Question with Solution
Consider two solid spherical asteroid of uniform density of mass M and radius R. In one asteroid a tunnel of very small size of depth R is bored to the centre and in other asteroid a spherical cavity of radius R/2 is made as shown in the figure. Now, identical particles of mass m dropped into the cavities of both asteroids from the top most point P. If force experienced by particle is and respectively in cavities of asteroids I and II, when they are x distance away from the centre of asteroids. If the time taken by particles to reach the centre of asteroids is and respectively then
Options:(select one or more)
Answer:
Solution:
Asteroid - I Asteroid - II
\text { Field } \mathrm{E}=\frac{\mathrm{GM}}{\mathrm{R}^{3}} \mathrm{x} & \frac{\mathrm{~F}_{\mathrm{I}}}{\mathrm{~F}_{\mathrm{II}}}=\frac{2 \mathrm{x}}{\mathrm{R}}
\mathrm{~T}_{\mathrm{I}}=\frac{2 \pi}{4} \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}} & \text { Field } \mathrm{E}_{\mathrm{II}}=\frac{\mathrm{GM}}{\mathrm{R}^{3}}\left(\frac{\mathrm{R}}{2}\right)
\mathrm{R}=\frac{1}{2} \frac{\mathrm{GM}}{\mathrm{R}^{3}}\left(\frac{\mathrm{R}}{2}\right) \mathrm{T}_{\mathrm{II}}^{2} & \mathrm{~T}_{\mathrm{II}}=2 \sqrt{\frac{\mathrm{R}^{3}}{\mathrm{GM}}}
