Question #7201d

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Question #7201d

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Answer 1 · 2015-05-10T22:18:57

$\frac{g_{e}}{g_{m}} = 6.05$ $(g_e)/(g_m)=6.05$

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 $m$ $m$ on the surface of the earth with an object of mass $m$ $m$ on the surface of the moon.

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

$F = \frac{G m M}{R^{2}}$ $F=(GmM)/(R^(2)$

For this problem:

$F_{e}$ $F_e$ = force of attraction on the earth

$F_{m}$ $F_m$ = force of attraction on the moon

$R_{e}$ $R_e$ = radius earth

$R_{m}$ $R_m$ = radius moon

$m$ $m$ = mass object

So:

$F_{e} = \frac{G m M_{e}}{R_{e}^{2}}$ $F_e=(GmM_e)/(R_e^(2)$ Eqn $(1)$ $color(red)((1))$

$F_{m} = \frac{G m M_{m}}{R_{m}^{2}}$ $F_m=(GmM_m)/(R_m^(2))$ Eqn $(2)$ $color(red)((2))$

We know that:

$R_{m} = 0.273 R_{e}$ $R_m=0.273R_e$

and:

$M_{m} = \frac{M_{e}}{81.3}$ $M_m=M_e/81.3$

So we can substitute these values into $(2)$ $color(red)((2))$ $\Rightarrow$ $rArr$

$F_{m} = \frac{G m [\frac{M_{e}}{81.3}]}{{(0.273 R_{e})}_{e}^{2}}$ $F_m=(Gm[(M_e)/(81.3)])/((0.273R_e)^(2))$

$F_{m} = \frac{G m [\frac{M_{e}}{81.3}]}{0.0745 R_{e}^{2}}$ $F_m=(Gm[(M_e)/(81.3)])/(0.0745R_e^(2))$ Eqn $(3)$ $color(red)((3))$

Now divide $(1)$ $color(red)((1))$ by $(3)$ $color(red)((3))$ $\Rightarrow$ $rArr$

$\frac{F_{e}}{F_{m}} = \frac{G m M_{e}}{R_{e}^{2}} \times \frac{0.0745 R_{e}^{2}}{G m [\frac{M_{e}}{81.3}]}$ $(F_e)/(F_m)=(GmM_e)/(R_e^(2))xx(0.0745R_e^(2))/(Gm[[M_e)/(81.3)]$

This cancels down to:

$\frac{F_{e}}{F_{m}} = \frac{0.0745}{\frac{1}{81.3}} = 6.05$ $(F_e)/(F_m)=(0.0745)/((1)/(81.3))=6.05$

Since $F = m a$ $F=ma$ :

$\frac{m g_{e}}{m g_{m}} = 6.05$ $(mg_e)/(mg_m)=6.05$

So:

$\frac{g_{e}}{g_{m}} = 6.05$ $(g_e)/(g_m)=6.05$

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.

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