Acceleration due to gravity at the surface of | vedclass

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(B) The acceleration due to gravity $g$ at the surface of a body of radius $R$ and density $\rho$ is given by:$g = \frac{GM}{R^2} = \frac{G(\rho \cdot

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$\frac{2}{3} R$

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(B) The acceleration due to gravity $g$ at the surface of a body of radius $R$ and density $\rho$ is given by:$g = \frac{GM}{R^2} = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3} \pi G \rho R$Given that $g_p = g_e$ and $\rho_p = 1.5 \rho_e = \frac{3}{2} \rho_e$.Equating the gravity expressions for the planet and the Earth:$\frac{4}{3} \pi G \rho_p R_p = \frac{4}{3} \pi G \rho_e R_e$$\rho_p R_p = \rho_e R_e$Substituting $\rho_p = \frac{3}{2} \rho_e$ and $R_e = R$:$(\frac{3}{2} \rho_e) R_p = \rho_e R$$R_p = \frac{2}{3} R$Thus,the radius of the planet is $\frac{2}{3} R$.

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

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