Two stars, each of mass M and separated by a distance, orbit about their centre of mass. A planetoid of mass m(m<<M) moves along the axis of this system perpendicular to the orbital plane. Let $T_{P}$ be the period of simple harmonic motion for the planetoid for small displacement from the center of mass along the z–axis and let $T_{S}$ be the period of motion for the two stars. Determine the ration $\frac{T_{P}}{T_{S}}$ .

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The ratio of $\frac{T_{P}}{T_{S}} = \frac{1}{3}$

The ratio of $\frac{T_{P}}{T_{S}} = \frac{1}{2}$

The ratio of $\frac{T_{P}}{T_{S}} = \frac{1}{2 \sqrt{2}}$

The ratio of $\frac{T_{P}}{T_{S}} = 2$

answer is C.

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Detailed Solution

Force on m due to $M (F) = \frac{GMm}{(ℓ^{2} + x^{2})}$
Net downward force on $m = 2 Fcos θ$
Question Image
$= 2 \frac{GMm}{ℓ^{2} + x^{2}} \cdot \frac{x}{\sqrt{ℓ^{2} + x^{2}}} = \frac{2 GMmx}{{(ℓ^{2} + x^{2})}^{3 / 2}}$
If $x < < ℓ, F_{net} = \frac{2 GMmx}{ℓ^{3}} = ma$
$a = \frac{2 GMx}{ℓ^{3}} \Rightarrow T_{P} = 2 π \sqrt{\frac{ℓ^{3}}{2 GM}}$
For T_S

$\begin{array}{l} F = \frac{GMM}{(2 ℓ)^{2}} = {Mω}^{2} ℓ \Rightarrow \frac{GM}{4 ℓ^{2}} = ω^{2} ℓ \Rightarrow ω = \sqrt{\frac{GM}{4 ℓ^{3}}} \\ \Rightarrow T_{S} = 2 π \sqrt{\frac{4 ℓ^{3}}{GM}} \end{array}$

$Thus, \frac{T_{P}}{T_{S}} = \frac{1}{2 \sqrt{2}}$

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