Do you know that when a ball is thrown up into the air from the surface of the Earth by our hand, it does not have enough energy to escape? So it falls back down because of Gravity. The reason why space rockets fly in outer space is that they travel at a high velocity. This is called Escape Velocity.

The velocity required to maintain a circular orbit at the same altitude equals the square root of 2 (or around 1.414) times the escape velocity. Let’s study the escape velocity formula and its application in the article below.

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Earth’s Escape Velocity

• If you throw an object straight up, it will rise until the negative acceleration of gravity stops it, then returns it to earth. The gravitational force diminishes as the distance from the centre of the Earth increases. So if you can throw the object with enough initial upward velocity so that gravity’s decreasing force can never slow it to a complete stop, its decreasing velocity can always be just high enough to overcome gravity’s pull. This initial velocity needed to achieve that condition to escape is called escape velocity.
• In simple words, Escape Velocity can be defined as the minimum velocity of an object required to escape the Earth’s gravitational field without ever falling back. The object must have greater energy than its gravitational binding energy to escape from the earth’s gravitational field.

Escape Velocity Formula

The equation for the escape velocity can be derived by applying the law of conservation of energy. This law states that the sum of total potential and kinetic energy of the objects are constant.

The escape velocity equation is also a function separating the object’s centres and the celestial body from which it is escaping.

Derivation

The derivation starts with the initial gravitational potential energy and the object’s kinetic energy at the given altitude. This total energy is then compared with the potential and the kinetic energies at infinite separation to determine the escape velocity equation.

When a stone is thrown up, it goes up to a maximum height and then returns. As the particle goes up, the gravitational potential energy increases and the particle’s kinetic energy decreases. The particle will continue to go up till its kinetic energy becomes zero and will return from there.

Example:- A object of mass \(\left( m \right)\) is on the surface of the earth having mass \(\left( M \right)\) and radius \(\left( R \right)\), and its initial speed is equal to its escape velocity\(\left( {{v_e}} \right)\). Moreover, at its final state, it will be at an infinite distance away from the planet. Its speed will be negligibly low (for minimum initial velocity).

Now, after considering initial and final conditions:-

We have,

Initial:-

Let the initial velocity of the object is \(v_e\).

The initial kinetic energy of the object is \({K_i} = \frac{1}{2}mv_e^2\) , and

The gravitational potential energy of the earth–object system near the surface of the Earth is

\({U_i} = – \frac{{GMm}}{R}\)

Where

\(M = \) The Mass of the Earth,
\(m = \) The Mass of the Object
\(R = \) The Radius of the Earth.

Now,

Total Initial Energy \( = {\left( {{K_i} + {U_i}} \right)_{{\rm{initial }}}} = \frac{1}{2}mv_e^2 – \frac{{GMm}}{R}\quad \ldots (1)\)

Final:- When it reaches a height \(h\) above the earth’s surface,

The final speed becomes \(v\).

Then, the final kinetic energy \({K_f} = \frac{1}{2}m{v^2}\) and,

The gravitational potential energy \({U_f} = – \frac{{GMm}}{{R + h}}\)

Now,

Total Final Energy \(\left( {{K_f} + {U_f}} \right) = \frac{1}{2}m{v^2} – \frac{{GMm}}{{R + h}}\quad \ldots (2)\)

So, By conservation of energy  :-

We have : –

\({K_i} + {U_i} = {K_f} + {U_f}\)
\( \Rightarrow \frac{1}{2}mv_e^2 – \frac{{GMm}}{R} = \frac{1}{2}m{v^2} – \frac{{GMm}}{{R + h}}\)

Now for minimum initial velocity, \({K_f} = 0\), because the final velocity is arbitrarily small, and the final distance approaches infinity so, \({U_f} = 0\).

Then, we can write our equation as:-

\(\frac{1}{2}mv_e^2 – \frac{{GMm}}{R} = 0 + 0\)

\( \Rightarrow \frac{1}{2}mv_e^2 = \frac{{GMm}}{R}\)

\( \Rightarrow {v_e} = \sqrt {\frac{{2GM}}{R}} \quad \ldots .(3)\)

This critical initial velocity  is called the escape velocity

Orbital Velocity and its Derivation

Orbital Velocity:-Orbital Velocity is the minimum velocity required to put the satellite into its orbit around the Earth.

Derivation of Orbital Velocity Formula

For the derivation, let us consider a satellite of mass \((m)\) revolving around the earth having mass \((M)\) and the radius \((R)\) in a circular radius \((r)\) at a height \((h)\) from the surface of the Earth. To revolve around the Earth, the Gravitational force between the satellite and the Earth provides a centripetal force.

As we know that, According to the law of gravitation, the force of gravity on the satellite at an altitude of \(h\) is given by:-

\({F_g} = \frac{{GMm}}{{{{(r)}^2}}} = \frac{{GMm}}{{{{(R + h)}^2}}}\,\,\,(r = R + h) \ldots (4)\)

And, the centripetal force required by the satellite to keep it in its orbit is given as

\({F_r} = \frac{{mV_o^2}}{r} = \frac{{mV_o^2}}{{R + h}}\quad \ldots .(5)\)

So, the satellite will orbit the earth if,

\({F_{{\rm{Gravitational }}}} = {F_{{\rm{Centripetal }}}}\)

\( \Rightarrow \frac{{GMm}}{{{{(R + h)}^2}}} = \frac{{mV_o^2}}{{R + h}}\)

\( \Rightarrow V_o^2 = \frac{{GM}}{{R + h}}\)

\( \Rightarrow {V_o} = \sqrt {\frac{{GM}}{{R + h}}} \)

Now when the satellite revolves close to the surface of the Earth then,(\(h = 0\), on comparing with the radius of the earth), and the orbital velocity becomes:-

\( \Rightarrow {V_o} = \sqrt {\frac{{GM}}{R}} \ldots (6)\)

Where,

\(M = \) Mass of the Earth,

\(R = \) Radius of the Earth,

\(m = \) Mass of the Satellite

\({V_o} = \) Orbital Velocity of the Satellite

\(H = \) Height of the satellite above the earth’s surface

Relation Between Escape Velocity and Orbital Velocity

There exists a relationship between escape velocity and orbital velocity. The mathematical relation between the escape velocity and the orbital velocity will be:-

From the expression of the Escape velocity,

We have:- \({v_e} = \sqrt {\frac{{2GM}}{R}} \quad \cdots (3)\)

And from the expression of the orbital velocity,

We have:- \({V_o} = \sqrt {\frac{{GM}}{R}} \quad \cdots (4)\)

Now, On dividing  Eq \((3)\) and Eq \((4)\)

We have:-

\(\frac{{{v_e}}}{{{V_o}}} = \sqrt {\frac{{2GM}}{R}} \times \sqrt {\frac{R}{{GM}}} \)

\( \Rightarrow {v_e} = \sqrt 2 {V_o}\quad \cdots (7)\)

The above relation is valid for any orbit at a distance of \(r\) from the earth’s centre.

Calculation of Escape Velocity from the Surface of the Earth

To escape from the earth surface, we calculate the escape velocity as shown below:-

As we know that:-

The Mass of the Earth \( = 5.98 \times {10^{24}}\;{\rm{kg}}\)

The Radius of the Earth \( = 6.38 \times {10^6}\;{\rm{m}}\)

Newtons Gravitational Constant, \(G = 6.673 \times {10^{ – 11}}{\rm{N}}\,{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\).

Now we can find the escape velocity from the earth using the escape velocity formula:-

\({v_e} = \sqrt {\frac{{2GM}}{R}} \)

Now substituting all the above values, we have:-

\({v_e} = \sqrt {\frac{{2 \times \left( {6.673 \times {{10}^{ – 11}}} \right)\left( {5.98 \times {{10}^{24}}} \right)}}{{6.38 \times {{10}^6}}}} \)

\( \Rightarrow {v_e} = \sqrt {\frac{{\left( {7.981 \times {{10}^{14}}} \right)}}{{6.38 \times {{10}^6}}}} \)

\( \Rightarrow {v_e} = \sqrt {1.251 \times {{10}^8}} \)

\( \Rightarrow {v_e} = 11184\;{\rm{m}}\,{{\rm{s}}^{ – 1}}\)

\( \Rightarrow {v_e} = 11.2\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\)

Therefore, the escape velocity from the Earth is \(11.2\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\).

Factors on which Escape Velocity Depends

1. The Escape Velocity mainly depends on the gravitational potential at the point from where the body is launched. Since this potential depends on the height of the point from the earth’s surface, escape velocity depends on the height from which it is projected.
2. The escape velocity is independent of the body’s mass and the direction of projection of the object for the escape.

Summary

1. The escape velocity for any planet of mass \(M\) and radius \(R\) from its surface is given by, \({v_e} = \sqrt {\frac{{2GM}}{R}} \).
2. The orbital velocity for any planet of mass \(M\) and radius \(R\) in orbit near its surface is given by, \({V_o} = \sqrt {\frac{{GM}}{R}} \).
3. For any orbit, the value of escape velocity will be \(\sqrt 2 \) times of orbital velocity.

Solved Problems

Q.1. What is the escape velocity from the moon if the radius of the moon is given as \(1.74 \times {10^6}\;{\rm{m}}\)? And the mass of the moon is \(7.35 \times {10^{22}}\;{\rm{kg}}\).
Ans: Given:-    Radius of the moon \( = 1.74 \times {10^6}\;{\rm{m}}\)
Mass of the moon is \( = 7.35 \times {10^{22}}\;{\rm{kg}}\)
Newtons Gravitational constant, \(G = 6.673 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\)

We have:-

\({v_e} = \sqrt {\frac{{2GM}}{R}} \)

\( \Rightarrow {v_e} = \sqrt {\frac{{2 \times \left( {6.673 \times {{10}^{ – 11}}} \right)\left( {7.35 \times {{10}^{22}}} \right)}}{{1.74 \times {{10}^6}}}} \)

\( \Rightarrow {v_e} = 2374\;{\rm{m}}\,{{\rm{s}}^{ – 1}}\)

\( \Rightarrow {v_e} = 2.37\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\)

Therefore, the escape velocity from the Earth is \(2.37\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\).

Q.2. If the mass of a planet is eight times the mass of the earth and its radius is twice the Earth’s radius, what will be the escape velocity of that planet?
Ans: Given:-   Mass of the planet \({M_1} = 8 \times \) Mass of the Earth \(\left( {{M_2}} \right)\)
The radius of the planet \({R_1} = 2 \times \) Radius of the Earth \(\left( {{R_2}} \right)\)
The Mass of the Earth \( = 5.98 \times {10^{24}}\;{\rm{kg}}\)
The Radius of the Earth \( = 6.38 \times {10^6}\;{\rm{m}}\)
 Newtons Gravitational Constant, \(G = 6.673 \times {10^{ – 11}}\;{\rm{N}}\;{{\rm{m}}^2}\;{\rm{k}}{{\rm{g}}^{ – 2}}\)
The Mass of the planet \( = 8 \times 5.98 \times {10^{24}}\;{\rm{kg}}\)
The Radius of the planet \( = 2 \times 6.38 \times {10^6}\;{\rm{m}}\)
Now using the escape velocity formula,
Now we have:-\({v_e} = \sqrt {\frac{{2GM}}{R}} \)
\( \Rightarrow {v_e} = \sqrt {\frac{{2 \times \left( {6.673 \times {{10}^{ – 11}}} \right)\left( {8 \times 5.98 \times {{10}^{24}}} \right)}}{{2 \times 6.38 \times {{10}^6}}}} \)
\( \Rightarrow {v_e} = 22368.96\;{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
\( \Rightarrow {v_e} = 22.36\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\)
The escape velocity of the planet is \(22.36\;{\rm{km}}\;{{\rm{s}}^{ – 1}}\).

FAQs

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