# Falling object in gravitation field

1. Jul 30, 2013

### neerajareen

I am trying to study the motion of a falling body in the earth’s gravitational field however not assuming a constant acceleration. I want to work out the distance travelled in a given time.
We know that ￼
$$a = \frac{{GM}}{{{r^2}}}$$
Likewise, we can calculate that
￼$$\begin{array}{l} \frac{{da}}{{dr}} = \frac{{ - 2GM}}{{{r^3}}}\\ da = \frac{{ - 2GM}}{{{r^3}}}dr \end{array}$$
We also know that $$s = \frac{1}{2}a{t^2}$$
￼. Therefore I’m guessing we can say that
￼.$$s = \frac{{GM}}{{{r^3}}}{t^2}dr$$
I’m thinking we have to do some kind of an integral. How would one tackle this problem?

Thank you

2. Jul 30, 2013

### Simon Bridge

$s=\frac{1}{2}at^2$ is for a constant acceleration.

Start from:
$$\frac{d^2r}{dt^2}=\frac{GM}{r^2}$$ ... solve the DE to get r(t).

3. Jul 30, 2013

### siddharth23

Yup, you'll have to integrate. But you can't have 's', 't' and 'r' in the equation at the same time. You'll have to eliminate one. And in the last equation, it should be

ds = (-GMt^2dr)/r^2

not s =....

4. Jul 30, 2013

### Creator

THank you for bringing that up....I asked the very same question earlier ...2 days ago on this thread....
I think you give a much more succinct and tractable answer than I got before (as long as the acceleration & velocity is along the radius)........ however shouldn't your last step say:
￼.$$\frac{{ds}}{{dr}}= \frac{{GM}}{{{r^3}}}{t^2}dr$$ since you took the derivative of a on the right side wrt r ?
And that is simply the Keplerian formula if s is pointing along r so ds/dr becomes unity...?

If so then it reminds me of what Halls of Ivy said earlier that it is some sort of elliptical function...which probably is the more general formulation with lateral motion and eccentricity.

If we allow the object to drop through a hole in earth then it will follow Keplerian motion and oscillate at the same period as if it were orbiting from the same height, r.
I think.

Creator

(opps ; I see I crossed over your last post..Siddaharth; I see you are making my same point.)

Last edited: Jul 30, 2013
5. Jul 30, 2013

### siddharth23

Haha..that's alright. Could you tell me how to write the formulae in a post. I don't know so I have to write like ds/dr :(

6. Jul 31, 2013

### siddharth23

$\frac{ds}{dr}$ Gottit!

neerajareen, did you get the answer?

7. Jul 31, 2013

### neerajareen

Thank you all for your replies. I understand that we need to eliminate some variables according to what siddharth said. I realize that s is related to r because the distance fallen is along the radius as pointed. But i'm still not sure how to go about doing the integral because R is a function of time as the particle falls isn't it ?

8. Jul 31, 2013

### Simon Bridge

Thinking: isn't $\small s$ a radial displacement? (in context).
Then, $\small s = r_2-r_1=\Delta r$
... so an infinitesimal radial displacement at time $\small t$: $\small ds=r(t+dt)-r(t)=dr$ ???

@neerajareen: any of this any use?

9. Jul 31, 2013

### neerajareen

Yeah i meant that s and r are related but i still do not know how to solve the DE you said @simon

10. Jul 31, 2013

### Simon Bridge

A lot depends on your requirements - if you want to find r(t) then you need to learn how to solve DEs. :)

There are also approaches that involve finding the momentum as a function of radius.
It is a biggish subject ... see: Tan S. M. http://home.comcast.net/~szemengtan/ClassicalMechanics/SingleParticle.pdf [Broken] s1.7 (p12).

Last edited by a moderator: May 6, 2017
11. Jul 31, 2013

### rcgldr

A similar problem was done before. The Kepler based math is shown in post #11, and the math starting with

F = -G m1 m2 / r2

with the closure rate (velocity) between objects defined as v = dr/dt, and accleration as a = dv/dt, resulting in:

a = - G (m1 + m2) / r2