Orbital Velocity Derivation

Orbital velocity is defined as the velocity at which a body (satellite) revolves around the other body (earth).

The other ways to express orbital velocity are as follows:

  • The particle velocity due to wave motion
  • Production of orbital kinetic energy by the electron

Derivation of orbital velocity

Orbital velocity is derived in the following way:

\(F_{g}=\frac{GMm}{r^{2}}\) (gravitational force between the earth and the satellite) (eq.1)

\(F_{c}=\frac{mV^{2}}{r}\) (centripetal force acting upon the satellite) (eq.2)

Where,

M: mass of the earth

m: mass of the satellite

r: radius of uniform motion of the satellite around the earth

\(r=R+h\)

Where,

R: radius of the earth

h: height of the satellite

V: linear velocity of the satellite

Fc: centripetal force

Fg: gravitational force

\(F_{g}=F_{c}\) (from eq.1 and eq.2)

\(\frac{GMm}{r^{2}}=\frac{mV^{2}}{r}\)

\(V=\left [ \frac{GM}{r} \right ]^{\frac{1}{2}}\) (eq.3)

Therefore, this is the equation of orbital velocity.

Difference between orbital velocity and escape velocity

Escape velocity is defined as the minimum velocity required by a free object to escape from the gravitational force of a massive body.

It is calculated by the formula given below:

\(v_{e}=\sqrt{\frac{2GM}{r}}\)

Where,

G: universal gravitational constant

M: mass of the body to be escaped from

r: distance between the centre of mass of the body and the object

In order to break out from the orbit of the massive body, the object must have escape velocity square root of two times greater than the orbital velocity.

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A wire has a non-uniform cross-sectional area as shown in figure. A steady current i flows through it. Which one of the following statement is correct

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