Final Velocity of a Spaceship Approaching Earth from Infinity Due to Gravity

[Luke1121]

Homework Statement

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

Homework Equations

[itex]F= \frac{GmM}{x^2}[/itex]
Also using the chain rule [itex]a=v\frac{dv}{dx}[/itex]

The Attempt at a Solution

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

[itex]v\frac{dv}{dx}= \frac{GM}{x^2}[/itex]

seperating the variables gives [itex]vdv=\frac{GM}{x^2}dx[/itex]

then

[itex]\int vdv=GM\int x^{-2} dx[/itex]

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?

 

[haruspex]

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

 

[TSny]

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

 

[Luke1121]

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

but I'm not sure about the limits, would it be [itex]\frac{1}{2}v^2=-GM\int_{\infty}^{6370}x^{-2}[/itex]

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

 

[haruspex]

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

but I'm not sure about the limits, would it be [itex]\frac{1}{2}v^2=-GM\int_{\infty}^{6370}x^{-2}[/itex]

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

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.