# Gravitational potential energy between Earth and Moon

1. Nov 20, 2013

### fogvajarash

1. The problem statement, all variables and given/known data

Consider the earth's radius to be 6.40x103km, its mass to be 6.00x1024kg, the Moon's mass to be 7.36x1022kg, and the Moon's radius to be 1.74x103km. The average value for the Earth-Moon distance is 3.84x105kg. Neglect friction and rotation.

a. Sketch the potential energy as a function of the position on a line between the Earth's centre and the Moon's centre, for positions between the surfaces of the two bodies (ignore positions inside the bodies).
b. Find the position where the potential has an extremum (maximum or minimum) between the Earth and the Moon. Find the gravitational force on a test particle of mass m at that location.
c. Calculate the minimum work required to transport a mass of 1.0kg from the surface of the Earth to the surface of the moon, ignoring air resistance.

2. Relevant equations

3. The attempt at a solution

I'm not sure on how to start with this problem. I don't know how i should graph the potential energy as a function of the position on a line (firstly, how can we find the potential energy in a point between two masses? Calculate both potential energies with respect to the Moon and Earth at that point?). As well, how can we calculate the work required to transport the mass? I tried finding the difference in potential energy between a point in the surface to a point in the Moon, but it doesn't work (answer is 58.7 MJ).

Thank you in advance.

2. Nov 20, 2013

### Pythagorean

if you have a line, D, between two points, and you are traveling between them, choose r to be the distance between you and the left point. Then the distance between you and the right point is D-r.

The potential energy due to two masses is just the superposition of the potential from each individual mass.

3. Nov 20, 2013

### haruspex

I feel c) is ambiguous. Is it asking for net work, or the amount of work that has to be available to achieve the transfer?

4. Nov 20, 2013

### Pythagorean

I think they just want students, once they get force as a function of x, to integrate it along a straight line.

5. Nov 20, 2013

### fogvajarash

The thing is that supposedly we aren't required to use calculus for all of the exercises. I feel that b. as well requires some calc in order to optimize the function (and wouldn't it be negative everywhere, and have the extrema at the end points?

Edit: Wouldn't the function that gives the potential energy be:

$E_{p}=\displaystyle\frac{-Gm_{e}m_{p}}{t}-\displaystyle\frac{Gm_{m}m_{p}}{3.48x10^{8}-t}$

Where mp is a point mass?

Last edited: Nov 20, 2013
6. Nov 20, 2013

### Pythagorean

Yes, though t bothers me since it's often used for time. I would go with r, but that's just convention and not technically wrong.

I have no idea how you'd do it without calculus.

7. Nov 20, 2013

### haruspex

No integration needed. Part a) found the potentials.
Part b) suggests differentiation, but there might be another way.

8. Nov 21, 2013

### fogvajarash

I have tried differentiating but to no avail. The answers are here if anyone can find out what it is about:

Extrema positions: ($9.28x10^{6}m, 7.51x10^{8}m$)

9. Nov 21, 2013