A) \[g\propto \,\frac{1}{{{r}^{2}}}\]
B) \[g\propto \,\frac{1}{r}\]
C) \[g\propto \,\,r\]
D) \[g\propto \,\,{{r}^{2}}\]
Correct Answer: D
Solution :
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.
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