Calculating Work for Launching a Spacecraft to a Great Distance from Earth

[solarcat]

Homework Statement

Say you're launching a spacecraft of mass m from the surface of the Earth (mass Me and radius re) to a low height h (h is much smaller than r, so h is essentially negligible). How much work is required to move the spaceship from its low orbit to a great distance from earth? Ignore gravitational effects of other planets/moons/stars.

Homework Equations

Work = change in kinetic energy
Fc = mv^2/r
Fg = GmM/r^2

The Attempt at a Solution

When the spaceship is on the ground, PE=0. When the spaceship is in low orbit, PE=0 because h is essentially negligible. When the spaceship is very far away, my book says that PE will also be 0 (U = -GMem/r). So work depends only on kinetic energy.

I said the work required to get the spacecraft to height h would be GmMe/(2*re^2). This would also be its kinetic energy since kinetic energy on the ground is 0.
To find the change in kinetic energy:
KE when spaceship is in low orbit = GmMe/(2*re^2)
KE when spaceship is a great distance away from earth:
m*v^2/(re+d) = GmMe/(re+d)^2 (d=distance from earth)
Cancel m: v^2/(re+d) = GMe/(re+d)^2
v^2 = GMe/(re+d)
d is very large, so re+d approaches infinity, so v is zero. final kinetic energy = 0
Work = change in kinetic energy = - GmMe/(2*re^2).

But I don't think this can be correct because the next question essentially states that the energy of the spaceship at low orbit should be midway between the energy on Earth and the energy at a very great distance.

 

[Bandersnatch]

When the spaceship is on the ground, PE=0. When the spaceship is in low orbit, PE=0 because h is essentially negligible. When the spaceship is very far away, my book says that PE will also be 0 (U = -GMem/r).

When you substitute the radius of the Earth into the equation quoted above, do you get a 0?

I said the work required to get the spacecraft to height h would be GmMe/(2*re^2).

It looks like you compared the force of gravity to centrifugal force in order to obtain the kinetic energy in circular orbit at the surface. Which is a good approach. But you made a mistake. Review your algebra, or show your work step by step, if you can't find the error.

The general approach should be to write down the total mechanical energy (PE+KE) for each of the three cases and then compare them. It should make it easier to see what's going on if you separate the cases clearly (i.e. case 1: PE+KE = ...; case 2: =, and so on).
Remember that, intuitively, it takes energy to get higher or faster, and the solution should reflect that.

 

[solarcat]

This is what I did to find the work required to get the spacecraft to height h:
mv^2 / (re + h) = GmMe/(re+h)^2 (v is velocity at height h)
Cancel m: v^2/(re + h) = GMe/(re+h)^2
Multiply (re+h) by both sides; v^2 = GMe/(re+h)
h is negligible: v^2 = GMe/re
Final KE = 1/2 * m * GMe/re
Also, since h is negligible, the change in potential energy is negligible also. Hmm...

 

[Bandersnatch]

Final KE = 1/2 * m * GMe/re

This is correct. In the equation you wrote earlier you had the radius squared - which was incorrect.

Go back to potential energy in the first two cases. Why did you write it was 0?