Speed of a Satellite in Circular Orbit, Orbital Velocity, Period, Centripetal Force, Physics Problem
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- Опубліковано 7 лют 2025
- This physics video tutorial explains how to calculate the speed of a satellite in circular orbit and how to calculate its period around the earth as well. It uses the formula for centripetal force and gravitation to derive the equation for the orbital velocity of a satellite.
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Wow that's a really fast satellite.
This is about the speed of an actual satellite
@Jebediah Jr,how?l mean what is the importance of calculating the speed of a satellite if it isn't real?
@@hannalot_ how?
Sarah Abderrahman ISS moves in almost 8km/s, so that’s even faster. And what do you mean with not real? Being able to calculate these simple things are essential to understand the fundamentals of how an object would act when put into orbit. It’s important as hell
@@dominobuilder100 oh I just thought that a satellite with that high speed hasn't been made yet I mean it is still on paper
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The first part of this video helped me understand an extra credit question in my Geog605 class, remote sensing of the environment. We were asked how many times a certain satellite would orbit the earth in a day and we were only give the satellite altitude, earth's mass and radius, and the gravitational constant. The essential piece of missing info was the equation for the distance of a circular path.
One question though, how did you translate your velocity value into m/s?
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5:07 is the speed of the satellite angular velocity?
Thank you very much for this video
Thank you once again
when do we use m/s and km/s when solving for velocity?
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is this related in calculating quantities related to planetary or satellite motion?
thanks, great explanation!
Redirected here from a link in my book
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while on a circular orbit if the velocity of the object is increased then it will go further away but a new balance can be achieved only if distance decrease. (Higher velocity means higher centrifugal force and it can be balanced only by a higher gravitational force which will happen only if distance decrease) in reality an elliptical orbit is created. how do you explain this by the above simple physics?
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It really helped
Thank you so much for your help, this was extremely helpful to me.
What calculator are you using bro???
Sir, how you can cancel both R, In centripetal Force R is the radius of circle(i think it should be the distance between Surface of Earth and Satellite) and in Gravitational Force R is the separation between two masses (from center of Earth to the Satellite). Could you Please Explain
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To me nice lesson sir
Sir,for explaining which app u have used?
Tysm bro
could you maybe explain what the variables mean
he did.
so helpful, thanks!!!
thanks bro
What will happen if new mass added to the satellite while moving? affect of the linear momentum? and how it maintain the speed constant?
if you add new mass the same way rockets dock to ISS, then it won't affect anything because the new mass already achieve equal speed to the satellite
@@jo-oy4vj ok, thanks
Bless your soul
42,200,000 x 2pi / 86,400 = 3068
Seems a good deal easier for the last answer and we'll within reason given rounding. If we had more sigfigs for R it would be just as accurate.
Believe it or not, I'm having to perform many calculations for a singular mission that I'm planning in Kerbal Space Program
Thanks man
Can you do ESCAPE VELOCITY OF A SATELLITE
Im an average student and nim thankful for this but kinda lost in number 2(b) 😭
Isn't v the tangential velocity? So shouldn't T = 2pir/v/r since v/r = w which is the rotational velocity?
I'm not getting the calculation for r even though I put in all the values u have in for the cube root thingy
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I think he made a mistake in the calculations, I wasn't able to find the same answer as him for the radius between the center of the earth to the satellite.
This dude 100% got rid of his speech impediment! How tf do you do that?
This wonderful
Isn't it v=√(2GM/R) ? First example speed of satellite
pretty sure thats the escape velocity my guy
so satelies move at 11059.2km/h? constantly around earth?
Where did you get the 6254.3 m/s from. What did you do to get it.
He used the equation: v=square root (GM/r). This will give you the speed of the satellite which is in circular orbit. If it was in an elliptical orbit, you'd have to use a different equation.
It is probably too late now, but what he did was double parenthesis like this: ((6.67 X 10 ^ -11)(5.97 X 10^24) divided by 1.018 X 10^7))
thank you so much hocam
tq sir
Faster than speed of light orbiting earth to send videos to earth from the future
Thankkks
can u plz explain the variation of centripetal accl. and velocity of planets at different distances from sun?
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Hello can you solve this problem? For me please 😭
Venus orbits the sun at an average of 21.75 mile per second while earth orbits the sun at an average of 18.46 per second how much faster is the orbital speed of Venus?
Thanks in advance
I keep getting 6.2543E10 for the speed on my calculator. I was wondering what am I doing wrong? I did convert the km into m by multiply it by 1000 and got the same radius in the video. I plugged the exact number in also.
i got the same problem
What is the work of the Earth on the satellite and why is it equal to zero?
XxPlayMakerxX131 i dont fucking know but these are so fucking hard fr
Forces only do work if they have some vector component along the path of motion. This path of motion is the circular orbit, the force is the centripetal force directed towards the center of the Earth. These are perpendicular, so gravity does no work. It's a similar story for charged particles moving in a mass spectrometer, if you've seen that.
if the earth is inside the system it cannot do work. hence 0. If it is outside the system like external forces. it does work.
my prof has us set it up like so
work= change in energy
work= PE + KE ...
on the left is external forces that create work (b0th pos. and neg. work has to be parallel though. Work that is perpendicular is 0. Torque is when there are perpendicular force and distance )
on the right is internal energies, Gravitational potential and kinetic. If 0 work that means the internal forces are equal and opposite (for this scenario)
Hope this helps.
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amazing video, thank you!
You made mistake because on last video you said that V =4πsquareR\Tsquare not 4πsquareRsquare\Tsquare
please clarify me it's angular speed is equal to that of earth about its own axis
you have been clarified
thank you....