JEE MAIN - Physics (2017 (Offline) - No. 18)
The variation of acceleration due to gravity $$g$$ with distance d from centre of the earth is best represented by
(R = Earth’s radius):
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Explanation
When d < R means distance of a point is d from the center of the circle where d is inside the earth surface. In this case the value of acceleration, $$g = -{{GMd} \over {{R^3}}}$$
$$\therefore$$ $$g \propto d$$
So inside of the earth surface g - d graph is straight line.
When d > R means distance of a point is d from the center of the circle where d is outside of the earth's surface. In this case the value of acceleration, $$g = - {{GM} \over {{d^2}}}$$
$$\therefore$$ $$g \propto {1 \over {{d^2}}}$$
So outside of the earth surface g - d graph is hyperbolic.
$$\therefore$$ $$g \propto d$$
So inside of the earth surface g - d graph is straight line.
When d > R means distance of a point is d from the center of the circle where d is outside of the earth's surface. In this case the value of acceleration, $$g = - {{GM} \over {{d^2}}}$$
$$\therefore$$ $$g \propto {1 \over {{d^2}}}$$
So outside of the earth surface g - d graph is hyperbolic.
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