Gravitational Potential energy Problem

[SnakeDoc]

Homework Statement

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?

Homework Equations

K_F+U_F=K_I+U_F
U=-G(M₁m₂)/r²
K=1/2mv²
mass of Earth = 5.972E24
radius of Earth = 6371km
G=6.673E-11

The Attempt at a Solution

1/2mv²_F+(-G(M₁m₂)/r²_F=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)/r²_F=1/2v²₀+(-GM₁)/r²₀

Next I try and get the r_F by itself

(-GM₁)=(1/2v²₀+(-GM₁)/r²₀)*r²_F

(-GM₁)/(1/2v²₀+(-GM₁)/r²₀)=r²_F

√(-GM₁)/(1/2v²₀+(-GM₁)/r²₀)=r_F

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

 

[Dick]

Homework Statement

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?

Homework Equations

K_F+U_F=K_I+U_F
U=-G(M₁m₂)/r²
K=1/2mv²
mass of Earth = 5.972E24
radius of Earth = 6371km
G=6.673E-11

The Attempt at a Solution

1/2mv²_F+(-G(M₁m₂)/r²_F=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)/r²_F=1/2v²₀+(-GM₁)/r²₀

Next I try and get the r_F by itself

(-GM₁)=(1/2v²₀+(-GM₁)/r²₀)*r²_F

(-GM₁)/(1/2v²₀+(-GM₁)/r²₀)=r²_F

√(-GM₁)/(1/2v²₀+(-GM₁)/r²₀)=r_F

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

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

 

[SnakeDoc]

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

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!

 

[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: $$\frac{1}{2}v^2 -gR = -\frac{gR^2}{(R+r)}$$