Question

define orbital velocity of satellite. derive the experience of orbital velocity of satellite​

Answer 1

Here, (Vo = orbital velocity)

Answer:

Orbital velocity is the velocity given to artificial satellite so that it may start revolving around the earth. Expression for orbital velocity: Suppose a satellite of mass m is revolving around the earth in a circular orbit of radius r, at a height h from the surface of the earth.

The expression for orbital velocity is

√g( R+h) = √gr.

Answer 2

orbital velocity is the velocity given to artificial Satellite so that it may start revolving around the earth.

Derivation:-

suppose a satellite of mass m is revolving around the earth in a circular orbit of radius r, at a height h from the surface of earth. Let M be the mass of earth and R be the radius of earth.

$\large\rm { \therefore r = R + h}$

To revolve the satellite, centripetal force of $\large\rm { \frac{mv_{0}^{2}}{r}}$ is required ( where $\large\rm { v_{0}}$ is orbital velocity) which is provided by gravitational force $\large\rm { \frac{GMm}{r^{2}} }$between the earth and the satellite.

$\large\rm { \therefore \frac{mv_{0}^{2}}{r} = \frac{GMm}{r^{2}}}$

so $\large\rm { v_{0} = \sqrt{ \frac{ GM}{ R+h } } }$

But $\large\rm { GM = gR^{2}}$ where g is acceleration due to gravity.

so $\large\rm { v_{0} = \sqrt{\frac{gR^{2}}{r} } }$

$\large\rm { = R \sqrt{\frac{g}{r+h}}}$

let $\large\rm { g'}$ be acceleration due to gravity in the orbit at a height h from the surface.

$\large\rm { \therefore g' = - \frac{GM}{(R+h)^{2}}}$

$\large\rm { \therefore v_{0} = \sqrt{g'(R+h)} }$

$\large\boxed{\bf{ v_{0} = \sqrt{g'r}}}$