You take the velocity to be 2pir / T. Since the orbit is elliptical, does this approximation make a difference? I know Earth's orbit is not very eccentric, but what about for orbits with greater eccentricities?
Looks great! But this didn't occur to me before: is it worth pointing out that, for the constant of proportionality between T^2 and a^3 to be 1, G also has to be in non SI units, kg and AU and years instead of (newtons and) kg and meters and seconds? There might be some student who picks up a calculator and tries it out and finds that 4Pi^2/GM is NOT one. OR perhaps equally true: that constant is not just 1, but 1 year^2/AU^3. I don't think this is necessary, though, at least I always found these things highly unusual to talk about, but I always want to because I make such a huge deal in my classes about units.
Everything else is called either a planetary system or stellar system, since only our solar system has Sol in it, i.e. the sun. The answer is yes, as it is valid for any planetary system, given just one central body that is the significantly more massive than everything else in the system, and that interplanetary forces are negligible. This is usually what you would expect, because stars as so much more massive than planets. Kepler's laws are also valid for binary star systems, where there are two bodies in orbit around each other. Instead of one of the stars being at the focus point, the center of mass of the two of them will be at the focus point, due to the fact that gravity is a two way street per Newton's third law. You will also notice that systems of moons follow Kepler's laws. For a system of two stars, or interplanetary forces that cannot be neglected, it can be much more complicated, and it is a problem that cannot be solved in closed form. The solution will have some properties in common with Kepler's laws, but there inevitably will be irregularities not captured by Kepler's laws. Even our own solar system has interplanetary forces, which played an important role in the discovery of Neptune. The deviations from a perfect Kepler orbit observed for Uranus required an unknown planet to explain them, which enabled us to predict the existence of Neptune, and know where to look to discover it by sight.
Carl, I left this on a Google Doc script you left a bunch of edit suggestions on and I am just making sure you got the message. (Who knows if you would notice this left as a reply to a Google Doc suggestion.) I think it would be logical for you to sign up for my Quality Control Team Remind. The instructions are at the top of this page: www.flippingphysics.com/quality-control.html Right not my only way to communicate with you is through TH-cam comments or these Google Docs. Remind would provide us a way to communicate and for you to learn more about how this qulte Quality Control Team thing for Flipping Physics works.
![How the Bizarre Path of Mars Reshaped Astronomy [Kepler's Laws Part 1]](http://i.ytimg.com/vi/Phscjl0u6TI/mqdefault.jpg)
Thank you so much for making the video, it really helps me understand physics more!
Glad to help!
you are a genius and a life saver!
very clear explanation - at the same time also fun. I love your channel
Thanks for the love!
You take the velocity to be 2pir / T. Since the orbit is elliptical, does this approximation make a difference? I know Earth's orbit is not very eccentric, but what about for orbits with greater eccentricities?
Thanks for this vid dad!
Looks great! But this didn't occur to me before: is it worth pointing out that, for the constant of proportionality between T^2 and a^3 to be 1, G also has to be in non SI units, kg and AU and years instead of (newtons and) kg and meters and seconds? There might be some student who picks up a calculator and tries it out and finds that 4Pi^2/GM is NOT one. OR perhaps equally true: that constant is not just 1, but 1 year^2/AU^3. I don't think this is necessary, though, at least I always found these things highly unusual to talk about, but I always want to because I make such a huge deal in my classes about units.
Interesting point. I do not think it is logical to put in the video, however, it is an interesting extension. Thanks!
Does this has similarity with pendulum equation.
Sir ,,is Kepler's law's of planetary motion valied for any solar system in this universe?
Everything else is called either a planetary system or stellar system, since only our solar system has Sol in it, i.e. the sun. The answer is yes, as it is valid for any planetary system, given just one central body that is the significantly more massive than everything else in the system, and that interplanetary forces are negligible. This is usually what you would expect, because stars as so much more massive than planets. Kepler's laws are also valid for binary star systems, where there are two bodies in orbit around each other. Instead of one of the stars being at the focus point, the center of mass of the two of them will be at the focus point, due to the fact that gravity is a two way street per Newton's third law. You will also notice that systems of moons follow Kepler's laws.
For a system of two stars, or interplanetary forces that cannot be neglected, it can be much more complicated, and it is a problem that cannot be solved in closed form. The solution will have some properties in common with Kepler's laws, but there inevitably will be irregularities not captured by Kepler's laws. Even our own solar system has interplanetary forces, which played an important role in the discovery of Neptune. The deviations from a perfect Kepler orbit observed for Uranus required an unknown planet to explain them, which enabled us to predict the existence of Neptune, and know where to look to discover it by sight.
Carl, I left this on a Google Doc script you left a bunch of edit suggestions on and I am just making sure you got the message. (Who knows if you would notice this left as a reply to a Google Doc suggestion.)
I think it would be logical for you to sign up for my Quality Control Team Remind. The instructions are at the top of this page: www.flippingphysics.com/quality-control.html
Right not my only way to communicate with you is through TH-cam comments or these Google Docs. Remind would provide us a way to communicate and for you to learn more about how this qulte Quality Control Team thing for Flipping Physics works.
@@FlippingPhysics Thank you for reaching out, Mr P. I'll sign up for your QC team and look forward to the opportunity.
How the guy commented 1 month ago ???
Some people are faster than others.
The video went "public" today...
Why don't you teach modern physics
Modern Physics is not a part of any of the curricula I have made videos for so far.
mahirap