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Which of the following is the mass of the earth?

(Given gravitational constant $G$ is $6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ , acceleration due to gravity $g$ is $9.8$ ${ms}^{- 2}$ , radius of earth $R_{e}$ is $6370 km$ )

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5.96 \times 10^{24}

kg

59.6 \times 10^{24}

kg

5.96 \times 10^{30}

kg

5.96 \times 10^{20}

kg

answer is A.

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Detailed Solution

The mass of the earth with the given details is

5.96 \times 10^{24} kg

.
As per Newton's universal law of gravitation we know that force of attraction between 2 bodies is given by

F = \frac{G m_{1} m_{2}}{r^{2}} - - - (a)

G

is the gravitational constant

m_{1}

is the mass of 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 - - - (b)

gravitational constant

G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

Acceleration due to gravity

g = 9.8 {ms}^{- 2}

Radius of earth is

R_{e} = 6370 km

\Rightarrow R_{e} = 6370 \times 10^{3} m

We are asked to find the mass of the earth

M

.
Mass of any body can be considered as

m

.
From (a) and (b) the force of attraction between the body and earth can be written as

\frac{GMm}{{R_{e}}^{2}} = mg

Hence, we can write

M = \frac{gm {R_{e}}^{2}}{Gm}

\Rightarrow M = \frac{g {R_{e}}^{2}}{G} - - - - (i)

Substituting the values in (i)

\Rightarrow M = \frac{9.8 \times {(6370 \times 10^{3})}^{2}}{6.67 \times 10^{- 11}}

\Rightarrow M = 59618233.9 \times 10^{6} \times 10^{11}

\Rightarrow M = 5.96 \times 10^{24}

\Rightarrow M = 5.96 \times 10^{24} kg

Thus, the mass of the earth with the given details is

5.96 \times 10^{24} kg .

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