Question

If mass and radius of earth is $6.0\ast {10}^{24}$ kg and $6.4\ast {10}^6$ m respectively, calculate the force exerted by earth on a body of mass
1 kg. Also, calculate :
(i) acceleration produced in the body of mass
1 kg, and
(ii) acceleration produced in the earth

Answer 1

From Newton's law of gravitation, we know that the force of attraction between two bodies is given by

$F=\frac{G{m}_1{m}_2}{r^2}$

Here, m1 = mass of earth = $6.0\ast {10}^{24}$ kg;

m2 = mass of body = 1 kg

r = distance between the two bodies

= radius of earth = $6.4\ast {10}^6$ m

G = $6.67\ast {10}^{-11}$ N$\frac{m^2}{k{g}^2}$

$F=\frac{6.67\ast {10}^{-11}\ast 6.0\ast {10}^{24}\ast 1}{(6.4\ast {10}^6{)}^2}$ = 9.8 N

This shows that earth exerts a force of 9.8 N on a body of mass of 1 kg. The body will exert an equal force of attraction of 9.8 N on earth.

(i) Acceleration produced in the body of mass 1kg

Force = mass × 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 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.

(ii) Acceleration produced in the earth

Similarly, acceleration of earth is given by

= $\frac{Force}{Massofearth}$ = $\frac{9.8}{6.0\ast {10}^{24}}$

1.63 × ${10}^{-24}$ $\frac{m}{s^2}$

This shows that the acceleration produced in the earth by a body of mass 1 kg is
1.63 × ${10}^{-24}$ $\frac{m}{s^2}$​ which is very small and cannot be observed.