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.