Geosynchronous orbits and geostationary orbits are not the same thing. Technically speaking, geostationary orbits are a type of geosynchronous orbit, but not all geosynchronous orbits are geostationary.
FYI: It says this in the description of the video. "Geostationary orbit is a special case of geosynchronous orbit. A geosynchronous orbit simply has the same 24 hour period as the Earth, however, it is inclined relative to the equator and traces out an ellipse in the sky as seen from the Earth. (Sorry they are incorrectly identified as the same in the video.) Thank you to Dan Burns @kilroi22 and Christopher Becke @BeckePhysics for the correction!"
Thanks. It does take a fair amount of work to get that to all fit correctly. If you are curious, you can see the process here: ua-cam.com/users/videogTsrWnv7OgQ/
Hello there.... It is a quite happier experience for me to be taught by you ... A tremendous one present on this planet See your way of teaching is quite different and innovative... I literally enjoying all your stuff. Thanks for providing such an adorable material but could I ask you for a favor that may I get the notes of a few chapters for my exam ... If you don't mind
Please try to assist Me... As I love your way of teaching and really in a need for a solution... Please have me the honour to have the stuff and some easy ways to solve the problems.
Hi, I was wondering about calculating the orbit height by myself and while searching for the correct numbers I stumbled upon this video. It was very helpful and thank you! However, when putting in the numbers I found a radius of 42164,24 km from the centre of gravity or 35786,24 km above the equatorial surface of the earth. I took a sedereal day of 86164,091 seconds as the orbit time. Just a little difference but where does it come from? You used 24 hours for a day but whatever I do, I can't get to the 35900 km that you say NASA states.
The sidereal day is a "more correct" period to use in this calculation, than the 24 hour mean solar day. The sidereal day is Earth's period of rotation relative to the inertial reference frame, and relative to distant stars. If you do a precise measurement of the Coriolis effect or the Eotvos effect to make an inertial measurement of Earth's rotation instead of a visual measurement, you will measure this as the period of Earth's rotation, rather than 24 hours. Keep in mind that figures on NASA's website are likely rounded numbers, so that they don't overwhelm their audience with an excessive number of digits for what you'd actually have to know to put a spacecraft in geostationary orbit.
aaron - I thought asbout asking a question about why solar day was used instead of sideral day but I decided to first skim some of the other comments. Glad you beat me to it.
Hi, nice explanation. But I am having a doubt. Is it possible to put a geostationary satellite at a higher altitude than 35900 km by giving it a higher angular velocity?
The planet would have to rotate at a different rate, for a geostationary satellite to orbit at a different position. By definition, a geostationary orbit has to have a rate of orbit that matches Earth's rate of rotation. The altitude of a geostationary orbit is a special case where g=G*M/r^2 coincides with a=v^2/r, to make this happen. You would need an additional force of probably tension, to have a geostationary orbit at a higher altitude. This is one of the working principles of the hypothetical space elevator, where a tether to the ground holds a counterweight slightly higher than geostationary orbit, such that the center of mass between the counterweight and the elevator cab, is at geostationary orbit, as the elevator cab climbs the cable to meet it.
A stupid question: when you said the angular velocity (omega)= delta theta divided by delta t. Why didn't you equate the delta theta to the circumference of the satellites orbit. (it should have been 2 x pi x r). I understand that the answer it correct but I'd like someone to explain this to me. (Took me an hour to figure out that it'd have been the same thing this way, but why?!)
Change in angular position and circumference are two different things. They are related, however, they are different. I would suggest these two videos for you: www.flippingphysics.com/arc-length.html and www.flippingphysics.com/pi.html
Hi there again, since you have taught AP Physics 1 I was wondering if you could answer a question on scheduling. My teacher said that we are behind currently in Physics as we have barely finished momentum. He said that we should be somewhere in rotation, but because of a new schedule it's been hard to do. So I ask you again, is my physics class behind? Is it something that should concern me?
Not a question I will answer for you, however, all the information is there for you to provide your own answer. I have 11 topical review videos for AP Physics 1. www.flippingphysics.com/ap-physics-1-review.html The length of each video is a good estimate of the relative amount of time that probably should be spent on each topic. Therefore, you can decide which topics y'all have already covered, how much time you have left in the school year, and don't forget to leave some time for review before the exam. Good luck!
Hey Mr. Palmer, Wouldn't solving for r using Tearth^2=Tsatellite^2, 86400^2=(4pi^2r^3)/(G(5.97x10^24) be a little easier? I just finished a problem similar to this on a worksheet and just wanted to share.
Ah, yes. "Easier" what a fun word. Sure, that would be easier. However, that would require you have memorized Kepler's third law. I am very much against rote memorization. Memorizing equations and being able to plug in values is very different than understanding the physics. I am more concerned students understand what they are doing rather than just memorizing equations.
Thank you for the prompt reply! That makes a lot of sense. In our class, we did an example problem like this one right after learning the derivation of Kepler's third law, so that's what I jumped to when solving the worksheet problem.
Sure, but realize on the AP Physics exam you will likely have to be able to derive these equations from drawing a free body diagram and summing the forces like I have done here.
Sorry if it felt like I was harping on you. Memorization causing students to not understand physics is an issue I have run into quite a bit as a teacher. I even made a video about it: flippingphysics.com/understanding.html
*....Can someone please send me a Link to a Satellite Orbiting the Earth? I've been trying to find a Real Satellite for over 22 years, didn't think it would be that hard...?*
Nice... But the time for a geosynchronous satellite to make a 360 degree rotation is NOT the 24h solar day; it is the 23h 56m 4.0905s sidereal day - or the time between two consecutive meridian crossings of a star... If you wanna do a tutorial - do it right... - waw -
Geosynchronous orbits and geostationary orbits are not the same thing. Technically speaking, geostationary orbits are a type of geosynchronous orbit, but not all geosynchronous orbits are geostationary.
FYI: It says this in the description of the video.
"Geostationary orbit is a special case of geosynchronous orbit. A geosynchronous orbit simply has the same 24 hour period as the Earth, however, it is inclined relative to the equator and traces out an ellipse in the sky as seen from the Earth. (Sorry they are incorrectly identified as the same in the video.) Thank you to Dan Burns @kilroi22 and Christopher Becke @BeckePhysics for the correction!"
@@FlippingPhysics Oops, I didn't see that. My bad!
Dude, the timing of your "students" dialog is amazing. I'm not sure if many of the viewers understand how difficult a task that is.
Thanks. It does take a fair amount of work to get that to all fit correctly. If you are curious, you can see the process here: ua-cam.com/users/videogTsrWnv7OgQ/
This is kind of an academic sitcom. Laughed out loud with your impersonations. Great job !!!
Thanks! 😃
This method is just brilliant and seems much easier than using kepler's though it depends on the info in the question
haha thanks! Although I am not studying physics, I enjoy watching this video. I would recommend it to my remote sensing classmates.
And why are you not studying physics!!
Actually, a lot of people do not study physics. It makes me sad.
Thanks for recommending my channel!
Hello there.... It is a quite happier experience for me to be taught by you ... A tremendous one present on this planet
See your way of teaching is quite different and innovative... I literally enjoying all your stuff. Thanks for providing such an adorable material but could I ask you for a favor that may I get the notes of a few chapters for my exam ... If you don't mind
Please try to assist Me... As I love your way of teaching and really in a need for a solution... Please have me the honour to have the stuff and some easy ways to solve the problems.
Looking for a positive and prompt response
@@paritoshtomar2968 He has lecture notes on his website, linked in the description below each video.
Looks a lot like the first FRQ on the 2018 exam. Wish I had rewatched this vid :(
That's the good stuff right there
Glad you enjoyed it.
Hi, I was wondering about calculating the orbit height by myself and while searching for the correct numbers I stumbled upon this video. It was very helpful and thank you!
However, when putting in the numbers I found a radius of 42164,24 km from the centre of gravity or 35786,24 km above the equatorial surface of the earth. I took a sedereal day of 86164,091 seconds as the orbit time.
Just a little difference but where does it come from? You used 24 hours for a day but whatever I do, I can't get to the 35900 km that you say NASA states.
Marc van Oppen I believe your equation is correct.
The sidereal day is a "more correct" period to use in this calculation, than the 24 hour mean solar day. The sidereal day is Earth's period of rotation relative to the inertial reference frame, and relative to distant stars. If you do a precise measurement of the Coriolis effect or the Eotvos effect to make an inertial measurement of Earth's rotation instead of a visual measurement, you will measure this as the period of Earth's rotation, rather than 24 hours.
Keep in mind that figures on NASA's website are likely rounded numbers, so that they don't overwhelm their audience with an excessive number of digits for what you'd actually have to know to put a spacecraft in geostationary orbit.
aaron - I thought asbout asking a question about why solar day was used instead of sideral day but I decided to first skim some of the other comments. Glad you beat me to it.
Highly recommended ... The Physics Works...💝
Yes, it does!
Thank you!
You are welcome!
Is there a way to calculate the height of a geosynchronous satellite by adding up the time it takes to ping the satellite plus processing time?
Hi, nice explanation. But I am having a doubt. Is it possible to put a geostationary satellite at a higher altitude than 35900 km by giving it a higher angular velocity?
The planet would have to rotate at a different rate, for a geostationary satellite to orbit at a different position. By definition, a geostationary orbit has to have a rate of orbit that matches Earth's rate of rotation. The altitude of a geostationary orbit is a special case where g=G*M/r^2 coincides with a=v^2/r, to make this happen.
You would need an additional force of probably tension, to have a geostationary orbit at a higher altitude. This is one of the working principles of the hypothetical space elevator, where a tether to the ground holds a counterweight slightly higher than geostationary orbit, such that the center of mass between the counterweight and the elevator cab, is at geostationary orbit, as the elevator cab climbs the cable to meet it.
A stupid question: when you said the angular velocity (omega)= delta theta divided by delta t. Why didn't you equate the delta theta to the circumference of the satellites orbit. (it should have been 2 x pi x r).
I understand that the answer it correct but I'd like someone to explain this to me. (Took me an hour to figure out that it'd have been the same thing this way, but why?!)
Change in angular position and circumference are two different things. They are related, however, they are different. I would suggest these two videos for you: www.flippingphysics.com/arc-length.html and www.flippingphysics.com/pi.html
Hi there again, since you have taught AP Physics 1 I was wondering if you could answer a question on scheduling. My teacher said that we are behind currently in Physics as we have barely finished momentum. He said that we should be somewhere in rotation, but because of a new schedule it's been hard to do. So I ask you again, is my physics class behind? Is it something that should concern me?
Not a question I will answer for you, however, all the information is there for you to provide your own answer.
I have 11 topical review videos for AP Physics 1. www.flippingphysics.com/ap-physics-1-review.html The length of each video is a good estimate of the relative amount of time that probably should be spent on each topic. Therefore, you can decide which topics y'all have already covered, how much time you have left in the school year, and don't forget to leave some time for review before the exam. Good luck!
ok
Hey Mr. Palmer,
Wouldn't solving for r using Tearth^2=Tsatellite^2, 86400^2=(4pi^2r^3)/(G(5.97x10^24) be a little easier? I just finished a problem similar to this on a worksheet and just wanted to share.
Ah, yes. "Easier" what a fun word. Sure, that would be easier. However, that would require you have memorized Kepler's third law. I am very much against rote memorization. Memorizing equations and being able to plug in values is very different than understanding the physics. I am more concerned students understand what they are doing rather than just memorizing equations.
Thank you for the prompt reply! That makes a lot of sense. In our class, we did an example problem like this one right after learning the derivation of Kepler's third law, so that's what I jumped to when solving the worksheet problem.
Sure, but realize on the AP Physics exam you will likely have to be able to derive these equations from drawing a free body diagram and summing the forces like I have done here.
Flipping Physics Thank you, I’ll be sure to steer away from memorization like this in the future.
Sorry if it felt like I was harping on you. Memorization causing students to not understand physics is an issue I have run into quite a bit as a teacher. I even made a video about it: flippingphysics.com/understanding.html
*....Can someone please send me a Link to a Satellite Orbiting the Earth? I've been trying to find a Real Satellite for over 22 years, didn't think it would be that hard...?*
Hi flat earther
Are you guys 4-tuples?!?
Nice... But the time for a geosynchronous satellite to make a 360 degree rotation is NOT the 24h solar day; it is the 23h 56m 4.0905s sidereal day - or the time between two consecutive meridian crossings of a star... If you wanna do a tutorial - do it right... - waw -
Considering this video is not about the difference between a solar day and a sidereal day, I did not see the point in bringing that up.
Da Flip!?!
Hahahaha... well done!!!🙏🙏🙏🙏👏👏👏👏🤣😂🤣😂😂👍👍👍👍👍👌👌👌👌👌