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Chapter 7: Problem 83

Two particle of equal mass \(m\) go around a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is (A) \(\sqrt{\frac{G m}{R}}\) (B) \(\sqrt{\frac{G m}{4 R}}\) (C) \(\sqrt{\frac{G M}{3 R}}\) (D) \(\sqrt{\frac{G m}{2 R}}\)

Short Answer

Expert verified
The speed of each particle with respect to their centre of mass is (B) \(\sqrt{\frac{G m}{4 R}}\).

Step by step solution

01

Write the expressions for gravitational force and centripetal force

The expression for gravitational force between two masses \(m_1\) and \(m_2\) at a distance \(r\) is given by: \[F_g = G \frac{m_1 m_2}{r^2}\] Here, both masses are equal \(m = m_1 = m_2\) and the distance between them is twice the radius of the circle: \(r = 2R\). The expression for centripetal force acting on a mass \(m\) moving at a speed \(v\) in a circle of radius \(R\) is: \[F_c = m \frac{v^2}{R}\]
02

Equate the gravitational force and centripetal force

Next, we equate the gravitational force and centripetal force: \[G \frac{m^2}{(2R)^2} = m \frac{v^2}{R}\]
03

Solve for the speed v

Now, we can solve for the speed \(v\): \[\frac{G m^2}{4 R^2} = m \frac{v^2}{R}\] First, we can cancel one of the mass terms: \[\frac{G m}{4 R^2} = \frac{v^2}{R}\] And then, we can multiply both sides by \(R\) to isolate \(v^2\): \[v^2 = G \frac{m}{4R}\] Finally, we take the square root of both sides to find the speed: \[v = \sqrt{\frac{G m}{4 R}}\] So, the correct answer is (B) \(\sqrt{\frac{G m}{4 R}}\).

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Most popular questions from this chapter

From a solid sphere of mass \(M\) and radius \(R\), a spherical portion of radius \(\frac{R}{2}\) is removed, as shown in Fig. 7.17. Taking gravitational potential \(V=0\) at \(r=\infty\), the potential at the centre of the cavity thus formed is \((G=\) gravitational constant) (A) \(\frac{-G M}{R}\) (B) \(\frac{-2 G M}{3 R}\) (C) \(\frac{-2 G M}{R}\) (D) \(\frac{-G M}{2 R}\)

Given that mass of the earth is \(M\) and its radius is \(R\). A body is dropped from a height equal to the radius of the earth above the surface of earth. When it reaches the ground its velocity will be (A) \(\frac{G M}{R}\) (B) \(\left[\frac{G M}{R}\right]^{1 / 2}\) (C) \(\left[\frac{2 G M}{R}\right]^{1 / 2}\) (D) \(\left[\frac{2 G M}{R}\right]\)

The kinetic energy needed to project a body of mass \(m\) from the earth's surface (radius \(R\) ) to infinity is (A) \(\frac{m g R}{2}\) (B) \(2 m g R\) (C) \(m g R\) (D) \(\frac{m g R}{4}\)

The orbital velocity of an artificial satellite in a circular orbit just above earth's surface is \(v_{0} .\) For a satellite orbiting in a circular orbit at an altitude of half of earth's radius is (A) \(\sqrt{\frac{3}{2}} v_{0}\) (B) \(\sqrt{\frac{2}{3}} v_{0}\) (C) \(\frac{3}{2} v_{0}\) (D) \(\frac{2}{3} v_{0}\)

A satellite is revolving in a circular orbit at a height \(h\) from the earth's surface (radius of earth \(R . h \ll R\) ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere) (A) \(\sqrt{g R}\) (B) \(\sqrt{g R / 2}\) (C) \(\sqrt{g R}(\sqrt{2}-1)\) (D) \(\sqrt{2 g R}\)
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