An astronaut landed on a planet and found that his weight at the pole of the planet was one third of his weight at the pole of the earth. He also found himself to be weightless at the equator of the planet. The planet is a homogeneous sphere of radius half that of the earth. the duration of a day on the planet is……... Given density of the earth = d₀.

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$\sqrt{\frac{9 π}{2 G d_{0}}}$

$\sqrt{\frac{2 π}{9 G d_{0}}}$

$\sqrt{\frac{9 G d_{0}}{2 π}}$

$\sqrt{\frac{G d_{0}}{2 π}}$

answer is A.

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Detailed Solution

$\frac{1}{3} {(g_{pole})}_{earth} = {(g_{pole})}_{planet} \frac{1}{3} \frac{{GM}_{e}}{R_{e}^{2}} = \frac{{GM}_{P}}{R_{P}^{2}} \Rightarrow \frac{1}{3} \cdot \frac{\frac{4}{3} {πR}_{e}^{3} \cdot d_{0}}{R_{e}^{2}} = \frac{\frac{4}{3} {πR}_{P}^{3} \cdot d_{P}}{R_{P}^{2}} ∴ d_{0} R_{e} = 3 d_{P} R_{P} . . . . (1)$
For weightlessness at the equator
$ω^{2} R = g \begin{array}{l} ∴ ω^{2} R_{P} = \frac{{GM}_{P}}{R_{P}^{2}} \\ \Rightarrow ω^{2} = G \frac{4}{3} {πd}_{P} = \frac{4}{3} Gπ \frac{1}{3} \frac{d_{0} R_{e}}{R_{P}} \end{array}$
$= \frac{4}{9} {Gπd}_{0} (2) [\frac{R_{e}}{R_{P}} = 2] \begin{array}{ll} ∴ \frac{4 π^{2}}{T^{2}} = \frac{8}{9} {Gπd}_{0} \\ ∴ T^{2} = \frac{9 π}{2 {Gd}_{0}} \\ ∴ T = \sqrt{\frac{9 π}{2 {Gd}_{0}}} \end{array}$