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A certain planet completes one rotation about its axis in time $T$. The weight of an object placed at the equator on the planet's surface is a fraction $f(f$ is close to unity) of its weight recorded at a latitude of $60^{\circ}$. The density of the planet (assumed to be a uniform perfect sphere) is given by
$\left(\frac{4-f}{1-f}\right) \cdot \frac{3 \pi}{4 G T^2}$
$\left(\frac{4-f}{1+f}\right) \cdot \frac{3 \pi}{4 G T^2}$
$\left(\frac{4-3f}{1-f}\right) \cdot \frac{3 \pi}{4 G T^2}$
$\left(\frac{4-2f}{1-f}\right) \cdot \frac{3 \pi}{4 G T^2}$
Solution
(a)
Due to rotation of earth, acceleration due to gravity at latitude $\lambda$ is
$\quad g^{\prime}=g-\omega^2 R \cos ^2 \lambda$
$\text { At equator } \lambda=0^{\circ}$
$\Rightarrow \quad g_e=g-\omega^2 R$
$\text { At latitude of } 60^{\circ}, \lambda=60^{\circ}$
$\Rightarrow \quad g_\lambda=g-\omega^2 R\left(\frac{1}{2}\right)^2$
$\text { Given, }$
$\text { Weight at equator }=f \times \text { Weight at }$
$\Rightarrow \quad m g_e=f \times m g_\lambda \Rightarrow g_e=f g_\lambda$
$\Rightarrow \quad g-\omega^2 R=f\left(g-\frac{\omega^2 R}{4}\right)$
$\Rightarrow \quad g(1-f)=\frac{\omega^2 R}{4}(4-f) \quad \dots(i)$
$\text { Now, as } g=\frac{G M}{R^2} \text { and } \omega=\frac{2 \pi}{T}$
Also, $\quad M=\frac{4}{3} \pi R^3 \cdot \rho$
Substituting these in Eq. $(i)$, we get
$\frac{4}{3} \pi \rho G R(1-f)=\frac{4 \pi^2 R}{4 T^2}(4-f)$
$\Rightarrow \quad \rho=\left(\frac{4-f}{1-f}\right) \cdot \frac{3 \pi}{4 G T^2}$
