Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively If the speed of satellite \(\mathrm{A}\) is \(3 \mathrm{v}\), then speed of satellite \(\mathrm{B}\) is (A) \((3 \mathrm{v} / 2)\) (B) \((4 \mathrm{v} / 2)\) (C) \(6 \mathrm{v}\) (D) \(12 \mathrm{v}\)

In order to maintain a satellite's orbit around a planet, the gravitational force acting on the satellite must be equal to the centripetal force required for the satellite's circular motion. This can be expressed by the following equation: \[F_g = F_c\] The gravitational force is given by Newton's law of universal gravitation: \[F_g = G \frac{m_pm_s}{r^2}\] where: - \(F_g\) is the gravitational force, - \(G\) is the gravitational constant, - \(m_p\) is the mass of the planet, - \(m_s\) is the mass of the satellite, and - \(r\) is the distance between the satellite and the center of the planet (i.e., the radius of the satellite's orbit).

The centripetal force is given by the following formula: \[F_c = m_s a_c\] where: - \(F_c\) is the centripetal force, and - \(a_c\) is the centripetal acceleration. The centripetal acceleration can be expressed in terms of the satellite's speed as follows: \[a_c = \frac{v^2}{r}\]

Now let's equate the expressions for gravitational force and centripetal force: \[G \frac{m_pm_s}{r^2} = m_s \frac{v^2}{r}\] We can cancel out \(m_s\) from both sides: \[G \frac{m_p}{r^2} = \frac{v^2}{r}\] Now, let's solve for the speed of each satellite in terms of the radius of their orbits and the mass of the planet: \[v^2 = G \frac{m_p r}{r^2}\] \[v = \sqrt{G \frac{m_p}{r}}\]

Since we are given the speed of satellite A, we can find the speed of satellite B by finding the ratio of their speeds. Let A's speed be \(3v\), and the ratio of A's radius to B's radius is given as \(4R : R\). Now we can write the speed of satellite A and satellite B as follows: \[v_A = 3v = \sqrt{G \frac{m_p}{4R}}\] \[v_B = \sqrt{G \frac{m_p}{R}}\] Taking the ratio of their speeds: \[\frac{v_B}{v_A} = \frac{\sqrt{G \frac{m_p}{R}}}{\sqrt{G \frac{m_p}{4R}}}\] We can simplify this expression: \[\frac{v_B}{v_A} = \frac{\sqrt{4}}{\sqrt{1}} = 2\] Now, we can find the speed of satellite B: \[v_B = 2 \times v_A = 2 \times 3v = 6v\] Thus, the speed of satellite B is \(6v\). The answer is (C) \(6 \mathrm{v}\).