Question

The moon’s radius is $1/4$ times that of earth and its mass is $1/80$ times that of the earth. If g represents the acceleration due to gravity on the surface of the earth than on the surface of the moon its value is:
$
  (a){\text{ }}\dfrac{g}{4} \\
  (b){\text{ }}\dfrac{g}{5} \\
  (c){\text{ }}\dfrac{g}{6} \\
  (d){\text{ }}\dfrac{g}{8} \\
 $

Answer 1

Hint: In this question use the formula between the acceleration due to gravity (g), radius(r) and the mass (M) that is $g = G\dfrac{M}{{{r^2}}}$, here G is the universal gravitation constant. Apply this formula for earth and moon separately and then make use of the two questions formed along with the constraints of the question to reach the right answer.

Complete step-by-step answer:
As we know the direct relation of acceleration due to gravity (g), radius (r), mass (M) which is given as,

$ \Rightarrow g = G\dfrac{M}{{{r^2}}}$ m/s².
So acceleration due to gravity on earth is
$ \Rightarrow {g_e} = G\dfrac{{{M_e}}}{{r_e^2}}$ m/s²................... (1)
Where, ${g_e}$ = acceleration due to gravity on earth.
              G = Universal gravitational constant.
             ${M_e}$ = Mass of the earth.
              $r_e^2$ = radius of the earth.

And acceleration due to gravity on the moon is

 $ \Rightarrow {g_m} = G\dfrac{{{M_m}}}{{r_m^2}}$ m/s²..................... (2)

Where, ${g_m}$ = acceleration due to gravity on the moon.
              G = Universal gravitational constant.
             ${M_m}$ = Mass of the moon.
              $r_m^2$ = radius of the moon.

Now divide equation (2) with equation (1) we have,

$ \Rightarrow \dfrac{{{g_m}}}{{{g_e}}} = \dfrac{{G\dfrac{{{M_m}}}{{r_m^2}}}}{{G\dfrac{{{M_e}}}{{r_e^2}}}} = \dfrac{{{M_m}}}{{r_m^2}} \times \dfrac{{r_e^2}}{{{M_e}}}$................. (3)

Now it is given that the radius of the moon is (1/4) times that of earth.
$ \Rightarrow {r_m} = \dfrac{1}{4}{r_e}$

And it is also given that the mass of the moon is (1/80) times that of the earth.
$ \Rightarrow {M_m} = \dfrac{1}{80}{M_e}$

Now substitute this value in equation (3) we have,

\[ \Rightarrow \dfrac{{{g_m}}}{{{g_e}}} = \dfrac{{\dfrac{1}{{80}}{M_e}}}{{{{\left( {\dfrac{1}{4}{r_e}} \right)}^2}}} \times \dfrac{{r_e^2}}{{{M_e}}}\]

Now simplify this we have,

$ \Rightarrow \dfrac{{{g_m}}}{{{g_e}}} = \dfrac{{\dfrac{1}{{80}}}}{{\dfrac{1}{{{4^2}}}}} = \dfrac{1}{{80}} \times 16 = \dfrac{1}{5}$
$ \Rightarrow {g_m} = \dfrac{1}{5}{g_e}$

Now it is given that the gravitation at the earth is represented by g.

$ \Rightarrow {g_e} = g$
So we have,
$ \Rightarrow {g_m} = \dfrac{1}{5}g$

So this is the required value of the acceleration on the surface of the moon w.r.t the earth surface.
Hence option (B) is the required answer.

Note: Since earth is a bigger planet as compared to moon then eventually the mass of the moon is less as compared to earth, due to which the gravitational pull of moon over an object is less as compared to that of earth. This is the main reason we see objects flying in outer space. Humans even are weighted lighter or moon as compared to the earth due to this same reason.