Question

write a full derivation of orbital velocity of class 11




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Answer 1

Answer:

Derivation of Orbital Velocity

⇒v=√GMr.

⇒Fg=GMmr2.

⇒Fc=mv2r.

⇒mv2r.

⇒GMr=v2.

⇒v=±√GMr.

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Answer 2

Explanation:

Orbital velocity is defined as the minimum velocity a body must maintain to stay in orbit.

Due to the inertia of the moving body, the body has a tendency to move on in a straight line. But, the gravitational force tends to pull it down. The orbital path, thus elliptical or circular in nature, represents a balance between gravity and inertia. Orbital velocity is the velocity needed to achieve a balance between gravity’s pull on the body and the inertia of the body’s motion. For a satellite revolving around the Earth, the orbital velocity of the satellite depends on its altitude above Earth. The nearer it is to the Earth, the faster the required orbital velocity.

A satellite runs into traces of Earth’s atmosphere, at lower altitudes, which creates drag. This drag causes decay the orbit, eventually making the satellite to fall back into the atmosphere and burn itself up.

Derivation of Orbital Velocity

To derive the orbital velocity, we concern ourselves with the following two concepts:

Gravitational Force

Centripetal Force

It is important to know the gravitational force because it is the force that allows orbiting to exist. A central body exerts a gravitational force on the orbiting body to keep it in its orbit. Centripetal force is also important, as this is the force responsible for circular motion.

For the derivation, let us consider a satellite of mass m revolving around the Earth in a circular orbit of radius r at a height h from the surface of the Earth. Suppose M and R are the mass and radius of the Earth respectively, then r = R + h.

To revolve the satellite, a centripetal force of mv2or is needed which is provided by the gravitational force GMmr2 between the satellite and the Earth.

Therefore, equating both the equations, we get

mv20r=GMmr2

v20=GMr=GMR+h

Simplifying the above equation further, we get

v0=GMR+h−−−−√……(eqn 1)

But GM=gR2, where g is the acceleration due to gravity.

Therefore,

V0=gR2R+h−−−−√

Simplifying the above equation, we get

v0=RgR+h−−−−√

Let g’ be the acceleration due to gravity in the ( at a height h from the surface)

g′=GM(R+h)2

Simplifying further, we get

GM(R+h)=g′(R+h)=g′r……(eqn 2)

Substituting (2) in (1), we get

v0=g′r−−−√=g′(R+h)−−−−−−−−√

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