# How distance from CoG affects acceleration over time

Let's say an object is falling towards earth from a long distance away. As it gets closer and closer, the acceleration would increase, inversely proportional to the distance squared.

Is there any way I can graph this on an acceleration/time graph, or a distance/time graph?

The challenge is that the acceleration is dependent on distance from the centre of gravity, but to be able to find that distance at any given time, I first need to know the acceleration, so that I can double-integrate it.

If I know the acceleration is mG/D2 (assuming D to be distance), and I know that acceleration is the double-differential of distance, I can get:

D ' ' = mG/D2

To keep things simple, I'm going to assume mG to be 1:

D ' ' = 1/D2

But I'm not sure where to go from there, or if I'm even on the right track.

D ' ' = 1/D2

But I'm not sure where to go from there, or if I'm even on the right track.
I think you are on the right track. Now you just have to solve the differential equation. The variable is time.

May be you can use: (DD')' = (D')2 + DD'' , or similar trick.
Or elsewise open a diff.equ book or Mathematical methods in Phy ... , etc.

But I'm not sure where to go from there, or if I'm even on the right track.
a=(vdv)/dx gives velo as function of x
v=dx/dt gives distance as function of time
plug this function into mg/(x^2) gives acceleration as function of time

Stavros Kiri
You can also use Conservation of Energy (kinetic+potential energy = const), which is also known as "the first integral of motion" (or perhaps better [or equivallently] the "theorem of change of kinetic energy": ΔEk = total Work ...).

That would essentially reduce the 2nd order diff. equ. into 1st order (which you can also do via diff.equ. methods ... - cf. my other comment), and makes the math a lot simpler. However, be warned that you have to be good at 'integrals', to do it correctly (it may not be an easy integral). Try it out, by doing the math carefully.

Note: (Hint:) By cons. of energy you first find V = f(D). Then use V = dD/dt (assuming radial motion only) and you get 1st order differential equation (easy to integrate and solve by "separation by parts" [or "separation of parts"]* method, easily applied here [bring D only on one side, t only on the other ... (and integrate)]).

* or separation of variables - easy in this case

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a=(vdv)/dx gives velo as function of x
v=dx/dt gives distance as function of time
plug this function into mg/(x^2) gives acceleration as function of time
This is also another smart, correct and valid way! (Here x--> D [one dimensional radial motion] or distance from the center of gravity).

For others to follow I point out that your first equation a=(vdv)/dx comes from a = dv/dt = (dv/dx)(dx/dt) = (dv/dx)v = ...

But your method would also involve, and have to do with, 1st order diff.equ., easier to integrate though.

Note: In Newtonian dynamics, this method is essentially equivallent to the energy conservation method ['first integral of motion'] (see my other [previous] comment). This can be easily proved (for anyone that is interested).
The difficulty of the math involved to solve it this way is thus the same as by conserv. of ener. ...

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Folks, I think this is a perfect example to see how we can use the conservation of energy in mechanics, to simplify things, and it shows exactly why cons. of en. in mechanics is also called "the 1st integral of motion" (the 2nd would be to completely solve and integrate from that, in order to obtain a complete solution to the equation of motion, e.g. x = x(t) for 1-dim). [Usually with the "1st integral of motion" we obtain v = f(x) or f(t) etc. . Then we use v = dx/dt, etc. ...] [many problems in mechanics can be solved easily that way]

In other words, that way, we totally avoid the 2nd order differential equations or the double integrals (which in general come directly from Newton's 2nd law of motion - it is 2nd order in x, t ...) and we reduce them to 1st order, or simple integrals! [Now, of course, one can equivallently do the same with diff.equ. methods or tricks (such as 'change of variables', 'separation of variables', identities-tricks etc.).]

Thus we should thank the OP for bringing up that particular problem, as well as "hackhard" for showing a 3rd equivallent way to solve the problem.

You can safely use these methods for solving similar problems or homeworks.