The moon's radius is (1/4) th that of the earth its mass is (1/80) times that of earth, if g represents the acceleration due to gravity on the surface of earth, that on the surface of moon is -

Question

The moon's radius is (1/4) th that of the earth its mass is (1/80) times that of earth, if g represents the acceleration due to gravity on the surface of earth, that on the surface of moon is - (A) (g/4) (B) (g/5) (C) (g/6) (D) (g/8)

Tardigrade Admin · Accepted Answer

We know gravitational acceleration g=(G M/R2) for moon Rm=(1/4)R Mm=(M/80) so g'=(G (. (M/80) .)/(. (R)4 .)2)=(G M/5 R2)=(g/5)

Q. The moon's radius is $\frac{1}{4}$ th that of the earth & its mass is $\frac{1}{80}$ times that of earth, if $g$ represents the acceleration due to gravity on the surface of earth, that on the surface of moon is -

$\frac{g}{4}$

$\frac{g}{5}$

$\frac{g}{6}$

$\frac{g}{8}$

We know gravitational acceleration
$g = \frac{GM}{R ^{2}}$
for moon $R_{m} = \frac{1}{4} R & M_{m} = \frac{M}{80}$
so $g^{^{'}} = \frac{G ( \frac{M}{80} )}{( \frac{R}{4} ) ^{2}} = \frac{GM}{5 R ^{2}} = \frac{g}{5}$

Q. The moon's radius is 41​ th that of the earth & its mass is 801​ times that of earth, if g represents the acceleration due to gravity on the surface of earth, that on the surface of moon is -

4g​

5g​

6g​

8g​