Question 410: Gravitation

Question

The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in figures below
The correct figure is

Accepted Answer

The acceleration due to gravity at a depth d below surface of earth is
$$g' = {{GM} \over {{R^2}}}\left( {1 - {d \over R}} \right) = g\left( {1 - {d \over R}} \right)$$
g' = 0  at d = R.
i.e., acceleration due to gravity is zero at the centre of earth.
Thus, the variation in value g with r is
For,  r > R,
$$g' = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}} = {{g{R^2}} \over {{r^2}}} \Rightarrow g' \propto {1 \over {{r^2}}}$$
Here, R + h = r
For r < R, $$g' = g\left( {1 - {d \over R}} \right) = {{gr} \over R}$$
Here, $$R - d = r \Rightarrow g' \propto r$$
Therefore, the variation of g with distance from centre of the earth will be as shown in the figure.

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Gravitation

The dependence of acceleration due to gravity g on the distance r from the centre of the earth, assumed to be a sphere of radius R of uniform density is as shown in figures below The correct figure is

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