# Gravitational Potential energy Problem

1. Apr 16, 2015

### SnakeDoc

1. The problem statement, all variables and given/known data
A projectile is fired vertically from Earth's surface with an initial speed of 3.1 km/s. Neglecting air drag, how far above the surface of Earth will it go?

2. Relevant equations
KF+UF=KI+UF
U=-G(M1m2)/r2
K=1/2mv2
mass of earth = 5.972E24
G=6.673E-11
3. The attempt at a solution
1/2mv2F+(-G(M1m2)/r2F=1/2mv20+(-G(M1m2)/r20

I can cancel the m because they're in every term so and final kinetic energy since it should be zero at max height.

(-GM1)/r2F=1/2v20+(-GM1)/r20

Next I try and get the rF by itself

(-GM1)=(1/2v20+(-GM1)/r20)*r2F

(-GM1)/(1/2v20+(-GM1)/r20)=r2F

√(-GM1)/(1/2v20+(-GM1)/r20)=rF

then I subtract rF from the initial r0 to get my answer.
Did I do this right?

2. Apr 17, 2015

### Dick

I can tell you one thing that's wrong. $\frac{-G M m}{r}$ is potential energy. Your exponent on the $r$ is wrong.

3. Apr 17, 2015

### SnakeDoc

Oh wow I guess I was looking the wrong equation in my notes thanks. I think I got it from here then thanks a lot!!

4. Apr 17, 2015

### Simon Bridge

Well done:
Note: You could get it all in one go from: $$\frac{1}{2}v^2 -\frac{GM}{R} = -\frac{GM}{(R+r)}$$... and solve for r, where R and M the radius and mass values are for the Earth.

Of course, GM/R^2 = g which is a nicer number to handle.
So the relation becomes: $$\frac{1}{2}v^2 -gR = -\frac{gR^2}{(R+r)}$$