Question

Radius of earth is equal to $6 \times {10^6}m$. Acceleration due to gravity is equal to $9.8m/{s^2}$. Gravitational constant G is equal to $6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$. Then mass of the Earth is-
A. $6.0 \times {10^{24}}kg$
B. $5.3 \times {10^{24}}kg$
C. $5.9 \times {10^{24}}kg$
D. $6.6 \times {10^{24}}kg$

Answer 1

Hint: We will acknowledge all the values given in the question and put them respectively into the formula for acceleration due to gravity where M is mass, G is the gravitational constant and r is the radius, in this case, radius of the earth.
Formula used: $g = \dfrac{{GM}}{{{r^2}}}$

Complete Step-by-Step solution:
The acceleration due to gravity on the surface of the earth is given by the formula $g = \dfrac{{GM}}{{{r^2}}}$.
Where, g is the acceleration due to gravity and G is the gravitational constant.
The value of the radius of the earth given in the question is $r = 6 \times {10^6}m$
The value of the gravitational constant given in the question is $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
The value of acceleration due to gravity given in the question is $9.8m/{s^2}$
Putting all the values into the formula and the value of M from the above formula will be-
$
   \Rightarrow M = g\dfrac{{{R^2}}}{G} \\
    \\
   \Rightarrow M = \dfrac{{9.8 \times 36 \times {{10}^{12}}}}{{6.67 \times {{10}^{ - 11}}}} \\
    \\
   \Rightarrow M = 5.28 \times {10^{24}} \\
    \\
   \Rightarrow M \cong 5.9 \times {10^{24}} \\
$
Hence, option C is the correct option.

Note: The acceleration that an object is experiencing due to gravitational force is called the acceleration due to gravity. The unit SI is $m/{s^2}$. Gravity-based acceleration is a vector which means it has both a magnitude and a direction. The letter g represents the acceleration due to gravity at the Earth's surface. It has a standard value defined as 9.8 $m/{s^2}$ (32.1740 $ft/{s^2}$). However, in free fall, a body 's actual acceleration varies with the location.