# Fall of Object

1. Jan 25, 2006

### recon

If it takes an object at distance x from the surface of the Earth 17.5 hours to fall from space onto the surface of the Earth, how can we calculate x? The object initially has zero speed.

I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. Can anyone show me how to derive an equation of acceleration as a function of time?

2. Jan 25, 2006

### siddharth

You can use $$\frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr}$$
So I think from this you can first find v(r) and then r(t) and use that to find x.

3. Jan 25, 2006

### daveb

The first thing you may want to realize is that acceleration is a function of distance between the object and the center of the earth, and that the distance is a function of time.

4. Jan 25, 2006

### recon

I know $$\frac{dv}{dt} = \frac{GM}{r^2}$$ but I can't figure out how this may be used to obtain v as a function of r.
How is it possible to solve the equation $$\frac{d^{2}r}{dt^2}=\frac{GM}{(R+r)^2}$$, where R is the radius of the Earth? I have not had much calculus yet, so I have not touched on solving differential equations of this difficulty.

Last edited: Jan 25, 2006
5. Jan 25, 2006

### siddharth

You have

$$\frac{dv}{dt} = \frac{GM}{r^2}$$

Now, as I said in my previous post,

$$\frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr}$$

So,
$$v\frac{dv}{dr} = \frac{GM}{r^2}$$

Now, can you find v(r) from this diff equation?