Considering earth to be a sphere of radius ' $R$ ' having uniform density ' $\rho$ ', then value of…

Considering earth to be a sphere of radius ' $R$ ' having uniform density ' $\rho$ ', then value of acceleration due to gravity ' $g$ ' in terms of R, $\rho$ and $\mathrm{G}$ is
  1. $\mathrm{g}=\sqrt{\frac{3 \pi \mathrm{R}}{\rho \mathrm{G}}}$
  2. $g=\sqrt{\frac{4}{3} \pi \rho \mathrm{GR}}$
  3. $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}$
  4. $\mathrm{g}=\frac{\mathrm{GM}}{\rho \mathrm{R}^2}$

Solution

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{\mathrm{G} \times \frac{4}{3} \pi \mathrm{R}^3 \rho}{\mathrm{R}^2}=\frac{4}{3} \pi \rho \mathrm{GR}$

Asked in: MHT CET 2023 (14 May Shift 1)

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