Question

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

• A. $\left(\frac{4-f}{1-f}\right).\frac{3\pi }{4G{T}^2}$
• B. $\left(\frac{4-f}{1+f}\right).\frac{3\pi }{4G{T}^2}$
• C. $\left(\frac{4-3f}{1-f}\right).\frac{3\pi }{4G{T}^2}$
• D. $\left(\frac{4-2f}{1-f}\right).\frac{3\pi }{4G{T}^2}$

Answer 1

Answer: (A)

Due to rotation of earth, acceleration due to gravity at latitude $\lambda$ is
$g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$
At equator $\lambda =0^{\circ }$
$\Rightarrow g_e=g-{\omega }^{2}R$
At equator of $60^{\circ },\lambda =60^{\circ }$
$\Rightarrow g_{\lambda }=g-{\omega }^{2}R{\left(\frac{1}{2}\right)}^2$
Given,Weight at equator $=f\times$ Weight atlatitude $\lambda$
$\Rightarrow m{g}_e=f\times m{g}_{\lambda }$
$\Rightarrow g_e=f{g}_{\lambda }$
$\Rightarrow g-{\omega }^{2}R=f\left(g-\frac{{\omega }^{2}R}{4}\right)$
$\Rightarrow g\left(1-f\right)=\frac{{\omega }^{2}R}{4}\left(4-f\right)\dots \left(i\right)$
Now, as $g=\frac{GM}{R^2}$ and $\omega =\frac{2\pi }{T}$
Also, $M=\frac{4}{3}\pi R^3.\rho$
Substituting these in Eq. $\left(i\right)$, we get
$\frac{4}{3}\pi \rho GR\left(1-f\right)=\frac{4{\pi }^{2}R}{4T^2}\left(4-f\right)$
$\Rightarrow \rho =\left(\frac{4-f}{1-f}\right).\frac{3\pi }{4G{T}^2}$