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# 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 earth. If the radius of earth is R, radius of planet will be

$a)\; 2 R \\b) \; 4 R \\c)\;\frac{R}{4}\\ d)\;\frac{R}{2}$

$g= \large\frac{4}{3}$$\pi \rho GR$
$\large\frac{R_p}{R_E}=\bigg(\large\frac{g_p}{g_e}\bigg)\bigg(\large\frac{\rho _e}{\rho _p}\bigg)$
$\qquad= 1\times \large\frac{1}{2}$
$\qquad=> R_p =\large\frac{R_E}{2}=\large\frac{R}{2}$
Hence d is the correct answer.
edited Jun 26, 2014