Gujarati
Hindi
7.Gravitation
normal

A spherical part of radius $R/2$ is excavated from the asteroid of mass $M$ as shown in the figure. The gravitational acceleration at a point on the surface of the asteroid just above the excavation is

A

$\frac{{GM}}{{{R^2}}}$

B

$\frac{{GM}}{{2{R^2}}}$

C

$\frac{{GM}}{{8{R^2}}}$

D

$\frac{{7GM}}{{8{R^2}}}$

Solution

Acceleration due to gravitation

$=\frac{\mathrm{GM}}{\mathrm{R}^{2}}=\frac{\mathrm{GM} / 8}{\mathrm{R}^{2} / 4}=\frac{\mathrm{GM}}{2 \mathrm{R}^{2}}$

Standard 11
Physics

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