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

Answer 1

Answer: (A)

Due to rotation of earth, acceleration due to gravity at latitude $\lambda$ is
$g '=g-\omega^{2} R \cos^{2} \lambda$
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 $
$\Rightarrow mg_{e}=f\times mg_{\lambda}$
$\Rightarrow g_{e}=fg_{\lambda} $
$\Rightarrow g-\omega^{2}R=f\left(g-\frac{\omega^{2}R}{4}\right)$
$\Rightarrow g\left(1-f\right)=\frac{\omega^{2}R}{4}\left(4-f\right) \ldots\left(i\right)$
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) $
$\Rightarrow \rho=\left(\frac{4-f}{1-f}\right).\frac{3\pi}{4GT^{2}}$