Orbital Velocity Derivation

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 \(\frac{mv_{o}^2}{r}\) is needed which is provided by the gravitational force \(G\frac{Mm}{r^2}\) between the satellite and the Earth.

Therefore, equating both the equations, we get

\(\frac{mv_{0}^2}{r}=G\frac{Mm}{r^2}\)

\(v_{0}^2=\frac{GM}{r}=\frac{GM}{R+h}\)

Simplifying the above equation further, we get

\(v_{0}=\sqrt{\frac{GM}{R+h}}\)……(eqn 1)

But \(GM = gR^2\), where g is the acceleration due to gravity.

Therefore,

\(V_0=\sqrt{\frac{gR^{2}}{R+h}}\)

Simplifying the above equation, we get

\(v_0=R\sqrt{\frac{g}{R+h}}\)

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

\(g’=\frac{GM}{(R+h)^2}\)

Simplifying further, we get

\(\frac{GM}{(R+h)}=g'(R+h)=g’r\)……(eqn 2)

Substituting (2) in (1), we get

\(v_0=\sqrt{g’r}=\sqrt{g'(R+h)}\)

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