The density of a new planet is twice that of | vedclass

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(B) The acceleration due to gravity $g$ at the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$,where $\rho$ is the density

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$R / 2$

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(B) The acceleration due to gravity $g$ at the surface of a planet is given by the formula: $g = \frac{4}{3} \pi \rho G R$,where $\rho$ is the density,$G$ is the gravitational constant,and $R$ is the radius of the planet.Given that the density of the planet $\rho_p = 2 \rho_e$ and the acceleration due to gravity $g_p = g_e$.Equating the expressions for both the planet and the earth:$\frac{4}{3} \pi \rho_p G R_p = \frac{4}{3} \pi \rho_e G R_e$$\rho_p R_p = \rho_e R_e$Substituting $\rho_p = 2 \rho_e$ and $R_e = R$:$(2 \rho_e) R_p = \rho_e R$$2 R_p = R$$R_p = \frac{R}{2}$

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth,then the radius of the planet would be

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