If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?

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If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?

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Answer 1 · 2018-05-22T13:54:05

$\approx 19$ $~~ 19$ days

Explanation:

We can derive an expression for the time period of revolution by assuming that the moon is moving in a circular orbit around the earth.

The necessary centripetal acceleration is provided by the force of gravity and thus we have

$m_{m} ω^{2} r = G \frac{m_{e} m_{m}}{r^{2}} \Rightarrow$ $m_m omega^2 r = G(m_e m_m)/r^2 implies$

$ω^{2} = G \frac{m_{e}}{r^{3}} \Rightarrow$ $omega^2 = Gm_e/r^3 implies$

$T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{G m_{e}}} r^{\frac{3}{2}}$ $T = (2pi)/omega = (2 pi)/sqrt(G m_e )r^{3/2}$

Thus, under the conditions of the problem the period will decrease by a factor of $\sqrt{2}$ $sqrt2$ to become $\frac{27}{\sqrt{2}}$ $27/sqrt 2$ days $\approx 19$ $~~ 19$ days.

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If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?

1 Answer

Explanation:

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