Work required to increase Earth/Moon separation

[tony873004]

Homework Statement

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

Homework Equations

E_tot = -GMm/(2r)
W=ΔE

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 ΔE_tot and got the right answer. What am I missing?

Thanks!

 

[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.

 

[gneill]

What energies does your E_tot 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?

 

[BruceW]

Homework Equations

E_tot = -GMm/(2r)

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

 

[tony873004]

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

What energies does your E_tot include?...

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) = E_tot

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!

 

[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.