Gravitation Ques 38
- Four identical particles of mass $M$ are located at the corners of a square of side $a$. What should be their speed, if each of them revolves under the influence of other’s gravitational field in a circular orbit
circumscribing the square?
(a) $1.35 \sqrt{\frac{G M}{a}}$
(b) $1.16 \sqrt{\frac{G M}{a}}$
(c) $1.21 \sqrt{\frac{G M}{a}}$
(d) $1.41 \sqrt{\frac{G M}{a}}$
(Main 2019, 8 April I)
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Answer:
Correct Answer: 38.(b)
Solution:
- In given configuration of masses, net gravitational force provides the necessary centripetal force for rotation.
Net force on mass $M$ at position $B$ towards centre of circle is $F _{B O \text { net }}=F _{B D}+F _{B A} \sin 45^{\circ}+F _{B C} \cos 45^{\circ}$
$$ =\frac{G M^{2}}{(\sqrt{2} a)^{2}}+\frac{G M^{2}}{a^{2}} \frac{1}{\sqrt{2}}+\frac{G M^{2}}{a^{2}} \frac{1}{\sqrt{2}} $$
[where, diagonal length $B D$ is $\sqrt{2} a$ ]
$$ =\frac{G M^{2}}{2 a^{2}}+\frac{G M^{2}}{a^{2}} \frac{2}{\sqrt{2}}=\frac{G M^{2}}{a^{2}} \frac{1}{2}+\sqrt{2} $$
This force will act as centripetal force.
Distance of particle from centre of circle is $\frac{a}{\sqrt{2}}$.
Here, $F _{\text {centripetal }}=\frac{M v^{2}}{r}=\frac{M v^{2}}{\frac{a}{\sqrt{2}}}=\frac{\sqrt{2} M v^{2}}{a} \quad \because r=\frac{a}{\sqrt{2}}$
So, for rotation about the centre,
$F _{\text {centripetal }}=F _{B O \text { (net) }}$
$\Rightarrow \quad \sqrt{2} \frac{M v^{2}}{a}=\frac{G M^{2}}{a^{2}} \frac{1}{2}+\sqrt{2}$
$\Rightarrow \quad v^{2}=\frac{G M}{a} 1+\frac{1}{2 \sqrt{2}}=\frac{G M}{a}(1.35) \Rightarrow v=1.16 \sqrt{\frac{G M}{a}}$
BETA
