A $$90$$ kg body placed at $$2R$$ distance from surface of earth experiences gravitational pull of: ($$R$$ = Radius of earth, $$g = 10$$ m s$$^{-2}$$)

A 90 kg body is placed at a height $$2R$$ above Earth’s surface and the gravitational pull on it is to be determined.

The acceleration due to gravity at a height $$h$$ above Earth’s surface is given by $$ g' = g\left(\frac{R}{R + h}\right)^2 $$. This follows from Newton’s law of gravitation, since $$g' = \frac{GM}{(R+h)^2}$$ and $$g = \frac{GM}{R^2}$$.

For $$h = 2R$$, the distance from Earth’s center is $$R + h = 3R$$, so substituting into the formula yields $$ g' = g\left(\frac{R}{3R}\right)^2 = g\times\frac{1}{9} = \frac{10}{9}\,\text{m/s}^2 $$.

The gravitational force on the body is then $$ F = m\,g' = 90 \times \frac{10}{9} = 100\,\text{N} $$.

The correct answer is Option A: 100 N.