Question

If the axis of rotation suddenly becomes tangent at equator of the earth then the periodic time of a geostationary satellite:
A) 48 hrs
B) 96 hrs
C) 24 hrs
D) 84 hrs

Answer 1

Hint : Since the moment of inertia of the system is changing, Use the conservation of angular momentum and find out the change in angular velocity. By finding change in angular velocity, you will be able to find the change in time period by using the formula $w = \dfrac{{2\pi }}{T}$.

Complete step-by-step answer:
Given, the axis of rotation is being changed from the centre to the tangent.
Let the initial moment of inertia be I.
Let the final moment of inertia be ${I_1}$.
Since the axis of rotation is being changed from centre to the tangent. Hence new moment of inertia by parallel axis theorem we have,
${I_1} = {I_{COM}} + m{d^2} \\\ \Rightarrow {I_1} = \dfrac{2}{5}M{R^2} + M{R^2} \\\ \Rightarrow {I_1} = \dfrac{7}{5}M{R^2} \\\$
Putting the value of I (COM) as we know that moment of inertia of a sphere is $\dfrac{2}{5}M{R^2}$ from the central axis.
Now using the conservation of angular momentum we have,
$L = {L_1}$
$Iw = {I_1}{w_1} \\\ \Rightarrow \dfrac{2}{5}M{R^2}\,w = \dfrac{7}{5}M{R^2}\,{w_1} \\\ \Rightarrow {w_1} = \dfrac{2}{7}w \\\$ (putting the values)
Hence Time period becomes,
T = $\dfrac{{2\pi }}{{{w_1}}}$
$\Rightarrow$ T = $\dfrac{{2\pi }}{{\dfrac{2}{7}w}}$
$\Rightarrow$ T = 7 $\times \dfrac{{2\pi }}{w}$
$\Rightarrow$ T = 7 $\times$ 24 (Time period of earth = 24 hours)
$\Rightarrow$ T = 168 hr.

Note
i) We can only apply the law of conservation of angular momentum when there is no external torque acting on a body.
ii) Time period of a geostationary satellite is 24 hours.