Question

42. If the diameter of the Earth becomes two times its present value and its mass remains unchanged, then how would the weight of anobject on the surface of the Earth be affected?

Answer 1

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