Question

A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the satellite? A. T= 2.94 h, B. T= 3.23 h, v 5610 m/s C. T= 1.75 h, v = 5920 m/s D. T 1.12 h, v 4980 m/s E. T 2.58 h, v 6460 m/s

Answer 1

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

$v = \sqrt{\frac{G M }{r}}$

$v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24} }{3.18064\times 10^{10}}}$

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years