Can you please tell me about orbital velocity in detail ?? 

Asked by Sanket | 15th Oct, 2015, 06:56: PM

Expert Answer:

Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the Earth.
Suppose that a satellite of mass m has to be put into circular orbit around the Earth at a height x above its surface. Consider that Earth is a sphere of mass M and radius R. Thus, the radius of the orbit of the satellite will be R+x as shown in the figure.
 
 
Suppose that v is the required orbital velocity for the satellite.
The gravitational force of attraction between the satellite and the Earth will provide the necessary centripetal force to the satellite to move around the Earth in the circular orbit i.e.,
 begin mathsize 14px style fraction numerator Gmm over denominator left parenthesis straight R plus straight x right parenthesis squared end fraction equals fraction numerator mv squared over denominator straight R plus straight x end fraction or space space straight v equals square root of fraction numerator GM over denominator straight R plus straight x end fraction end root space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis end style
Thus, we find that the orbital velocity of a satellite is independent of the mass of the satellite but depends on the radius of the orbit (R+x) and mass of the planet (M) about which it revolves.
The acceleration due to gravity on the surface of the Earth is given by
begin mathsize 14px style straight g equals GM over straight R squared so space that space space GM equals straight g space straight R squared end style
Substituting the value of GM in the equation (1), we have,
begin mathsize 14px style straight v equals square root of fraction numerator straight g space straight R squared over denominator straight R plus straight x end fraction end root end style                  ... (2)
The equations (1) and (2) are used to find the orbital velocity for a satellite orbiting at a height x above the surface of Earth.
 
 
When the satellite is orbiting very close to the surface of the Earth:
In such a case, x≈0.
From equation (1), we have,
begin mathsize 14px style straight v equals square root of fraction numerator straight G space straight M over denominator straight R end fraction end root space space space space space space space space space... space left parenthesis 3 right parenthesis end style
Also, from the equation (2), we get,
begin mathsize 14px style straight v equals square root of straight g space straight R end root space space space space space space space space space space space... space left parenthesis 4 right parenthesis end style
By substituting, g = 9.8 m s-2 and R = 6.4 × 106 m in the equation (3), it can be found that a satellite requires a velocity of about 7.9 km s-1 to revolve in a orbit just near the surface of the Earth.

Answered by Faiza Lambe | 16th Oct, 2015, 04:08: PM

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