If I weigh 250 lbs on earth, how much do I weigh on the moon? One of my homework questions is to determine how much a 250 lb person weighs on the moon.  I first googled a calculator for this and found that the weight is 41.5 lbs.  So I tried to derive it myself and I cannot seem to get the correct answer.
Here is what I'm doing:
$$F=ma$$
I first converted $250$ lbs to Newtons:
$$250lb\frac{4.448 N}{1 lb}=1112N$$
So I then figured I'd plug values into the the formula $F=ma$
$$1112N=113.5kg(1.6\frac{m}{s^2})$$
But no matter how I solve this, I cannot seem to get the correct answer.  What am I doing wrong?
 A: $1112$ N is the force on earth: it’s (approximately) $$113.5\text{ kg}\cdot 9.8\frac{\text{m}}{\text{s}^2}\;.$$ To get the force on the moon you want
$$113.5\text{ kg}\cdot 1.625\frac{\text{m}}{\text{s}^2}\;,$$
which you’ll then have to convert to pounds. Of course you could simply multiply $250$ by the ratio of gravitational accelerations, $\dfrac{1.625}{9.8}$.
A: $$
{W_{\tiny earth} \over W_{\tiny moon}}\
=\
{\quad{\displaystyle{G\,{mM_{\it moon} \over R_{\tiny moon}^{2}}}}\quad
\over
{\displaystyle{G\,{mM_{\it earth} \over R_{\tiny earth}^{2}}}}}\
=\
\left(R_{\tiny earth} \over R_{\tiny moon}\right)^{2}
\,{M_{\tiny moon} \over M_{\tiny earth}}
$$

$$
\begin{array}{rcrrcr}
M_{\it moon} & = & 7.36 \times 10^{22}\ {\rm Kg}\,,
&\qquad
R_{\it moon} & = & 1737\ {\rm Km}\,,
\\
M_{\it earth} & = &6 \times 10^{24}\ {\rm Kg}\,,
&\qquad
R_{\it earth} & = & 6371\ {\rm Km}\,,
\end{array}
$$
$$
{W_{\tiny earth} \over W_{\tiny moon}}
=
\left(6371 \over 1737\right)^{2}\,{7.36 \over 600}
=
\color{#ff0000}{\Large 0.16502211055021}
$$

$W_{\tiny earth} = 250\ {\rm lb}$:
$$
W_{\tiny moon}
=
{W_{\tiny earth} \over 0.16502211055021}
=
{250\ {\rm lb}
 \over
 0.16502211055021}
=
{\large\color{#ff0000}{41.2555276375525}\ {\rm lb}}
$$
A: You just need to find the weight force the person creates when on the moon and then multiply the mass of the person by the ratio of $F_{moon}$ to $F_{earth}$
