Calc Fall of Object from Space: Derive Eqn of Accel as Fxn of Time

[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?

 

[siddharth]

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?

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.

 

[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.

 

[recon]

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.

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.

 

[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?