# Question 7201d

May 10, 2015

$\frac{{g}_{e}}{{g}_{m}} = 6.05$

The data given in the question is way out. The mass of the moon is 1/81.3 of the earth so I'll work through with that.

I'll also use the radius of the moon to be 0.273 x the radius of the earth.

We'll compare an object of mass $m$ on the surface of the earth with an object of mass $m$ on the surface of the moon.

Newton says that the force of attraction between 2 objects $M$ and $m$ is:

F=(GmM)/(R^(2)

For this problem:

${F}_{e}$ = force of attraction on the earth

${F}_{m}$ = force of attraction on the moon

${R}_{e}$ = radius earth

${R}_{m}$ = radius moon

$m$ = mass object

So:

F_e=(GmM_e)/(R_e^(2) Eqn $\textcolor{red}{\left(1\right)}$

${F}_{m} = \frac{G m {M}_{m}}{{R}_{m}^{2}}$ Eqn$\textcolor{red}{\left(2\right)}$

We know that:

${R}_{m} = 0.273 {R}_{e}$

and:

${M}_{m} = {M}_{e} / 81.3$

So we can substitute these values into $\textcolor{red}{\left(2\right)}$ $\Rightarrow$

${F}_{m} = \frac{G m \left[\frac{{M}_{e}}{81.3}\right]}{{\left(0.273 {R}_{e}\right)}^{2}}$

${F}_{m} = \frac{G m \left[\frac{{M}_{e}}{81.3}\right]}{0.0745 {R}_{e}^{2}}$ Eqn $\textcolor{red}{\left(3\right)}$

Now divide $\textcolor{red}{\left(1\right)}$ by $\textcolor{red}{\left(3\right)}$ $\Rightarrow$

(F_e)/(F_m)=(GmM_e)/(R_e^(2))xx(0.0745R_e^(2))/(Gm[[M_e)/(81.3)]#

This cancels down to:

$\frac{{F}_{e}}{{F}_{m}} = \frac{0.0745}{\frac{1}{81.3}} = 6.05$

Since $F = m a$:

$\frac{m {g}_{e}}{m {g}_{m}} = 6.05$

So:

$\frac{{g}_{e}}{{g}_{m}} = 6.05$

This agrees with the generally accepted value of the acceleration due to gravity of the moon being about 1/6 of that of the earth.

You can substitute the values given in the question if you want to but I would check their source.