# Calculating the gravity on the moon

1. May 23, 2014

### Blacky372

Hey guys, we just had a physics test and one exercise was the following:

The earth pulls down on stuff with 9.81 N/kg.
How strong is this force on the moon?

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

Radius of the earth re = 6,370km
Ratio between re and rm = 11:3
Ratio between me and mm = 81:1

2. Relevant equations

Gravitation formula. $F=\gamma*\frac{mM}{r^{2}}$

3. The attempt at a solution

In the test I got the correct result ~1,622$\frac{m}{s^{2}}$ but now I cant figure out how I did it.
This is my attempt:
http://imgur.com/4TfyD2G
My formula was basically this:
$F=6.67\cdot10^{-11}\frac{Nm^{2}}{kg^{2}}\cdot\frac{1kg\cdot\frac{9.81N\cdot6.37\cdot10^{6}m}{6.67\cdot10^{-11}\frac{Nm^{2}}{kg^{2}}}:81}{(6.37\cdot10^{6}m\cdot\frac{11}{3})^{2}}$

EDIT: Okay this is kind of illegible. This one is better: http://i.imgur.com/O6nEQyr.png

I first thought I was right but when i calculate the result its something like 2,72N instead of 1,622N.

How can I get to the right solution?

Last edited: May 23, 2014
2. May 23, 2014

### haruspex

Please post your working. How else can we determine what you're doing wrong?

3. May 23, 2014

### Blacky372

Generally it is better to solve problems symbolically, only plugging in the numbers at the very end. That is definitely true for this problem. Notice that two of the data you are given are ratios of the Earth and Moon radii and masses. That suggests that taking the ratio of the force on the Moon to the force on the Earth, $F_m/F_e$, is the most direct route to a solution. Do that symbolically and the problem becomes very simple.