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This question already has an answer here:

Is it possible to find how gravity changes with time? That is, height is not part of the equation.

For example, $g$ can be expressed as a function of height that does not involve time with this relation:

$g_i=g_0 (r /(r+h))^2$

where $g_i$ is $g$ at any height and $g_0$ is on the Earth's surface.

I also found through the above relation that $g_i = g_{i-1} ((r+h_{i-1})/(r+h_i))^2$

By the same token, there should be a relation between $g_0, g_i, t$ that does not involve distances. An object in free fall will be going through changes in $g$ as time progresses. Is it possible to create such an expression?

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marked as duplicate by DilithiumMatrix, ACuriousMind, AccidentalFourierTransform, Qmechanic May 20 '16 at 13:06

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    $\begingroup$ Yes it is possible, but it requires calculus. Are you up for it? $\endgroup$ – Pranav Hosangadi Jan 12 '14 at 6:04
  • $\begingroup$ Yes, most definitely. $\endgroup$ – Luis Jan 12 '14 at 6:18
  • $\begingroup$ Okay, here's a hint: Consider the acceleration to be constant as the object falls over a height of dh in time dt and write the relation. Then integrate it from h = $h_0$ to h = 0 $\endgroup$ – Pranav Hosangadi Jan 12 '14 at 22:43
  • $\begingroup$ Perhaps I'm not getting it. If I integrate from $h_0$ to $0$ then I end up with an equation that has $h_0$ again. But my goal is to end up with an equation that has only $t$ and $g$ at another point in time and there is no $h$ or $h_0$ involved. It would really help if you can show what you mean. $\endgroup$ – Luis Jan 13 '14 at 7:10
  • $\begingroup$ That is not possible - because the time taken to fall down depends on the height it is released from, since its acceleration depends on the height it is at. However, that $h_0$ will be a known parameter for any application, so it doesn't count as a variable. $\endgroup$ – Pranav Hosangadi Jan 14 '14 at 0:16