Question

Question #7201d

Answer 1

`(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.