Question 616: Centre of Mass and Rotational Motion

Question

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

Accepted Answer

When considering the Sun as a solid sphere, the formula for the moment of inertia is given by:
$ I = \frac{2}{5} m R^2 $
Here, $ m $ is the mass and $ R $ is the radius of the Sun.
To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum. The principle states that angular momentum before and after the expansion must be equal:
$ l' \omega' = l \omega $
Substituting the expression for angular momentum $ (I 	imes \omega) $ for both initial and expanded states, we get:
$ \frac{2}{5} m (2R)^2 	imes \frac{2\pi}{T'} = \frac{2}{5} m R^2 	imes \frac{2\pi}{T} $
This simplifies to:
$ \Rightarrow 4mR^2 	imes \frac{2\pi}{T'} = mR^2 	imes \frac{2\pi}{T} $
Cancelling out common terms, we find:
$ \Rightarrow 4 	imes \frac{1}{T'} = \frac{1}{T} $
Thus:
$ \Rightarrow T' = 4T = 4 	imes 27 = 108 	ext{ days} $
Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.

Answer

115 days

Answer

108 days

Answer

100 days

Answer

105 days

Centre of Mass and Rotational Motion

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.

Choose the correct answer: