volcity or will this simply keep you at a cirtian orbital distance/radius until you accelatrate faster to either send you into a higher orbit or then to or past escape volcity to break free from the earth's orbit.
It seems very clear, not accounting for air friction that once escape volocity is achieved no further energy is required to escape the earth's gravitional pull.
Although it seems very confusing as to when a continuous force is applied to send you away from earth but not achieve anywhere near escape volocity what exactly would happen.
If you could maybe do a video explaining this that would be helpful to alot regular people like me who are confused about it
I am googling just now to find the answer to this . it's something I've wondered since I ever heard of escape velocity
Yes, you don't technically need to ever achieve escape velocity to escape the earth's gravity. At least not surface escape velocity, which is 11.2km/s. But this decreases with distance from earth, so at the altitude where the moon orbits, it's already only 1.4km/s (www.wolframalpha.com/input/?i=sqrt%282*mass+of+earth*G%2Fradius+of+moon%27s+orbit%29). So it is sufficient to just continuously push with higher acceleration than whatever the current g is, until you get to about the moon's orbit, achieve velocity 1.4km/s and you have left the earth's gravitational pull. But of course, you will have spent much more fuel with this approach, so it's not really practical.
Your videos are clear and your explanation is so good thanks for that great effort..you saved plenty of students all over the world.
I'm no physicist, in fact a lot of these equations are over my head!; but i just want to talk from an observation stand point. When i see a rocket launching into space i know that its not travelling 25,000 mph yet its still beating earth's gravity (even with all the weight its carrying!!). Just to understand Escape velocity, is it an equivalent of the force needed from a rocket to generate such speed? and also does the EV need to be maintained till it escapes the earth's gravity or just initially so that there is enough kenetic energy to reach orbit?
The Apollo rockets had to go ~25,000 miles per hour in order to get to the Moon.
Zim Zimmer, that observation is probably correct when the rocket is getting 'up to speed' initially upon launch. But you have to consider that the velocity will increase to the point where it really does need to travel that fast to leave earths orbit! I find it hard to believe too, but when you look at a spacecraft moving in space, you don't really have a reference point to see just how fast it's really travelling! Kind regards.
interesting thanks
Only has to achieve escape velocity for an instant then engines can shut off and craft will still escape
But surely, theoretically, if a rocket launches and maintains a speed of 100mph it would eventually leave earth's gravitational pull so it would never have to reach that escape velocity?
I don't understand escape velocity. The further from the centre of the Earth, the less the Earth's gravitational force therefore the escape velocity should be constantly changing. As your distance gets further and further from Earth, you're required less and less velocity to escape because Earth's gravitational force keeps getting weaker and weaker as you get further and further therefore how can we define 1 specific excape velocity? I've watched 6 escape velocity videos now and this seemingly simple question/confusion/misunderstanding of mine is yet to be answered...
The escape velocity by definition, is referenced to the surface of a planet, moon, asteroid, stat, etc.
The escape velocity decreases as you move away from Earth, so your intuition is correct. The escape velocity of a celestial body if no distance from that celestial body is explicitly stated is simply from the surface of that celestial body. The 11.2 km/s escape velocity of the Earth is from the surface. The escape velocity of the Earth from 10,000 km away from Earth is less than that. In fact, you should be able to imagine that if you launched a projectile straight up, it would immediately start slowing down due to gravity of Earth pulling it back down. If it were launched from the surface at exactly escape velocity - or about 11.2 km/s, then obviously it would start slowing down, but never reach zero and start falling back down. At any point along the way, you would note that the velocity is decreasing all the time, and ignoring all other factors including other gravitational sources and drag, you could say that the projectile is at all times traveling at the exact escape velocity of Earth from that location relative to Earth. Obviously, this must necessarily be the case since the projectile will indeed escape, and it doesn't travel at a constant 11.2 km/s as it is escaping. It is always slowing down. So, when it is traveling 5.5 km/s, the escape velocity from that distance from Earth is exactly 5.5 km/s. When it is traveling 1 cm/s, the escape velocity from that distance from Earth is exactly 1 cm/s, and so on.
There is a common misconception that the escape velocity of 11.2 km/s is some magical speed a rocket has to reach in order to escape the Earth, and that's not what escape velocity is. Escape velocity is just a theoretical velocity at which a projectile will escape the Earth if given this velocity instantaneously and no other forces will act on it (other than gravity with respect to the body to which the escape velocity is attributed). In other words, if you wanted to fire a bullet so that it escaped the Earth, and you fired it from the surface (also ignoring things like drag), then you'd have to fire that bullet at 11.2 km/s. Because the bullet will never receive any other force. However, if you wanted to put that bullet in a rocket and send that rocket so far away that the escape velocity is only 5 km/s, then you could shoot that bullet at 5 km/s from that distance and it would still escape. Rockets are under constant power (or can be), so it is possible to escape the Earth while never reaching anywhere close to 11.2 km/s - because you can just ride that rocket at 5 km/s constantly until you reach a point where the escape velocity has decreased to 5 km/s and turn the engines off. This would be highly inefficient, but doable. In fact, in theory - you could ride a rocket that travels no faster than 1 m/s and just ride that rocket so high that you reach a point where the escape velocity is under 1 m/s. You will have escaped Earth having never traveled any faster than 1 m/s.
This guy is under rated
Dear Michel van Biezen, just one doubt:
If, hypothetical we build a indestructible sphere (with total mass of just 100 mg, I don't know made of carbyne or still don't known material) of 1m^3 "fill" with interestelar vaccum (10^-17 Pa), how higher these sphere can go out with boyound forces? Like the dream theorically of vaccum airships or evacuated airships of NASA. If we build these sphere at space at same interestelar vaccum would be easier, like fill a ballon in air than the botton of ocean, these sphere could be use to lift things from earth to space (assume even the rope is made of carbyne)? How weight it could support?
The higher you go in the atmosphere, the less buoyancy you will experience. That is why balloons can only go so high before they stop rising.
But, theorically, these special ballon wich I mentioned, how higher could be reach? Nasa say that much higher than Helium balon, but not specify how much. And, if we make a "ballon" outer space, could be used to anchor things to move up?
The height depends on the size of the balloon and on the size of the payload. You could not reach space since there is no atmosphere to push the balloon any higher.
A Quick and (Somewhat) Dirty Way to Calculate Escape Velocity - I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius of each of these bodies, ASSUMING THE ACCELERATION DUE TO GRAVITY REMAINED CONSTANT DURING THE FALL. Escape velocity is the minimum velocity needed to escape a gravitational field.
For example, escape velocity on earth, Vesc = 40,270 km/h (given).
Using the simple formula V2 = (2ar)^0.5, where
V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of earth (initial velocity = 0), and
a = 9.80665 m/s^2
r = 6,378,000 m
V2 = (2*9.80665*6378000)^0.5 = 11,184.53 m/s = 40,264.29 km/h
This value of 40,264 km/h compares very closely to the escape velocity of 40,270 km/h
I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body.
Moon:
Escape velocity = Vesc = 8,533.6 km/h
a = 1.62 m/s^2
r = 1,737,150 m
V2 = (2*1.62*1737150)^0.5 = 2,372.418 m/s = 8,540.7 km/h
Again, a very close fit.
Mars:
Escape velocity = Vesc = 18,108 km/h
a = 3.72761 m/s^2
r = 3,389,500 m
V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s = 18,097 km/h
Another very close approximation.
I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth).
I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known.
Sources:
www.livescience.com/50312-how-long-to-fall-through-earth.html
keisan.casio.com/exec/system/1360310353
www.wolframalpha.com/input/?i=escape+velocity+Mars
Kral adam.
Thank you. 🙂
What does the orbital velocity of Earth has to do with anything?
It shows the relationship between the orbital velocity and escape velocity.
At 3:03 he says "orbital velocity of the earth" , that would be the speed of the earth in orbit around the sun, right?
That is correct
So... What does the earth speed around the sun have to do with the speed of an object that we want to escape earth's gravity?
@@Alexandru.Popescu Nothing. You are confused. Orbital velocity _of a particular body_ is related to the escape velocity _of that same body._ The orbital velocity of the Earth itself around the sun is not related to the escape velocity of Earth. However, the orbital velocity of the Earth around the sun is related the SUN'S escape velocity from that same distance.
Regardless of the rockets weighing?
Yes the escape velocity is not dependent on the mass of the rocket
Thank you! So clear and thanks for tying in relationship to orbital velocity
So what's with air resistance while in the atmosphere?
Most equations and derivations in physics at the lower division level, ignore air resistance. For applications in space flight, air resistance must be taken into account.
Fair enough...
Can someone explain why he substituted radVM/R as 7900? Is that orbital velocity at earth's surface? Because I'm confused why that's a known number. Why would an object ever being orbiting earth at its surface?
+MutatedGamer The orbital velocity = sqrt (GM/R) = sqrt (6.67x 10^-11 * 5.98 x 10^24 / 6.378 x 10^6) = 7900 m/sec
Yes I understand that, but isn't that only the orbital velocity for that SPECIFIC radius? As in, 7900m/sec is only the orbital velocity at the surface of earth? I'm confused how that's possible since nothing orbits earth at its surface
+MutatedGamer Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
So escape velocity is the initial speed one needs to have from the surface of an object to escape the gravitational pull of that object (assuming you are in a vacuum and only factor in gravity). However if one could fly an aircraft in a vacuum at Mach1 you could still reach the moon from Earth it would just take an insane amount of energy because of gravitational drag?
We are assuming the when the initial velocity (escape velocity) is reached, the engines are turned off. (There isn't enough fuel to keep the engines going the entire time)
1) You do not have to escape the Earth to reach the moon. The moon clearly hasn't escaped the Earth (since it is in orbit around Earth), ergo reaching the moon wouldn't require escaping Earth.
2) Your intuition is correct. You would have to go a lot farther away than just the moon, but there is nothing preventing a rocket or anything that has some sort of fuel supply that can propel itself from escaping the Earth at any speed. As long as you still have fuel. You still have to reach "escape velocity", but escape velocity is not a constant. The velocities quoted for various celestial bodies are from their surface. In reality, there is an escape velocity from any arbitrary distance from any body, and that escape velocity decreases with distance, so if you got on a rocket and went 1 mph away from Earth, you would eventually reach a point where the escape velocity is under 1 mph, and you can turn off the engines as you are exceeding escape velocity, and never went faster than 1 mph.
My favourate teacher..
Can u please tell me how to calculate the time of an object if it has a velocity that is comparable to the escape velocity ?
We have many videos on the channel on special relativity. PHYSICS 62 RELATIVITY PHYSICS 62.1 UNDERSTANDING RELATIVITY & SPACE
Just use the Velocity formula.... v = distance ÷ time
Why is potential energy on the earths surface negative?
By definition. The zero potential energy point in space is at infinity, thus it must be less than zero when you get closer to the surface of Earth.
Yeah right!! Thanks .
escape velocity speed, applies to projectiles, not rockets or other chemically fueled object like space shuttles.
This is kind of misleading. Escape velocity still applies to rockets and the like. I mean, rockets must still reach escape velocity to escape the Earth. The issue is that escape velocity doesn't allow for powered flight. The calculated escape velocity simply assumes all that velocity is given instantaneously, and then the rest is just a ballistic trajectory. Rockets can burn their engines and get themselves into a position where the escape velocity is lower. They still need to reach that escape velocity though. For example, a rocket could escape the Earth at 1 mph. Eventually, it would be far enough away that the escape velocity is under 1 mph, and it could then cut off the engines. It still had to reach escape velocity, so it very much applies to rockets.
I understand the way that you found velocity using kinetic and potential energy, but thinking about this problem from a kinematics perspective has me confused. From my logic, in order to escape the pull of gravity, your force applied would have to be greater than the force of gravity right? So because force is equal to mass times acceleration, as long as your acceleration is greater that the acceleration of gravity g you should escape the pull of gravity, which i know is wrong. Why is it instead an escape velocity and not an escape acceleration? I'm trying to find the flaw in my logic of thinking because i know its wrong.
Your way of thinking about it is correct. If you continue to apply a force greater than your weight (which would diminish as you move farther away from Earth), you will eventually get away from the gravitational force of the Earth. But you would have to apply that force for a very long time and distance. Better to speed up the the escape velocity and then turn off the engines (or you will run out of fuel).
The notion of escape velocity assumes that some theoretical projectile is given some instantaneous velocity and no other forces (other than gravity) are ever exerted on it. That's escape velocity. From the surface of the Earth (and ignoring drag from the air), the velocity you would need to give some object after which you leave that object at the sole mercy of gravity alone is 11.2 km/s. This does not in any way mean that something MUST reach 11.2 km/s to escape the Earth. Obviously, we don't launch our rockets with instantaneous velocities; we accelerate them up to speed. There is nothing preventing you from propelling yourself out of Earth's gravity under constant power - and doing so at speeds much lower than escape velocity. A "rule" of escape velocity is that once given that initial velocity, no further propulsion is allowed.
But the denominator of the orbital velocity is R which equals the radius of Earth and the height, which means that orbital velocity is not constant and the escape velocity is not equal root two times orbital velocity, am I right?
No. The orbital velocity depends on the height of the orbit, that is correct. The escape velocity will always be sqrt(2) times the orbital velocity from the location of the orbit.
Is gravitational potential energy not equal to -GM/r?? Why did you use -GMm/r?
Because that is the correct equation.
@@MichelvanBiezen lmao🤣🤣🤣 ofcourse its the correct equation
sir you are the best sir😊😊
Thank you. Glad you are enjoying the videos.
Michel van Biezen Does/can(without intentionally attempting to) helium reach escape velocity? I'm sure by the time you've collected the information you need you'll be able to formulate questions that I wouldn't have dreamed of asking. So just remember to right down the questions that come to mind during this endeavour because I’m sure we'd love to hear it all, that is if you are willing and up to the challenge...
Sir can you make video on Variation in acceleration due to gravity inside the earth at different points
Well based on pure calculations I have a formula to find the variation in 'g' with depth
For a body Above the surface of the earth at height h, g is given by- 9.81/ {(Radius of earth + h) ÷ Radius of earth}^2
For a body inside the earth at depth d, g is given by- 9.81 x (Radius of earth - d) ÷ Radius of the earth
9.81 is the value of 'g' at the surface of the earth
I believe we already have a video on that.
@@MichelvanBiezen Can you provide me the link, also how do I calculate the time period of a body in space which stops rotating in elliptical path and is falling into sun
Awesome video..... My question was why we neglect air resistance?.... As air resistance will be there when the rocket is launching
All physics problems at the undergraduate level ignore air resistance. (which is typically fine as it offers a good approximation and it makes the problem a lot easier). The plan is to cover higher level physics after we complete many of the topics we are still working on.
Thnk u
This makes no sense to me. If I was able to fly straight up for any amount of time eventually I will "escape" the earth's gravitational pull no matter how fast or slow I go as long as I keep putting distance between me the surface. Am I wrong here?
If you continue to provide fuel to the engines in order to keep pushing the rocket further out. Spaceships don't have enough fuel to do that. They just carry enough fuel to give the spaceship enough speed to escape Earth. They then turn the engines off and float through space. (That is how the Apollo missions made it to the Moon).
Catalyst - your logic is sound. Escape velocity is merely a theoretical threshold. It is a property of Earth's gravity. If you had the means, you could indeed propel yourself out of Earth's gravitational well at any speed. The quoted "escape velocity" of Earth is merely the instantaneous velocity a projectile would need if launched from the _surface_ of Earth - to never fall back down. Since Earth's gravity decreases as you get farther away, this escape velocity decreases as well.
About the average peak velocity for the Apollo missions was about 34,200 ft/s (23,318 mph or 10.4 km/s). This is the fastest they every went on the way to the moon, and while they aren't on the surface when they reached this speed, they were only a little over 100 miles away, so technically the escape velocity from that altitude is only 11.0 km/s. Apollo missions never reached escape velocity; they didn't need to. A mission to the moon does not require escaping Earth gravity. Even while standing on the surface of the moon, you are still very much under the influence of Earth's gravity; the moon's influence is just stronger. The moon is still well within Earth's gravitational well, so a visit to the moon does not require escaping Earth.
Back to the original point, if you had the fuel, you could fire up a rocket and go 10 m/s and never go any faster. You would eventually reach a distance where the escape velocity was under 10 m/s at which point you could turn off the engine. This would be a horribly inefficient rocket, and it couldn't actually exist, but in principal, your logic is correct.
Why you don't make full lectures?
The vast majority of the viewers have expressed that they like this format. My students also prefere the small lectures where each one covers a specific topic or problem. So we stuck with this approach.
Hello Michel
Excellent videos, nice opportunity for me to practice some maths
wait isnt PE supposed to be GmM/r²?
That is the equation for the force of gravity between two masses, not the equation for gravitational potential.
It's gravitational force. Just integrate dis force, den u'll get the potential energy.
integrate it
-(Gmm/r^2)
If a navy ship's canon were able to luanch a projectile into space what speed it need to start at?
It is not a matter of what speed it needs to start at, but what speed you can build up to. To reach orbital speed you must reach a speed of 18,000 miles per hour (29,000 km/hour). Theoretically (ignoring wind resistance), you can reach space if you had that initial velocity.
If "space" is 100 km up, you ignore air resistance and assume the projectile starts with the instantaneous speed it needs :
v = √(2gh) = √(2 * 9.8 m/s² * 100,000 m) = √1,960,000 = 1,400 m/s (3,132 mph)
whats that in mph ?
That is about 25,000 miles / hour.
Michel van Biezen orrrrr ffs i cant run even 80% that speed what am i gonna do now?!??
thanks for the video..
i have a few questions that pop to mind if you could help me with them...
...why couldnt the rocket keep at the same speed as TAKE-OFF?? obviously its enough to overcome gravity FROM THE SURFACE... which, im assuming, stronger than a few 100 miles up?? no?. if the rocket keeps propeling upwards, gravity shouldnt be getting stronger, and pulling back down?! no?
....if escape velocity is around 40,000 kph.... and re-entry is 20,000 kph.. why do things burn up coming down, but not burn when going up... AT TWICE THE SPEED.. IN THE SAME ATMOSPHERE???
thanks in advance..
Re-entry is about 40,000 km/h. By the time the rocket accelerates to higher velocities, it has reached the upper regions of the atmosphere where it is really thin and doesn't cause extreme temperatures like during re-entry.
Also when taking off, the rocket takes a long time to reach the escape velocity. (F = ma)
Think of it in terms of escape energy. Sure, you could escape earth's gravity going two miles per hour constantly, that would require less force per period of time over a longer time interval. Integrating would yield the same escape energy. That's how I think about it anyway
Escape velocity is the speed at which you can turn off your engines and not be pulled back to earth
I have a question about escape velocity. No matter how fast you go you have a limited amount of kinetic energy and even though gravity weakens as you go farther away from Earth the gravitational force never goes to zero which means after a long enough time wouldn't anything traveling away from Earth eventually come back unless it was moving at an infinite speed?
That is the concept of the escape speed. If you are traveling at a speed greater than the escape speed, the rocket will continue to slow down forever, but will never come to a complete stop, as it will continue to move away forever, although at a slower and slower speed.
@@MichelvanBiezen is it like an asymptote where it never goes quite to zero?
If the speed is equal to the escape speed, yes. if the speed is much greater than the escape speed, then the final velocity will be much greater than zero
Why are you able to say theres no kinetic nor potential energy in the final state. Whered all the energy go? Surely there’s still velocity at the final state right? KE?
By definition, the escape velocity is the velocity necessary to break free from the Earth's gravitational force, in such a way that when you reach an infinite distance the velocity will be zero. Therefore, there will be NO KE and NO PE.
Why is the potential energy at infinity 0 shouldn't it be infinity as potential energy=mgh
At infinity h=infinity??
That is by definition. PE is taken as zero at infinity and less than zero everywhere else.
Because small g is actually decreasing the further away you are from the center of gravity. Small g = GM/(r^2). Big G is the gravitational constant, big M is the mass of the source of gravity (Earth, in our case), and r is the distance between the object and the center of gravity. As you can see, as the distance r tends toward infinity, small g gets smaller and smaller and approaches zero, which makes PE also approaches zero.
Please help I am not able to understand why escape velocity don't depend on angle of projection
Amandeep Kumar because both move away from the center of the planet, it does not matter if it was 90 degrees or 10 degrees from the horizon. both bodies will be moving away from the center.
Why would the potential energy be equal to 0 as the distance goes to infinity?
That is by definition when calculating the gravitational potential energy.
Sir, why is the potential energy equation equal to GmM/RE?
That is explained in this video: ua-cam.com/video/louI_Ncp6gY/v-deo.html
Might be late, but does it affect the velocity if i stand on a higher ground, say Mount everest?
You would still need to reach escape velocity, but it would take a little less energy to do so.
Yes it does affect the velocity - the higher you go, the lower the escape velocity from that height. Mt. Everest won't make much of a difference though. You don't need to reach escape velocity to escape the Earth - this is a misconception. You only need to reach escape velocity if you will never give yourself any further propulsion. If you have a way to power yourself (like with a rocket), you can escape the Earth at any speed you wish. Theoretically, you could build a ladder to infinity and climb away from Earth at 1 mph and you could successfully escape the earth. The idea of "escape velocity" is that instantaneous velocity needed that - if you were to never apply any other propulsive force to it every again - would still escape.
@@willoughbykrenzteinburg first and foremost, I'm an idiot, and I've been drinking. I have a hypothesis that if you could construct an aircraft that could provide the initial thrust to put a spacecraft into the upper atmosphere, then eject said spacecraft (in upper earth atmo) that can propel itself using some type of ion drive that works both in atmo and vacuum, would it be possible that such a craft to exit earths gravity? Again, I've been drinking and I don't know what my brain is up to at the moment lol
how to know about the velocity the earth hitting the sun, if velocity of the earth in orbit is zero? please show me ......teacher
That would be equal to the escape velocity of the Sun. Ve = sqrt(2) * sqrt(GM/R) where M is the mass of the Sun and R is the radius of the Sun (in meters).
@@MichelvanBiezen That's not right. If something fell into the sun from an infinite distance away starting at zero velocity, it would impact the sun at the sun's surface escape velocity, but from just 150,000,000 km away, it would be less than the sun's escape velocity. Just logically, you could deduce that something that fell from a distance farther from Earth would hit the sun harder than the Earth would by simply pointing out that by the time that object reached the Earth-distance from the sun, it would already have some non-zero velocity.
If the distance from the Earth to the rocket at infinity is infinite shouldn't we also have infinite potential energy? Why does it goes to 0?
Like if potencial energy = m g h shouldn't that be an insanely large number at the very least?
As "h" increases, "g" decreases. As h approaches infinity, g approaches zero. It doesn't matter how big h is. If it's all multiplied by zero, the answer is zero.
@@willoughbykrenzteinburg thank you so much
sir,how to find gravitational force due to a ring??
I haven't made those types of videos. (That gives us a good idea to make some like that). It would be similar to finding the electric field due to a ring of charge. Take a look at: Physics - E&M: Electric Field (8 of 16) Ring of Charge
excellent video, thank you.
I don't get it, shouldn't gravity be constantly slowing down the object no matter how far away it is?
Ray Kay
Ray,
You are correct. The force of gravity will always slow down an "escaping" object no matter how far away.
However sine the force of gravity is proportional to the square of the distance, the farther the object the smaller the force of gravity.
When the force is very small, the acceleration is very small, and if it is too small compared to its velocity, the object will never stop.
+Ray Kay Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
Why is the KEf 0 ? I got that it is because it reaches an infinite point... but why ?
That is done so we can calculate the minimum velocity required to escape the gravitational pull of the Earth.
So this formula is abstract ? We cant apply this in real cases, because we are in a solar sistem, not just 2 bodys interacting with eachother, we have the moon, the sun... right ?
With this can I find the escape velocity of any planet?
Yes, you can apply this to any planet.
Why is the initial potential energy (GMm)/R^2 instead of mgh?
+afroN1gga nvm I figured it out
+Tom Siddle I made a mistake in my previous comment. The initial (gravitational) potential energy is actually (GMm)/Re... where Re is the distance of the object from the centre of the earth.However, the "g" (9.8m/s^2) variable in the simplified potential energy equation (mgh) is actually the number you get when you multiply GM/Re^2. Pretty cool!!
I'm not the too knowledgable on the subject but I hope that helps!!
man, i wish you were my physics teacher. I woulda wooped ass. :)
Why didn’t we use PE=mgh
Because the strength of gravity is not constant and is inversely proportional to distance squared.
Michael Douglas
Thank you
Why are we saying it is going to an infinite distance thats the only thing i dont get
Are we saying until the Ke and Pe reach that infinite point where they become 0
I heard the energies required to allow escape velocity is when K=U . Is this true?
Thank you.
You're welcome!
Can you explain how you found 7900 m/s?
LooL I watched all your video!!!
it was fantastic.
Wow sir how are you make this type videos
i love u
I am glad you are enjoying the videos. 🙂
A weather balloon filled with helium can reach space, as demonstrated in a famous youtube video of a legoman going into space, so what is the relevance of escape velocity, when you can clearly escape at a lower speed?
oh, seems like this was already answered below. Escape velocity is the speed needed if there is no additional energy added to the system. Anyhow, we should be taking helium balloons into space and then turning on the rockets later. We would save a ton of fuel.
If a balloon gets you to space your velocity would be very small. To get away from the pull of the Earth, you would still need to reach a velocity of 25,000 miles per hour.
Actually, a weather balloon does not reach space (which starts at about 65 miles (100 km). At that altitude the air is too thin to provide the buoyancy force needed to push a balloon that high.
1) The helium balloon didn't reach space
2) Getting to space is not "escaping the Earth". The MOON is 240,000 miles away, and it hasn't escaped the Earth. The Apollo astronauts that went to the moon did not escape the Earth.
3) The only reason rockets go to space is to avoid most of the drag air causes. Otherwise, they'd be able to orbit anywhere higher than any obstacles on Earth (higher than the mountains). Rockets need all that fuel NOT so they can get high - but so they can get fast. They need to reach upwards of 17,000 miles per hour LATERALLY across the surface to get into an orbit. Launching from a higher altitude does not add any horizontal speed, so they'd need just about the same amount of fuel. Certainly any savings in fuel would be more than offset by the cost of the gargantuan mechanism it would take to reliably and safely lift those rockets up to that altitude in the first place.
The guy that jumped out of the redbull stratos, Felix Baumburger or something, was nowhere near the edge of space, even though it looked that way. He was only about a quarter the way.
Love this channel.
Why can you not travel at a low speed (all the way up) to escape earth's gravity? Say if you had a ship that could ascend at a rate of 50 feet a second or something, and do it all the way till earth's gravity was zero. Why is that not possible??
It is theoretically possible, but it would require so much fuel that the rocket would be so heavy you would barely get it off the ground. It is much more efficient to build up a lot of speed quickly and that will require much less fuel.
It is possible, and in fact - you wouldn't have to do this all the way until Earth's gravity was zero. Though, it's important to point out that there is no point where Earth's gravity is zero - since it technically reaches to infinity, BUT obviously there would be a point where the gravity of some other body was stronger than Earth's.
Escape velocity is simply the instantaneous velocity something needs to never fall back down from a specific body. This varies depending on how far away from the body you already are. From the surface of Earth, it is about 11.2 km/s and this completely ignores drag. This is the instantaneous velocity something would need to escape Earth without further propulsion. There is nothing preventing something from simply traveling much slower than this under constant power. You could climb a ladder out of Earth's gravity well theoretically.
As for your example, you wouldn't have to travel 50 ft/s forever. If you had some mechanism that constantly powered you and maintained this 50 ft/s, you would eventually reach a point where this 50 ft/s exceeded the escape velocity. For the Earth, this would be about 2.1 billion miles (distance somewhere between Uranus and Neptune's distance to the sun). Once you got this far away, you could shut your engine down and you'd never return. As has been pointed out though, this would take an incredible amount of fuel. You're better off just building up as much velocity as quickly as possible.
I have a question...how come potential energy at infinity be zero?
That is by definition. In space that becomes our reference point and thus everywhere else, the potential energy is less than zero.
why we can't escape from earth even if we do not move with escape velocity. we can get away from earth easily since gravitational pull will be decreasing. so why we need escape velocity. please answer
man i wish some 1 answered this i came here thinking the same thing
The escape velocity is the theoretical velocity needed to hurl an (inanimate) object away from a larger object (e.g. a rock from the surface of the earth) out to where gravity is no longer acting. To fire a rock from the surface to space without any additional on-board propulsion you need to achieve Mach 33 at sea level. It is therefore much easier to send a rocket up with its own on-board propulsion to reach an altitude where its orbital velocity has significantly increased and the escape velocity has diminished to where it can move to a place where it can remain without needing further propulsion to stop it tumbling back to Earth (i.e. "escaped" Earth gravity).
Thanku sirb😁
Why does infinite distance mean that potential energy is zero?
That is by definition. We define the potential energy to be zero at infinity.
in order to make the faulty logic work...
the real question would be.. why do we need escape velocity?? escape velocity is the force needed to GET OFF GROUND.. for the MASS of the given object.. and CONTINUE to apply that same propulsion until far enough from the surface...
one would think that force would DECREASE the further the rocket got from the center of the surface/planet.. NOT INCREASE TO 25,000 miles/ hr!!!!
Why did the m got cancelled u did not take lcm
sir please upload a video of condition of weightlessness on earth
So uuhm...what exactly is gravity again?
Lol
Why is PE equal to gravitational force
Notice the difference in the equation: there is a negative sign and there is only a 1/R in the denominator
how we know that escape radius equals to earth's radius?
Because we are trying to escape from the surface of the Earth.
Legend
Glad you like our videos. 🙂
good
bruh do u teach in America
Yes I do in the Los Angeles area.
So basically it would take a rocket to move 7mi/s to escape earths surface without falling down?
That is correct.
wow sir...
Escape velocity isnt a difficult, complex scientific calculation. It is actually very basic. If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity. To achieve escape velocity, you just have to maintain the objects rate of climb while ensuring the rate does not succumb to gravitational pull. So effectively, a ball can achieve escape velocity so long as its climb rate does not fall below its initial launch speed. It could be 1mph, 5mph or 60mph. In fact any speed above 0mph would see an object escape the earths gravity so long as it is maintained.
" If you throw a ball up in the air, it has achieved escape velocity because it is stronger than the pull of the earths gravity."
You're comparing velocity (v) to the force of gravity (F = ma, where a is the rate of change of v)
The flash
thanksssssss
How the potential energy could be zero at a height?
Ur proof contains some mistakes..
That is by definition. PE is considered negative everywhere except at infinity where it is zero.
first and foremost, I'm an idiot, and I've been drinking. I have a hypothesis that if you could construct an aircraft that could provide the initial thrust to put a spacecraft into the upper atmosphere, then eject said spacecraft (in upper earth atmo) that can propel itself using some type of ion drive that works both in atmo and vacuum, would it be possible that such a craft to exit earths gravity? Again, I've been drinking and I don't know what my brain is up to at the moment lol
i guess what im asking is does this formula only affect relative vertical EV or can it be reduced by following the curvature of the earth using centripetal force?
Escape velocity from Sun: 642 miles/s; Voyagers 1 & 2 are nowhere near that velocity.
Yes, but they are also not near the surface of the Sun. Farther away from the Sun, the escape velocity is much lower.
@@MichelvanBiezen So escape velocity is related to the distance of the object from the Sun, not the radius of the Sun. Similarly for the Earth.
Actually both, but like with the Earth, when you are far away, the escape velocity is much smaller.
My G 1:37
so is the earth flat ? just kidding.i am troubled by all of the flat earth people . we really need some flights over the south pole ,so these people can understand .i think it is the only way they would believe .although some claim that flight would just be flying in a circle. i point out the north star and none have replied yet . i feel bad for them.they have been lied to and trick so much that even the shape of the planet is hard for them to believe .what is a simple way to really set them straight in a nice way?
omg that makes sense thank you
In other words, damn fast...
escape velocity has NOTHING to do with REALITY as we know it..
it is SOLELY for a HYPOTHETICAL object.. THROWN out to space... not SELF-PROPELLED rockets and machines.
"Escape Velocity' for a rocket, is whatever speed it has when LIFTING OFF the Earth!!! or CENTIMETERS per Second!!!!
in other words, it doesnt need to go fast... it just needs to KEEP GOING..... at ANY VELOCITY!!!!!
The explanation about escape-speed that the initial speed can take an object to as a far distance as infinity away from a gravitational field or influence of the Earth is totally nonsensical
It is so obviously absurd to say so simply because the Earth's gravitational force can never reach anywhere farther beyond this solar system. By saying that an object from the Earth has to get such an initial speed that it can continue traveling to infinity to escape the Earth's gravitational force , you guys are automatically proving yourselves to be so ignorant as idiots that you guys cannot even know that the Earth's gravitational force can influence only objects within a certain limited space within the solar system , but not even the big lion share of the solar system, and the only form of existence in the solar system which has influence on all other forms of existence within the solar system is the Sun
Each of the forms of existence within the solar system alone can never ever have any influence on any other form of existence outside the solar system within this galaxy. Only the whole solar system with the interreactions among the Sun and all other forms of existence , and the interactions among all other forms of existence in a continuous chain of interreactions can have a certain influence on the nearby systems within this galaxy, and the whole galaxy system can have interractions with nearby galaxies which are still within the limit of distance from this galaxy
It seems that all you guys have been doing is repeating after, like a monkey or parrot , whatever were said so long ago by scientists of the previous centuries when human knowledge of science is far much poorer than it is today , instead of activating your idle minds to see the flaws and defective holes of such obsolete notions of the old worlds
Huh
WTF? I mean, thank you for teaching something about 'pure physics' that wasn't clear to me yet. I mean that.
But still, WTF?????????? I am pretty sure that you don't have to travel to infinity away from the earth before another gravity well (i.e. another planet, a star, a black hole .... a tooth brush another traveler dropped.....) *Sigh .... but hey .. in the mean time maybe you will find out if there are truly no white crows.
Great video until you started changing things into miles.
Thomas Nesmith naaa living in la miles is preferred