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- #1

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How much maths do you know? Differential equations?

- #3

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Unfortunately not yet, I am still in highschool. Can you show me the formula and explain it briefly?

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Unfortunately not yet, I am still in highschool. Can you show me the formula and explain it briefly?

Take a look at this and see whether you can follow the maths:

https://www.physicsforums.com/threa...ith-varying-acceleration.866218/#post-5437928

- #5

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It is too advanced for me.

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It is too advanced for me.

Getting a formula for the height (or speed) of a falling object as a function of time is mathematically very difficult (as you can see).

You can, however, calculate the speed of the object at each height using conservation of energy. Do you know about potential and kinetic energy under gravity?

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yes of course, calculating the velocity is not a problem at all.

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yes of course, calculating the velocity is not a problem at all.

If you are looking for an interesting problem, you might like to try this. If a planet is in an elliptical orbit round the sun and you know its nearest point (perigee) ##r_1## and its furthest point (apogee) ##r_2##, can you calculate its speed at either of these points in terms of ##G##, ##M## - the mass of the sun - and ##r_1## and ##r_2##?

- #9

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If your problem involves elliptical motion knowledge, as you can guess, I will not be able to solve it.

The original question was asked because of pure curiosity. Is there any ellegent proof for that? and can you express the equation by only the distance between the objects, not the surface level?

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As far as I understand, at the points of perigee and apogee there is no radialic speed, so, it is calculated using the circular motion, isn't it?

If your problem involves elliptical motion knowledge, as you can guess, I will not be able to solve it.

Yes, that there is no radial component of the velocity is key. You don't need to know anything else about elliptical orbits. You need conservation of energy and one other thing. Can you work out what that other thing is? It's another conservation law.

The original question was asked because of pure curiosity. Is there any ellegent proof for that? and can you express the equation by only the distance between the objects, not the surface level?

You mean you think my solution in the above thread is not elegant enough? Some problems admit solutions in terms of energy conservation but don't give up exact solutions in terms of time easily or, sometimes, not at all.

Those equations in the above post are all based on distance between the centres of mass. The surface level ##R## simply represents the point at which the free fall abrupty comes to an end!

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[tex]h=\dfrac{1}{2}gt^{2}[/tex]

Source: http://formulas.tutorvista.com/physics/free-fall-formula.html

- #12

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is it √GM/r ?

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is it √GM/r ?

That's for a circular orbit, yes.

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I thought you were first considering a scenario of a free falling object?

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Others may correct me if I am wrong, but I don't think there is a direct solution for finding distance as a function of time in this situation. You can find the time to fall a given distance by single equation, but not the other way around. You get the same problem with elliptical orbits. You can directly solve for the time it takes to travel from one point of an orbit to another, but you can't do so going the other way (except for when the two points are the periapis and apoapis.)

This doesn't mean that you can't solve the problem, but it generally involves a process of "narrowing down" to the answer rather than a direct solution.( one solution method involves iteration, where you plug a number into a equation, take the answer and plug it back into the equation and then repeat. ( how many times you repeat will determine the accuracy of your final answer.)

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Yes, but a planet in an elliptical orbitI thought you were first considering a scenario of a free falling object?

This formula only is accurate for situations where g does not change with height. The original question dealt with the situation where it does, and this is something that must be accounted for with elliptical orbits.I was in high school last year, I only have half of my notes. Is this the formula? Unless your high school is more advanced, I don't remember using perigee and apogee.

h=12gt2h=\dfrac{1}{2}gt^{2}

- #17

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Yes, but a planet in an elliptical orbitisa free falling object.

This formula only is accurate for situations where g does not change with height. The original question dealt with the situation where it does, and this is something that must be accounted for with elliptical orbits.

Oh, I could only assume we used the same formula, I don't remember too much, but thanks for that.

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