Question

A satellite of earth revolves in a circular orbit of height of 300 km above the earth's surface. Calculate: a. Orbital velocity, b. Period of revolution of satellite. Given : R= 6.4*10^6m and g= 10m/s sq

Answer 1

Given:

Height of satellite = 300 km

Radius of earth = 6.4*10^6

g = 10 m/s^2

To find:

1. Orbital velocity
2. Period of revolution

Solution:

As we know that, The orbital velocity of the satellite is v=  (GM/R+h)^0.5

where M= mass of earth

R= radius of earth

h = height of satellite

v = (6.67*10^-11 * 6*10^24 / (6.4*10^6 + 0.3*10^6))^1/2

v = 59.7 km/s

Period of revolution T =2π/R ((R+h)³/g)^1/2

On putting the values we get:

T = 2*22/7/ 6.4*10^6 ((6.4*10^6 + 0.3*10^6)³/10 )^1/2

T = 5367.68 s

Therefore the orbital velocity is 59.7 km/s and period of revolution = 5367.68 s.