Question

If the diameter of the earth becomes twice of the present value but its mass remains unchanged, then how would be the weight of an object on the surface of the earth affected?

Answer 1

Dear student,
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\mathrm{is}\mathrm{related}\mathrm{with}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{as}, \\ \mathrm{g}=\frac{\mathrm{G}\mathrm{M}}{{\mathrm{R}}^2}----\left(1\right) \\ \mathrm{When}\mathrm{R}\mathrm{becomes}2\mathrm{R}\mathrm{then}, \\ \mathrm{g}\text{'}=\frac{\mathrm{GM}}{4{\mathrm{R}}^2}=\frac{\mathrm{g}}{4}---\left(2\right) \\ \mathrm{Now}\mathrm{the}\mathrm{body}\mathrm{which}\mathrm{weighed}\mathrm{W}=\mathrm{mg}\mathrm{on}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{earth}\mathrm{now}\mathrm{has}\mathrm{weight}, \\ \mathrm{W}\text{'}=\mathrm{mg}\text{'}=\mathrm{m}\times \frac{\mathrm{g}}{4}=\frac{\mathrm{W}}{4}. \\ \mathrm{i}.\mathrm{e}.\mathrm{the}\mathrm{body}\mathrm{now}\mathrm{has}\mathrm{weight}1/4\mathrm{of}\mathrm{the}\mathrm{previous}\mathrm{case}. \end{gathered}$$
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