Question

derive expression for orbital speed and time period for a satellite revolving at height 'h' above earth surface. Find time period of a satellite just revolving above the earth surface??
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Answer 1

formula #1

The Gravitational Force between the earth and the satellite = Fg = (G.M.m)/r2 ……………… (1)

The centripetal force acting on the satellite = Fc = mV2/r ……………………….. (2)

Here, M is the mass of earth and m is the mass of the satellite which is having a uniform circular motion in a circular track of radius r around the earth. V is the linear velocity of the satellite at a point on its circular track.

Now this r is the sum of the radius of the earth(R) and the height(h) of the satellite from the surface of the earth.

r = R + h

Now equating, equation 1 and 2 we get,

Fg = Fc

=> (G.M.m)/r2 = mV2/r

V = [(GM)/r]1/2 ……………………………….. (3)

This is the first equation or expression of Orbital Velocity of a satellite. Here r = R +h

Answer 2

Near the earth's surface time period of a satellite is 1.4 hrs.Find its time period if it is at the distance 4R from the centre of earth.

Orbital Velocity is expressed in meter per second (m/s). Question 1: Calculate the orbital velocity of earth so that the satellite revolves round the earth if radius of earth R = 6.5 × 106 m, mass of earth M = 5.5 × 1024 kg and Gravitational constant G = 6.67 × 10-11 m3/s2 kg.

Satellites that are further away actually travel slower. The International Space Station has a Low Earth Orbit, about 400 kilometers (250 miles) above the earth's surface. Objects orbiting at that altitude travel about 28,000 kilometers per hour (17,500 miles per hour).

The factors are : the mass of the Earth, the mass of the satellite, and the distance between the center of the Earth and the center of the satellite. Also, the gravitational constant, which is involved in a set of equations that relate these factors to the orbital speed of the satellite.

here is your answer mate