# Final velocity due to gravtiy

1. Nov 1, 2013

### Luke1121

1. The problem statement, all variables and given/known data
Assuming the Earth is at rest and alone in the universe what speed would a spaceship arriving at the surface of Earth from an infinitley large distance be if the speed at infinity= 0 and acceleration is caused only by gravity

2. Relevant equations
$F= \frac{GmM}{x^2}$
Also using the chain rule $a=v\frac{dv}{dx}$

3. The attempt at a solution

well I set ma equal to the force due to gravity so

$v\frac{dv}{dx}= \frac{GM}{x^2}$

seperating the variables gives $vdv=\frac{GM}{x^2}dx$

then

$\int vdv=GM\int x^{-2} dx$

would the limits of integration need to be from infinity to the surface of the earth, or am i on the wrong track all together?

2. Nov 1, 2013

### haruspex

You're doing fine. Just do the integral as you suggest.

3. Nov 1, 2013

### TSny

You're on the right track. But you might need to be careful with signs. [Edit: I see haruspex beat me ]

4. Nov 1, 2013

### Luke1121

Ok, then doing the indefinite integral gives $v=\sqrt{\frac{2GM}{x}} + C$

but i'm not sure about the limits, would it be $\frac{1}{2}v^2=-GM\int_{\infty}^{6370}x^{-2}$

where 6370 is the radius of the earth in km?
Thanks

5. Nov 1, 2013

### haruspex

The limit is the radius of the Earth. What units you use are up to you, so long as you are consistent. I would think your value for G assumes m, not km.
I think you have a sign wrong above. The minus sign should only appear after performing the integral.