Question #7201d

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Answer 1 · 2015-05-10T22:18:57

$(g_e)/(g_m)=6.05$

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 $color(red)((1))$

$F_m=(GmM_m)/(R_m^(2))$ Eqn $color(red)((2))$

We know that:

$R_m=0.273R_e$

and:

$M_m=M_e/81.3$

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

$F_m=(Gm[(M_e)/(81.3)])/((0.273R_e)^(2))$

$F_m=(Gm[(M_e)/(81.3)])/(0.0745R_e^(2))$ Eqn $color(red)((3))$

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

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

This cancels down to:

$(F_e)/(F_m)=(0.0745)/((1)/(81.3))=6.05$

Since $F=ma$ :

$(mg_e)/(mg_m)=6.05$

So:

$(g_e)/(g_m)=6.05$

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.