Question

A satellite orbits in a nearly circular orbit, 30,000 m above the earth's surface. It orbits once every day. Find it's orbital speed.

Answer 1

given

altitude of the satellite, h = 30000 m

we know, radius of the earth, Re = 6370 km = 6.37*10^6 m
mass of the Earth Me = 5.97*10^24 kg

radius of orbit of the satellite, r = Re + h

= 6.37*10^6 + 30000

= 6.40*10^6 m

from the given data time period, T = 24 hours

= 24*60*60 s

orbital speed, v = 2*pi*r/T

= 2*pi*6.40*10^6/(24*60*60)

= 465 m/s

Note :
The above answer is correct if the satellite revolves once every day. But usually it does not happen.

using the actual equation for orbital speed of a satellite,

v = sqrt(G*Me/r)

= sqrt(6.67*10^-11*5.97*10^24/(6.4*10^6))

= 7888 m/s (or) 7.888 km/s <<<<<<<<-----------------Answer