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AIIMS 2017
Exam
AIIMS
Year
2017
Subject
Physics
Chapter
Rotational Motion

AIIMS 2017 · Physics · Rotational Motion

A communication satellite of 500 ~kg revolves around the earth in a...

A communication satellite of 500 kg500 \mathrm{~kg} revolves around the earth in a circular orbit of radius 4.0×107 m4.0 \times 10^7 \mathrm{~m} in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

Options

  1. A∼0.13×1014 kg m2 s−1\sim 0.13 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}
  2. B∼1.3×1014 kg m2 s−1\sim 1.3 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}
  3. C∼0.58×1014 kg m2 s−1\sim 0.58\times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}
  4. D∼2.58×1014 kg m2 s−1\sim 2.58 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}

Correct answer

C. ∼0.58×1014 kg m2 s−1\sim 0.58\times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}

Step-by-step solution

As the satellite is moving in equatorial plane with orbital radius 4×107 m4 \times 10^7 \mathrm{~m}. ∴\therefore \quad Satellite is geostationary satellite. Hence, the time taken by satellite to complete its one revolution, T=24 h=86400 sT=24 \mathrm{~h}=86400 \mathrm{~s} Velocity of satellite, v=2πrTv=\frac{2 \pi r}{T} Angular momentum, L=mvrL=m v rL=m(2πrT)r=2πmTr2∴L=2×3.14×50086400×(4×107)2=0.58×1014 kg m2 s−1\begin{aligned} & L=m\left(\frac{2 \pi r}{T}\right) r=\frac{2 \pi m}{T} r^2 \\ & \therefore \quad L=\frac{2 \times 3.14 \times 500}{86400} \times\left(4 \times 10^7\right)^2 \\ & =0.58 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\ & \end{aligned}$$
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