Question:
Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is \(2 \times 10^{3}\) km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is
\(\dfrac{\pi}{x}\ \mathrm{rad\,h^{-1}}\)
where \(x\) is ____________.
Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is \(2 \times 10^{3}\) km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is
\(\dfrac{\pi}{x}\ \mathrm{rad\,h^{-1}}\)
where \(x\) is ____________.
\(\dfrac{\pi}{x}\ \mathrm{rad\,h^{-1}}\)
where \(x\) is ____________.
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Use \(T^2\propto r^3\) to get the second radius, then find the relative velocity perpendicular to the line joining them when they are collinear with the planet.
Updated On: Nov 4, 2026
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Correct Answer: 3
Solution and Explanation
Kepler's third law: \(\dfrac{r_2}{r_1} = \left(\dfrac{T_2}{T_1}\right)^{2/3} = 8^{2/3} = 4\), so \(r_2 = 8000\) km.
\(\omega_1 = \dfrac{2\pi}{1} = 2\pi\) rad/h, \(\omega_2 = \dfrac{2\pi}{8} = \dfrac{\pi}{4}\) rad/h.
Speeds: \(v_1 = \omega_1 r_1 = 4000\pi\) km/h, \(v_2 = \omega_2 r_2 = 2000\pi\) km/h, in the same direction (perpendicular to the line joining them) when both are closest.
Distance between them: \(d = 8000 - 2000 = 6000\) km.
Relative speed of farther w.r.t. nearer: \(4000\pi - 2000\pi = 2000\pi\) km/h, perpendicular to the line joining them.
Angular speed as observed: \(\omega = \dfrac{v_{rel}}{d} = \dfrac{2000\pi}{6000} = \dfrac{\pi}{3}\) rad/h.
So \(x = 3\).
Answer: 3
\(\omega_1 = \dfrac{2\pi}{1} = 2\pi\) rad/h, \(\omega_2 = \dfrac{2\pi}{8} = \dfrac{\pi}{4}\) rad/h.
Speeds: \(v_1 = \omega_1 r_1 = 4000\pi\) km/h, \(v_2 = \omega_2 r_2 = 2000\pi\) km/h, in the same direction (perpendicular to the line joining them) when both are closest.
Distance between them: \(d = 8000 - 2000 = 6000\) km.
Relative speed of farther w.r.t. nearer: \(4000\pi - 2000\pi = 2000\pi\) km/h, perpendicular to the line joining them.
Angular speed as observed: \(\omega = \dfrac{v_{rel}}{d} = \dfrac{2000\pi}{6000} = \dfrac{\pi}{3}\) rad/h.
So \(x = 3\).
Answer: 3
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