Question

a satellite which is at a height h from earths surface complete one revolution of the earth in 90 minute. if the radius of the earth be 6370 km and acceleration due to gravity 9.8 ms^-2 , then find the value of h.

Answer 1

R = radius of earth = 6370 km = 6370,000 m

h = height of satellite above the earth's surface

g = acceleration due to gravity = 9.8 m/s²

T = time period of the satellite for one revolution around earth = 90 min = 90 x 60 sec = 5400 sec

time period of the satellite for one revolution around earth is given as

T = (2π) √(R + h)³/(gR²)

taking square both side

T² = (2π)² (R + h)³/(gR²)

T² (gR²)/ (2π)² = (R + h)³

inserting the values

(5400)² ((9.8) (6370,000)²)/ (2(3.14))² = ((6370,000) + h)³

h = 279532 m = 279.532 km