Calculating the gravity on the moon

[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?

Homework Statement

Radius of the Earth r_e = 6,370km
Ratio between r_e and r_m = 11:3
Ratio between m_e and m_m = 81:1

Homework Equations

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

The Attempt at a Solution

In the test I got the correct result ~1,622\frac{m}{s^{2}} but now I can't 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?

 

[haruspex]

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

 

[Blacky372]

Oh sorry, the picture isn't loading.
I will try to fix that.

 

[tms]

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}}

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.