A satellite moves in a circular orbit around Earth at a speed of 4705 m/s.
(a) Determine the satellite\'s altitude above the surface of Earth.
m
(b) Determine the period of the satellite\'s orbit.
h
.
Solution
r = G M / v2 = (6.673 * 10-11 * 5.97 * 1024) / (47052)
= 1.799 * 107 m
The satellite\'s altitude above the Earth\'s surface is h = r - R
= (1.799 * 107) - (6.378 * 106) = 1.16 * 107 m
The satellite\'s orbital period T = 2 pi r /v
= (2 pi * 1.799 * 107) / 4705
= 24024 sec = 6.67 hrs.
A satellite moves in a circular orbit around Earth at a speed of 470.pdf
1. A satellite moves in a circular orbit around Earth at a speed of 4705 m/s.
(a) Determine the satellite's altitude above the surface of Earth.
m
(b) Determine the period of the satellite's orbit.
h
.
Solution
r = G M / v2 = (6.673 * 10-11 * 5.97 * 1024) / (47052)
= 1.799 * 107 m
The satellite's altitude above the Earth's surface is h = r - R
= (1.799 * 107) - (6.378 * 106) = 1.16 * 107 m
The satellite's orbital period T = 2 pi r /v
= (2 pi * 1.799 * 107) / 4705
= 24024 sec = 6.67 hrs
