Question

Find the period, T, of a communications satellite in a circular orbit 22,300 miles above the earths surface, given that the radius of the earth is 4000 miles, that the period of the moon is 7.3 days, and that the orbit of the moon is almost circular with a radius of 239,000 miles

Answer 1

The period of a communication satellite will be given by -

we know that, T = 2r³ / G m_E

where, G = gravitational constant = 6.67 x 10^-11 Nm²/kg²

m_E = mass of the earth = 5.972 x 10²⁴ kg

r = distance from the center of earth = r_E + h

then, we get

T = 2 (3.14) [(4000 mi) + (22300 mi)]³ / [(6.67 x 10^-11 Nm²/kg²) (5.972 x 10²⁴ kg)]

T = (6.28) (26300 mi)³ / (3.98 x 10¹⁴ Nm²/kg)

converting miles to meter in above eq.

T = (6.28) [(42325747 m)³ / (3.98 x 10¹⁴ Nm²/kg)]

T = (6.28) 1.905 x 10⁸ s²

T = [(6.28) (13802.7 sec)]

T = 86680.9 sec