Question

The value of acceleration due to gravity g at distance r from earths centre such that r < R depend on r according to relation (R = radius of earth)

• A. $g\propto \frac{1}{r^2}$
• B. $g\propto \frac{1}{r}$
• C. $g\propto r$
• D. $g\propto r^2$

Answer 1

Answer: (D)

Key Idea: The acceleration due to gravity arises due to force of attraction acting on body due to earth. Gravitational force of attraction is $F=\frac{G{M}_{e}m}{r^2}$ ?(i) where G is gravitational constant, $M_e$ is mass of earth and r the distance from earths centre. From Newtons law, F = mg ?(ii) Equating Eqs. (i) and (ii), we get $\frac{G{M}_{e}m}{r^2}=mg$ $\Rightarrow$ $g=\frac{G{M}_e}{r^2}$ $\Rightarrow$ $g\propto \frac{1}{r^2}$ Note: The value of g decreases on going below the surface of earth and finally becomes zero at the centre.