# Work required to increase Earth/Moon separation

1. Apr 8, 2013

### tony873004

1. The problem statement, all variables and given/known data
The Moon and Earth have masses of 7.4 × 1022 kg and 6.0 × 1024 kg, respectively, and the distance between their centers is 3.8 × 108 m. The amount of work required to pull them further apart so that their separation increases by a factor of 1.6 is most nearly

2. Relevant equations
Etot = -GMm/(2r)
W=ΔE

3. The attempt at a solution
ΔE=(-GMm/2)(1/(1.6*3.8 x 108) - (1/3.8 x 108)) = 1.44 x 1028 J

Answer key says 2.9 x 1028 J
That's twice my answer. Is it possible that the person who made the answer key forgot to factor out the /2. Or should I be considering only ΔU? I did a similar problem from this chapter using ΔEtot and got the right answer. What am I missing?

Thanks!

2. Apr 8, 2013

### mukundpa

Is the moon still in a stable circular orbit? I think if the motion of the moon is not considered than we have to take only change in potential energy.

3. Apr 8, 2013

### Staff: Mentor

What energies does your Etot include?

Why don't you do a check of the result by performing the work integration (integral of F.dr) for the increase in separation?

4. Apr 8, 2013

### BruceW

Why is the 2 here? The potential energy of the system is just -GMm/r

5. Apr 8, 2013

### tony873004

That formula is for total energy, not PE
U=-GMm/r
KE=0.5mv2, v=sqrt(GM/r)
KE=0.5 GMm/r
KE=GMm/(2r)
U=-2GM/(2r) --add the 2's so U and KE have a common denominator
KE+U=(GMm/(2r)) - (2GMm/(2r))
KE+U=-GMm/(2r) = Etot

I think mukundpa solved it. The problem did not state that the Moon achieved a circular orbit at the new distance. So with its velocity constant, all that changed is the PE. Doing the problem that way, I get the answer the key is looking for.

Thanks everyone!

6. Apr 8, 2013

### BruceW

another way to interpret it, is that they wanted the work done, given that the kinetic energy of the moon and earth do not change.