A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 x 10⁷ m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

∼ 0.13 × 10¹⁴ kg m² s^-1

∼ 1.3 × 10¹⁴ kg m² s^-1

∼ 0.58 × 10¹⁴ kg m² s^-1

∼ 2.58 × 10¹⁴ kg m² s^-1

Solution

∼ 0.58 × 10¹⁴ kg m² s^-1

As the satellite is moving in equatorial plane with orbital radius 4 x 10⁷ m.

∴ Satellite is geostationary satellite.

Hence, the time taken by satellite to complete its one revolution

T = 24 h

= 24 × 3600 s

T = 86400 s

Velocity of satellite

v = $\frac{2 πr}{T}$

Angular momentum, L = mvr

L = m $(\frac{2 π r}{T})$ r

= $(\frac{2 π m}{T})$ r²

∴ L = $\frac{2 \times 3.14 \times 500}{86400}$ × ( 4 × 10⁷ )²

L = 0.58 × 10¹⁴ kg m² s^-1

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