# Escape and Orbital Velocity

If we want to define escape and orbital velocities in simple words then we can say that orbital velocity is the speed required by an object to orbit around another object. Whereas, escape velocity is the minimum speed required by an object to break free from the gravitational pull of another mass. We will look at the concepts of escape and orbital velocities and their relationship in detail below.

## What is Escape Velocity?

We have studied in kinematics that the range of a projectile depends upon the initial velocity as ${{R}_{max}}\propto {{u}^{2}}\Rightarrow {{R}_{max}}=\frac{{{u}^{2}}}{2g},$ which implies that for a certain initial velocity provided to the particle, the particle flies away from the gravitational influence of the earth.

This minimum amount of velocity for which the particle escapes the gravitational sphere of influence of a planet is known as escape velocity (ve). When escape velocity is provided to a body it goes to infinity theoretically.

As the gravitational force is a conservative force, the law of conservation of energy holds good. Applying the law of conservation of energy for the particle which is provided with a required minimum velocity to go to infinity

${{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}$

At infinity, the particles experience no interaction so the final potential energy and we know from motion in 1D chapter the final velocity of a body becomes zero after reaching its maximum height so from this we can deduce the final kinetic energy of the particle.

Then ,

${{U}_{i}}+{{K}_{i}}=0~\text{and }\!\!~\!\!\text{ we }\!\!~\!\!\text{ know }\!\!~\!\!\text{ that},{{U}_{i}}=\frac{-GMm}{R}~,{{K}_{i}}=~\frac{1}{2}m{{v}_{e}}^{2}$

We get,

$\frac{1}{2}m{{v}_{e}}^{2}+\left( \frac{-GMm}{R} \right)=0\Rightarrow \frac{1}{2}m{{v}_{e}}^{2}=\frac{GMm}{R}$

That implies,

${{v}_{e}}=\sqrt{\frac{2GM}{R}}$ ……………(1)

From the above formula, it is clear that escape velocity does not depend upon the test mass (m).

If the source mass is earth then the escape velocity has a value of 11.2 km/s.

If $v={{v}_{e}}$ the body escapes the planet’s gravitational sphere of influence, if $0\le v<{{v}_{e}}$ the body either falls back onto the earth or continues to orbit around the planet within the sphere of influence of the planet.

### Orbital Velocity

The velocity with which the test mass orbits around a source mass is known as orbital velocity $({{v}_{o}}),$ when the test mass is orbiting around the source mass in a circular path of radius ‘r’ having a centre of the source mass as the centre of the circular path, the centripetal force is provided by the gravitational force as it is always an attracting force having its direction pointed towards the centre of a source mass.

$\frac{m{{v}_{o}}^{2}}{r}=\frac{GMm}{{{r}^{2}}}$

$\frac{{{v}_{o}}^{2}}{r}=\frac{GM}{{{r}^{2}}}$

⇒ ${v}_{o}=\sqrt{\frac{GM}{r}}$

If the test mass is at small distances from the source mass $r\approx R\left( radius~of~the~source~mass \right)$

Then,

$v_{o}=\sqrt{\frac{GM}{r}}$ …………..(2)

The above formula suggests that the orbital velocity is independent of the test mass ( the mass which is orbiting).

## Relationship Between Escape And Orbital Velocity

The relationship between escape velocity and orbital velocity can be mathematically represented as:

Ve =√2V0

Meanwhile, if we divide Eq.(1) and Eq (2) we get,

$\frac{{{v}_{e}}}{{{v}_{o}}}=\frac{\sqrt{\frac{2GM}{R}}}{\sqrt{\frac{GM}{R}}}$

$\frac{{{v}_{e}}}{{{v}_{o}}}=\sqrt{2}$

It shows that escape velocity is $\sqrt{2}$ times greater than orbital velocity.

Certain conditions need to be taken into consideration. The main one is that the escape velocity should be square-root of 2 times larger than the orbital velocity to be free.

• When the velocities are the same, the object will be in constant orbit and the same elevation.
• If escape velocity is less than orbital then the orbit will diminish which will result in the object crashing.
• If it is more then the object will be free in the orbit and will likely float into space.

### Motion Of Satellites Around Earth

Let us consider a satellite assumed to be revolving around the earth in a circular orbit of radius ‘r’ having the centre of the earth as its centre, let us see the kinematics of the satellite moving around the earth.

Here the test mass is a satellite and the source mass is earth. The velocity with which a satellite orbit around the earth is given by the orbital velocity,

${{v}_{o}}=\sqrt{\frac{GM}{r}}$

### Time Period Of A Satellite

The time taken by the satellite to complete one revolution around the earth is known as the time period of a satellite. The time period of a satellite is the ratio of the total distance travelled by the satellite around the earth to the orbital velocity.

$T=\frac{Total~distance~travelled~by~the~satellite}{Orbital~velocity}$

$T=\frac{2\pi r}{{{v}_{O}}}$

$T=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}$

$T=\frac{2\pi }{\sqrt{GM}}{{\left( r \right)}^{{}^{3}/{}_{2}}}$

Squaring ${{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{\left( r \right)}^{3}}$

which is Kepler’s 3rd law.

The constant of proportionality in the above equation depends only on the source mass but not on the test mass.

### Kinetic Energy Of A Satellite

Kinetic energy is the energy possessed by the body in motion ( whether translational or rotational or a combination of both) in the case of the satellite orbiting around the earth. The satellite makes a pure rotational motion. In a pure rotation, the kinetic energy of the satellite is given by;

$K=\frac{1}{2}m{{r}^{2}}{{\omega }^{2}}$

Where is the angular velocity of a satellite which is related to the time period of a satellite by a formula $\omega =\frac{2\pi }{T}$.

Substituting $K=\frac{1}{2}m{{r}^{2}}{{\left( \frac{2\pi }{T} \right)}^{2}}$

From Kepler’s 3rd law we know the time period of a satellite, substituting it in the above formula we get

$K=\frac{GMm}{2r}$

Kinetic energy can never be negative for any force.

### Potential Energy Of A Satellite

Potential energy is the energy possessed by the body in a particular position. Potential energy changes when the position of the body changes. To study potential energy we require two bodies, one is source mass(M) which provides the gravitational force and the other is test mass(m) which experiences the gravitational force by the source mass. In our case satellite is the test mass and earth is the source mass. The potential energy possessed by the satellite at a distance ‘r’ from the centre of the earth is given by;

$U=\frac{-GMm}{r}$

### Total Energy Of A Satellite

The total energy of the satellite is the sum of all energies possessed by the satellite in the orbit around the earth. As only the mechanical motion of the satellite is considered, it has only kinetic and potential energies.

Total energy of the satellite = kinetic energy of the satellite + potential energy of the satellite

$E=K+U$

$E=\frac{GMm}{2r}+\left( \frac{-GMm}{r} \right)$

$E=~-\frac{GMm}{2r}$

From the above equation,

Total energy of the satellite = – (kinetic energy of the satellite)

Total energy of the satellite = $\left( \frac{\text{potential }\!\!~\!\!\text{ energy }\!\!~\!\!\text{ of }\!\!~\!\!\text{ the }\!\!~\!\!\text{ satellite}}{2} \right)$

### Virial Theorem

Virial theorem gives us the relationship between kinetic energy and potential energy. According to the virial theorem, if potential energy is proportional to nth power of position (r) then the average kinetic energy is equal to (n/2) times potential energy at that position.

If $U\propto {{r}^{n}}$ then, $K=\frac{n}{2}\left( U \right)$

### Binding Energy Of A Satellite

Binding energy is the energy required to be given to the satellite to escape the gravitational pull of the earth. It is numerically equal to the negative of total energy possessed by the satellite.

Binding energy of a satellite = $\frac{GMm}{2r}$ .

The binding energy of a system of two bodies is the amount of minimum energy required to separate them by infinite distance or it is simply the amount of energy required to make the potential energy of the system equal to zero, which is numerically equal to the kinetic energy of the body under motion.

### Angular Momentum Of A Satellite

When a satellite of mass ‘m’ is orbiting with an angular speed on the orbital path of radius ‘r’, its angular momentum is given by $L=m{{r}^{2}}\omega$

We know that $\omega =\frac{2\pi }{T}$, substituting this in the above equation we get

$L=m{{r}^{2}}\frac{2\pi }{T}$

Using $T=\frac{2\pi }{\sqrt{GM}}{{\left( r \right)}^{{}^{3}/{}_{2}}}$ we get,

$L=m\sqrt{GMr}$

From the above equation, it is clear that the angular momentum of a satellite depends on both the mass of a satellite and mass of earth. It also depends upon the radius of the orbit of the satellite.

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