Question 186: Gravitation

Question

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :

Accepted Answer

The given situation is shown below
The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre.
Assuming 2 m at origin, the centre of mass of the system lies at
$$x = {{2m 	imes 0 + m 	imes d} \over {2m + m}} = {d \over 3}$$
Hence, $${F_G} = {F_C}$$
where, F
G
is gravitational force between them and F
C
is centripetal force.
$$ \Rightarrow {{G{m_1}{m_2}} \over {{r^2}}} = 2m{\omega ^2}x$$
$$ \Rightarrow {{G(2m)(m)} \over {{d^2}}} = 2m{\omega ^2} 	imes {d \over 3} \Rightarrow {\omega ^2} = {{3Gm} \over {{d^3}}}$$
$$ \Rightarrow \omega  = \sqrt {{{3Gm} \over {{d^3}}}} $$
We know that,
$$\omega  = {{2\pi } \over T}$$
$$	herefore$$ $$T = {{2\pi } \over \omega }$$
$$\sqrt {{{3Gm} \over {{d^3}}}} $$
$$ = 2\pi \sqrt {{{{d^3}} \over {3Gm}}} $$ [using Eq. (i)]

Answer

$${1 \over {2\pi }}\sqrt {{{{d^3}} \over {3Gm}}} $$

Answer

$$2\pi \sqrt {{{3Gm} \over {{d^3}}}} $$

Answer

$${1 \over {2\pi }}\sqrt {{{3Gm} \over {{d^3}}}} $$

Answer

$$2\pi \sqrt {{{{d^3}} \over {3Gm}}} $$

Gravitation

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :

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