Radius of a planet with twice Earth’s density

The density of a newly discovered 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 the earth. If the radius of the earth is $R$, the radius of the planet would be:

(2004)

$4R$

$\frac{1}{4}R$

$\frac{1}{2}R$

$2R$

Solution:

Using $g = \frac{4}{3}\pi G \rho R$, we have $\rho_P R_P = \rho_E R_E$ since $g$ is the same for both. Given $\rho_P = 2\rho_E$, we get $2\rho_E R_P = \rho_E R$, which gives $R_P = \frac{1}{2}R$.

Radius of a planet with twice Earth’s density

Solution:

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