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If a particle's acceleration due to gravity on earth is $9.8 m / s^{2}$ , what is the particle's acceleration due to gravity on the moon? (Given the mass of the earth is $6 \times 10^{22} kg$ , the mass of the moon is $7.3 \times 10^{22} kg$ , the radius of the earth is $6400 km$ and the radius of the moon is $1740 km$ .)

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0.16 m / s^{2}

1.57 m / s^{2}

4.9 m / s^{2}

None of these

answer is B.

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Detailed Solution

The value of g on the surface of the moon is

1.57 m / s^{2}

.
The acceleration due to gravity on the moon is

a = \frac{G M_{m}}{{R_{m}}^{2}} . . . . . . . . . . . . . . \dots (1)

Where,

M_{m}

is the mass of the moon and

R_{m}

is the radius of the moon.
The corresponding value of

g

on the surface of earth is

g = \frac{G M_{e}}{{R_{e}}^{2}} \dots . . . . . . . . . . . . . . . . . . . . . . . . . (2)

Where,

R_{e}

is the radius of earth and

M_{e}

is the mass of earth. Taking the ratios (1) and (2);

\frac{a}{g} = \frac{M_{m}}{M_{e}} \times \frac{{R_{e}}^{2}}{{R_{m}}^{2}} . . . . . . . . . . . . . . . . . . . \dots (3)

Given:
Mass of the earth is

M_{e} = 6 \times 10^{22} kg

,
Mass of the moon

M_{m} = 7.3 \times 10^{22} kg

,
The radius of the earth

R_{e} = 6400 km

The radius of the moon

R_{m} = 1740 km

.
Substituting the above values in (3):

\Rightarrow a = \frac{7.3 \times 10^{22}}{6 \times 10^{24}} \times ({\frac{6400}{1740})}^{2}

\Rightarrow a = 0.16 a

\Rightarrow a = 0.16 \times 9.8

\Rightarrow a = 1.57 m / s^{2}

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