# A trip to the Moon!

1. Sep 27, 2011

### aszymans

1. The problem statement, all variables and given/known data
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem:

mearth = 5.9742 x 1024 kg
rearth = 6.3781 x 106 m
mmoon = 7.36 x 1022 kg
rmoon = 1.7374 x 106 m
dearth to moon = 3.844 x 108 m (center to center)
G = 6.67428 x 10-11 N-m2/kg2

2. Relevant equations
On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?

2.
Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!

3. The attempt at a solution
In our lecture concerning this topic we discussed distance= v(not) /2(μ)g, but there is no μ here clearly, so I tried not using the μ and that did not work and I dont know if I should use the formula for PE gravity close to earth or not close to earth, -G(m1m2/r)+U(not) or -G(m1m2/r^2)r. I dont understand how to get my distance without using μ.

2. Sep 27, 2011

### Staff: Mentor

Use the -GMm/r formula; it's universal.

Use conservation of energy to determine the interchange of PE and KE. Remember that both the Earth and Moon will be trading energy with the ship

What speed will the first "unsuccessful" shot have when it reaches its maximum distance from the Earth? What then will be its KE?

3. Sep 27, 2011

### TMO

I, too have the same question. However, my question is with the second part of the problem:

Here is my attempt at a solution:

I know that when the traveler is at the surface of the Earth, there are two potentials at work: the potential of the earth and the potential of the moon. Let $d$ be the distance between the centers of the moon and the earth, and let $v_0$ be the initial velocity. Then the relation of kinetic energy equals the following equation:

$\frac{1}{2} mv_0^2 - \frac{GM_em}{r_e} - \frac{GM_mm}{d-r_e} = \frac{1}{2} mv^2 - \frac{GM_em}{d-r_m} - \frac{GM_mm}{r_m}.$

However, solving for v does not give the right solution. What gives?

4. Sep 27, 2011

### Staff: Mentor

What value did you calculate?

5. Sep 27, 2011

### TMO

I got 2263.52 m/s. Apparently, this is the correct answer (I didn't square the right value for the initial velocity).

6. Sep 27, 2011

### aszymans

what did you use for me "m" in 1/2mv^2

7. Sep 27, 2011

### aszymans

also in the first part of the question when solving for "D" I am still confused, in the equation you told me to use "-GMm/r, is Mm the mass of the moon or did you mean mass of moon times mass of earth?