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Which of the following is the mass of the moon if acceleration due to gravity on the moon is $1.62$ $m s^{- 2}$ ,, gravitation constant G is $6.67 \times 10^{- 11}$ $N m^{2} {kg}^{- 2}$ and the radius of the moon is $1.74 \times 10^{6} m$ )

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7.35 \times 10^{22} kg

74 \times 10^{6}

kg

7.4 \times 10^{16}

kg

7.4 \times 10^{20}

kg

answer is A.

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Detailed Solution

The mass of the moon is

7.35 \times 10^{22} kg

if acceleration due to gravity on the moon is

1.62

m s^{- 2}

,, gravitation constant G is

6.67 \times 10^{- 11}

N m^{2} {kg}^{- 2}

and the radius of the moon is

1.74 \times 10^{6} m

.
As per Newton's universal law of gravitation we know that force of attraction between two bodies is given by

F = \frac{G m_{1} m_{2}}{r^{2}} . . . . . . . . . . (i)

G

is the gravitational constant.

m_{1}

is the mass of the first body.

m_{2}

is the mass of the second body.

r

is the distance between two bodies.
At the same time, we know that

F = mg . . . . . . . . . (ii)

On the surface of the moon, the distance to the centre of mass is the same as the radius of the moon

R = 1.74 \times 10^{6} m

m

is the mass of the earth.

g

is the acceleration due to gravity.
Here we need to find the mass of the moon

M

.
From (i) and (ii)
The force of attraction between the body and earth can be written as,

\frac{GMm}{R^{2}} = mg . . . . . . . . . . . . . . . (iii)

From equation (iii) we can write

GMm = {mgR}^{2}

\Rightarrow M = \frac{{mgR}^{2}}{Gm}

\Rightarrow M = \frac{{gR}^{2}}{G} . . . . . . . . . . . . (iv)

Substituting the values in (iv)

∴ M = \frac{1.62 \times {(1.740 \times 10^{6})}^{2}}{6.67 \times 10^{- 11}}

\Rightarrow M = \frac{4.904 \times 10^{12}}{{6.67 \times 10}^{- 11}}

\Rightarrow M = \frac{0.735 \times 10^{12}}{10^{- 11}}

\Rightarrow M = 0.735 \times 10^{23}

\Rightarrow M = 7.35 \times 10^{22}

kg

Thus, the mass of the moon is

7.35 \times 10^{22}

kg .

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