Question

If mass and radius of the Earth are $6.0\times {10}^{24}$ kg and $6.4\times {10}^6$ m respectively, calculate the force exerted by the Earth on a body of mass 1 kg. Also, calculate acceleration produced in the body of mass 1 kg and acceleration produced in the Earth.

• A. 9.8 and 1 respectively
• B. 9.8 and respectively
• C. and 9.8 respectively
• D. 9.8 for both

Answer 1

The correct option is B

9.8 and respectively

From Newton's law of gravitation, we know that the force of attaraction between two bodies is given by:
F = $\frac{G{m}_1{m}_2}{r^2}$
Here, $m_1$ = mass of Earth = $6.0\times {10}^{24}$ kg and $m_2$ = mass of the body = 1 kg
r = distance between the two bodies = radius of the Earth = $6.4\times {10}^6$ m
G = $6.67\times {10}^{-11}$ N$\frac{m^2}{k{g}^2}$
F = $\frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}\times 1}{{(6.4\times {10}^6)}^2}$ = 9.8 N
This shows that the Earth exerts a force of 9.8 N on a body of mass 1 kg. The body will exert an equal attractive force of 9.8 N on the Earth.
Acceleration produced in the body of mass 1 kg:
Force = mass $\times$ acceleration
Therefore, Acceleration, a = $\frac{F}{m}$ = $\frac{F}{1}$ = 9.8 $\frac{m}{s^2}$
Thus, the acceleration produced in a body of mass 1 kg due to attraction of the Earth is 9.8 $\frac{m}{s^2}$, which is quite large. Thus, when a body is released, it falls towards the Earth with an acceleration of 9.8 $\frac{m}{s^2}$, which can be easily observed.
Acceleration produced in the Earth:
Similarly, acceleration of the Earth is given by $\frac{Force}{MassofEarth}$ = $\frac{9.8}{6.0\times {10}^{24}}$ = $1.63\times {10}^{-24}$ $\frac{m}{s^2}$
This shows that the acceleration produced in the Earth by a body of mass 1 kg is $1.63\times {10}^{-24}$ $\frac{m}{s^2}$ which is very small and cannot be observed.