Question 12

NUMERICALMEDIUM

A geostationary satellite above the equator is orbiting around the earth at a fixed distance r1r_1 from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance r2r_2 from the center of the earth, such that r1=1.21r2r_1 = 1.21 r_2. The time period of the second satellite as measured from the geostationary satellite is 24p\frac{24}{p} hours. The value of pp is ____

Correct Answer: 2.33

Detailed Solution

  1. For the geostationary satellite, the time period T1=24T_1 = 24 hours.
  2. According to Kepler's third law, T2∝r3T^2 \propto r^3. T1T2=(r1r2)3/2=(1.21)3/2=(1.1)3=1.331\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = (1.21)^{3/2} = (1.1)^3 = 1.331
  3. Thus, T2=T11.331=241.331T_2 = \frac{T_1}{1.331} = \frac{24}{1.331} hours.
  4. Angular velocities are ω1=2πT1\omega_1 = \frac{2\pi}{T_1} and ω2=2πT2\omega_2 = \frac{2\pi}{T_2}. Since they orbit in opposite directions, the relative angular velocity is: ωrel=ω1+ω2=2π(1T1+1T2)\omega_{rel} = \omega_1 + \omega_2 = 2\pi \left(\frac{1}{T_1} + \frac{1}{T_2}\right)
  5. The relative time period Trel=2πωrel=11T1+1T2=T1T2T1+T2T_{rel} = \frac{2\pi}{\omega_{rel}} = \frac{1}{\frac{1}{T_1} + \frac{1}{T_2}} = \frac{T_1 T_2}{T_1 + T_2}.
  6. Substituting T2=241.331T_2 = \frac{24}{1.331} and T1=24T_1 = 24: Trel=24⋅241.33124+241.331=241.331+1=242.331T_{rel} = \frac{24 \cdot \frac{24}{1.331}}{24 + \frac{24}{1.331}} = \frac{24}{1.331 + 1} = \frac{24}{2.331}
  7. Comparing with 24p\frac{24}{p}, we get p=2.331p = 2.331. Final Answer: 2.331 (Acceptable range 2.3 to 2.4)
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