Question

A communication satellite of $500 \,kg$ revolves around the earth in a circular orbit of radius $4.0 \times 10^7$ $m$ in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

• A. $\sim$ 0.13 × 10¹⁴ kg m² s^-1
• B. $\sim$ 1.3 × 10¹⁴ kg m² s^-1
• C. $\sim$ 0.58 × 10¹⁴ kg m² s^-1
• D. $\sim$ 2.58 × 10¹⁴ kg m² s^-1

Answer 1

Answer: (C)

As the satellite is moving in equatorial plane with orbital radius 4 × 10⁷ m.
$\therefore \,\,\,\,\,$ Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution, T = 24 h = 86400 s
Velocity of satellite, v = $\frac{2\pi r}{T}$
Angular momentum, L = mvr
L $=m\left(\frac{2\pi r}{T}\right)r=\frac{2\pi m}{T}r^{2}$
$\therefore \quad L=\frac{2\times3.14\times500}{86400}\times\left(4\times10^{7}\right)^{2}$
$=0.58\times10^{14}\,kg \,m^{2}\,s^{-1}$