Escape Velocity and Orbital Velocity of Satellite

Hello guys..!! Hope you all must be waiting eagerly for our next article. Here goes an explanation and derivation of escape velocity and its applications.

Escape Velocity:

It is the minimum velocity with which a body must be protected from the surface of the earth so that it escapes from the gravitational field of the earth. We can also say that a body, projected with escape velocity, will be able to go to a point which distance from the earth.

Let us imagine what happens to a body of mass m if it is thrown from the earth with a velocity (escape velocity). As the body moves away from the earth, it’s slow sown hence, its kinetic energy is converted into gravitational potential energy of the mass-earth system.

KE lost by mass m=gain in gravitational potential energy of mass-earth system


\(\frac{1}{2}m{{v}_{e}}^{2}\,=\,\,0\,-\,\left( -\frac{GmM}{R} \right)\).



Substituting the values of g = 9.81m/s2 and R = 6400km we get ve = 11.2 km/s

Application of concepts of escape velocity:

The following are the applications of the concept of escape velocity. The maximum velocity attained by a particle, orbital velocity and time period of satellites can be found if we know the escape velocity.

(a) Maximum velocity attained by a particle: Suppose a particle of mass m is projected vertically upwards with speed v and we want find the maximum height h attained by the particle. Then, we can use conservation of mechanical energy.

Decreases in kinetic energy=increases in gravitational potential energy



(b) Orbital velocity: The velocity of a satellite in its orbit is called orbital velocity. Let v be orbital velocity of satellite, then



Or \({{v}_{0}}\,=\,\sqrt{\frac{GM}{R\,+\,h}}\).

Hence orbital velocity decided by the radius of its orbit or its height above the earth surface


(c) Time period of satellites: The time taken to complete revolution is called the time period. It is given by

\(T\,=\,\frac{2\pi r}{{{v}_{0}}}\,\,=\,2\pi r\sqrt{\frac{r}{GM}}\).

\(T\,=\,\frac{2\pi {{r}^{3/2}}}{\sqrt{GM}}\).

\({{T}^{2}}\,=\,\frac{4{{\pi }^{2}}}{GM}{{r}^{3}}\).