The 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 $2 \times 10^{3} \mathrm{~km}$. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{\pi}{x} \mathrm{rad} \mathrm{h}^{-1}$ where x is ______.

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Gravitation

The 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 1 0^{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

\frac{π}{x} rad h^{- 1}

where x is ______.

Answer: 3

Solution

Kepler's third law gives

\frac{r _{2}}{r _{1}} = (\frac{T _{2}}{T _{1}})^{2/3} = 8^{2/3} = 4

, so

r_{2} = 8 \times 1 0^{3} km

. At closest approach the satellites share a radial line and move tangentially with speeds

v_{1} = ω_{1} r_{1} = 2 π \times 2000

and

v_{2} = ω_{2} r_{2} = \frac{π}{4} \times 8000 km/h

. The angular speed seen from the nearer satellite is the relative transverse speed over the separation:

\frac{v _{1} - v _{2}}{r _{2} - r _{1}} = \frac{4000 π - 2000 π}{6000} = \frac{π}{3} rad/h

, so

x = 3

.
Note: "as observed from the nearer satellite" means a frame that translates with it but does not rotate. The angular speed is then the transverse component of the relative velocity divided by the separation.

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Numerical answer

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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Gravitation

Solution

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