The value of g on the earth's surface is 9.8 | vedclass

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(B) The acceleration due to gravity $(g)$ on the surface of the earth is given by the formula:$g = \frac{GM}{R^2}$where $G$ is the gravitational const

Vedclass · Accepted Answer

$88.2$

Vedclass · Answer

(B) The acceleration due to gravity $(g)$ on the surface of the earth is given by the formula:$g = \frac{GM}{R^2}$where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.If the earth shrinks to one-third of its present size,the new radius $R'$ becomes $\frac{R}{3}$.The mass $M$ remains constant.The new acceleration due to gravity $g'$ is:$g' = \frac{GM}{(R')^2} = \frac{GM}{(\frac{R}{3})^2} = \frac{GM}{\frac{R^2}{9}} = 9 	imes \frac{GM}{R^2}$Since $g = \frac{GM}{R^2} = 9.8 \, m s^{-2}$,we have:$g' = 9 	imes 9.8 \, m s^{-2} = 88.2 \, m s^{-2}$.

The value of $g$ on the earth's surface is $9.8 \, m s^{-2}$. Suppose the earth suddenly shrinks to one-third of its present size without losing any mass. What is the value of $g$ on the surface of the shrunk earth (in $, m s^{-2}$)?

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