Time for freefall in a varying gravitational field?

1. Jan 21, 2009

Fr33Fa11

I have been trying to calculate the time it takes for an object of negligible mass to free fall towards an object with a large mass, taking into account that the gravitational acceleration experienced by the small object increases as it moves closer to the big object.
The first thing I tried was setting up a diffeq:

n''=GM/(h-n)^2
Where G is the gravitational constant, M is the mass of the object, h is the initial height, n is the distance traveled, and ' denotes a derivative. After trying for a few days to solve this equation for n(as a function of t, time), I gave up and tried to solve it using change in kinetic energy, which also didn't work. Any ideas? Thanks.

2. Jan 21, 2009

turin

This problem has been discussed somewhere on this forum, but I don't remember the title of the thread. Anyway, the differential equation (not exactly yours, but basically the same idea) has been solved. I'll give you a hint:

$$\frac{d^2x}{dt^2}=v_x\frac{dv_x}{dx}$$

You should prove this to yourself. (because it is actually not always true, so you should understand why it is true for your particular problem)

3. Jan 22, 2009

Fr33Fa11

I'm not sure I understand why that is true. Is v(x) velocity with respect to x, where x is distance traveled? Is v(x) just an arbitrary name for a function which fits the relationship?

4. Jan 22, 2009

turin

Yes, v_x (not v(x), BTW) is the velocity component in the x direction:

$$v_x\equiv{}\frac{dx}{dt}$$

You can replace x with whatever coordinate you want, but wierder coordinates will require more care in determining when this relationship is true. Are you familiar with the chain rule?

5. Jan 22, 2009

Fr33Fa11

sure, f(g(x))'=f'(g(x))g'(x)
Ok, so far I am following you. The expression rewrites the second derivative of position(acceleration) as the first derivative of position(velocity) times the derivative of velocity with respect to position. Using gravitational potential energy I have figured out the equation for velocity with respect to space, and I know how to take the derivative of that. That leaves just a single order diffeq. A single order diffeq of the form a'=a*k should have a general solution of the form e^(tk), so then acceleration should equal ce^kt where k is the derivative of velocity with respect to position right?

6. Jan 24, 2009

turin

Are you sure it has that form?

7. Jan 27, 2009

Sedunov

Hi, Fr33Fa11!
Your attempt to use change in kinetic energy should take into accout the change in potential energy, equal to GMm/(h - n) - GMm/h. Basing on it you can find the velocity dependence on distance: V = f(n). The time of flight can be found by integration of the expession dn/f(n).

Last edited: Jan 27, 2009
8. Feb 9, 2009

Fr33Fa11

I've solved it, and I have calculated the constants that come out of the problem. Do you know if this has been done(it would be really cool if it hasn't)

9. Feb 10, 2009

turin

I thought we already said that. Sorry, no Nobel prize today.

10. Feb 10, 2009

Fr33Fa11

I was expecting that something like this would have been done before. Do you have a link to the paper/post where it was solved, I want to see if they have a faster way of solving it, because my work is kind of messy, as is the answer.

11. Feb 10, 2009

D H

Staff Emeritus
This exact topic was discussed in the threads [thread=243444]Two particles and gravity; non-constant acceleration[/thread] and [thread=246833]Time for two bodies to collide under gravity[/thread].

12. Feb 10, 2009

Fr33Fa11

They did not arrive at a fully solved solution. I'm also trying to solve it for any number of dimensions.(Where GM/r^(n-1)) is the formula for gravitational acceleration with any number of dimensions.

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