Gravitation 5 Question 1

1. 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)

Show Answer

Answer:

Correct Answer: 1. (b)

Solution:

  1. 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,

$F _{\text {centripetal }}=F _{B O \text { (net) }}$

$\Rightarrow \quad \sqrt{2} \frac{M v^{2}}{a}=\frac{G M^{2}}{a^{2}} \frac{1}{2}+\sqrt{2}$

$\Rightarrow \quad v^{2}=\frac{G M}{a} 1+\frac{1}{2 \sqrt{2}}=\frac{G M}{a}(1.35) \Rightarrow v=1.16 \sqrt{\frac{G M}{a}}$



NCERT Chapter Video Solution

Dual Pane