Velocity of mass falling from height h above a star of radius R

[Karol]

Homework Statement

Mass m is left to fall from height h above the surface of a star with radius R. derive an expression for the velocity it hits the ground, using R, h and the acceleration g

Homework Equations

The energy: $E=\frac{1}{2}mv^2-\frac{GMm}{r}$
The acceleration: $g=\frac{GM}{r}$

The Attempt at a Solution

$$\frac{GMm}{R+h}=\frac{1}{2}mv^2-\frac{GMm}{R}$$
$$\frac{GM}{R+h}=\frac{v^2}{2}-g$$
This expression doesn't give the desired one: $\sqrt{\frac{2Rgh}{R+h}}$

 

[SammyS]

Homework Statement

Mass m is left to fall from height h above the surface of a star with radius R. derive an expression for the velocity it hits the ground, using R, h and the acceleration g

Homework Equations

The energy: $E=\frac{1}{2}mv^2-\frac{GMm}{r}$
The acceleration: $g=\frac{GM}{r}$

The Attempt at a Solution

$$\frac{GMm}{R+h}=\frac{1}{2}mv^2-\frac{GMm}{R}$$

The term on the left above is missing a negative sign.

$$\frac{GM}{R+h}=\frac{v^2}{2}-g$$
This expression doesn't give the desired one: $\sqrt{\frac{2Rgh}{R+h}}$

What does your expression give for a result?

 

[Karol]

oo

$$-\frac{GM}{R+h}=\frac{v^2}{2}-g$$
$$-\frac{RGM}{R(R+h)}=\frac{v^2}{2}-g$$
$$-\frac{Rg}{(R+h)}=\frac{v^2}{2}-g$$
$$v^2=2g\left( \frac{R}{R+h}+1 \right)$$

 

[Karol]

I solved it thanks to your remark, bye

 

[Orodruin]

*removed* posted at same time as Karol solved it.