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I want to have a function that describes where a falling object is. Like this one: h(t) = -g*t²/2 But this one is for the usual close to the surface case, where there is no variation of gravity due to distance to the center of the planet. But I want a formula that takes that variation into account, for distances close to several radius. It is hypothetical situation where the only bodies involved are the the falling one and the big one. And I'm talking only about Newton physics here. I've searched for it but found nothing.

Ramon Griffo
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  • Unless I'm mistaken, if you lift the "close to the surface" approximation, you simply get an elliptic trajectory, following Kepler's first law. – Miyase Jun 26 '22 at 22:18
  • If the object has no initial velocity it has no reason to form an elliptic trajectory, that would happen if it was in orbital velocity, and it could even happen close to the surface, if it is fast enough for that and if there is no atmosphere. – Ramon Griffo Jun 26 '22 at 22:20
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    I don't think that problem has an analytical solution, that is, there is no single formula of position as a function of time. –  Jun 26 '22 at 22:24
  • In the special case where initial velocity is zero, conservation of angular momentum (assuming gravitation is the only force) implies that the trajectory is a straight line, which a special case of elllipsis. Since the force follows the inverse square law, it's very likely that no analytical solution exists and that a numerical resolution is required. – Miyase Jun 26 '22 at 22:24
  • What are your variables? Especially what is $h(x)$ – Dale Jun 26 '22 at 22:31
  • h(t) is the travelled distance (edited the original, should be h(t)). – Ramon Griffo Jun 26 '22 at 22:34
  • I see, I thought there could be an analytical solution. – Ramon Griffo Jun 26 '22 at 22:35
  • from wikipedia: "The following formula approximates the Earth's gravity variation with altitude:

    {\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}}{\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}} Where

    gh is the gravitational acceleration at height h above sea level. Re is the Earth's mean radius. g0 is the standard gravitational acceleration.

     – niels nielsen Jun 26 '22 at 22:36
  • @RamonGriffo Outside the linear case, very few differential systems have analytical solutions. Sometimes we simply haven't found any, and sometimes it's been proven that no analytical solution can be written (although I have no example of that, it's just something that stuck in my mind after a discussion with a mathematician working on something related, years ago). – Miyase Jun 26 '22 at 22:37
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    Constant-gravity parabolae are secretly high-eccentricity ellipses. The problem of motion in a $1/r^2$ gravitational field is solved, though the gravitational three-body problem is chaotic. – rob Jun 26 '22 at 22:49
  • I guess the two body problem will work! The time needed fore two bodies to meet under the influence of gravitational attractive force! – Fardin Jul 20 '22 at 19:18

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You can solve the differential equation with varying $g$, analytically in mathematica/python/wolframAlpha to get exact time!

But if you want the time of Falling, I guess the two body problem will work! The time needed for two bodies to meet under the influence of gravitational attractive force!

$T_{meet}=\frac{π}{2\sqrt{2}}\sqrt{\frac{(R_{e}+h)^{3}-R_{e}^{3}}{G(M_{e}+m)}}$

Fardin
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