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. 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.

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

Derivation of Escape Velocity

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.

Let the initial velocity of the object is \(v_e\).
The initial kinetic energy of the object is \(K_i=\frac{1}{2} m v_e^2\), and
The gravitational potential energy of the earth-object system near the surface of the Earth is
\(
U_i=-\frac{G M m}{R}
\)
Where
\(M=\) The Mass of the Earth,
\(m=\) The Mass of the Object
\(R=\) The Radius of the Earth.
Now,
Total Initial Energy \(=\left(K_i+U_i\right)_{\text {initial }}=\frac{1}{2} m v_e^2-\frac{G M m}{R} \quad \ldots(1)\)

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 \(U_f=-\frac{G M m}{R+h}\)
Now,
Total Final Energy \(\left(K_f+U_f\right)=\frac{1}{2} m v^2-\frac{G M m}{R+h} \quad \ldots(2)\)
So, By conservation of energy
We have
\(
\begin{aligned}
&K_i+U_i=K_f+U_f \\
&\Rightarrow \frac{1}{2} m v_e^2-\frac{G M m}{R}=\frac{1}{2} m v^2-\frac{G M m}{R+h}
\end{aligned}
\)
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
\(
\begin{aligned}
&\frac{1}{2} m v_e^2-\frac{G M m}{R}=0+0 \\
&\Rightarrow \frac{1}{2} m v_e^2=\frac{G M m}{R} \\
&\Rightarrow v_e=\sqrt{\frac{2 G M}{R}} \quad \ldots(3)
\end{aligned}
\)
This critical initial velocity is called the escape velocity. Putting the values of \(G, M\) and \(R\), the escape velocity from the earth comes out to be \(11 \cdot 6 \mathrm{~km} \mathrm{~s}^{-1}\).

Thus, the particle will never return to the earth if

\(v_e \geq \sqrt{\frac{2 G M}{R}} \dots(4)\)

Equation (4) is valid for any celestial object. For example, if something is thrown up from the surface of the moon, it will never return to the moon if the initial velocity is greater than \(\sqrt{\frac{2 G M}{R}}\), where \(M\) is the mass of the moon and \(R\) is the radius of the moon.

If the object was thrown initially with a speed \(v_e\) from a point at a distance \(\left(h+R\right)\) from the centre of the earth \(\left(R=\right.\) radius of the earth), the minimum speed required for an object to reach infinity (i.e. escape from the earth) corresponds to
\(
\frac{1}{2} m\left(v_e^2\right)_{\min }=\frac{G m M}{h+R} \dots(5)
\)

If the object is thrown from the surface of the earth, \(h=0\), and we get
\(
\left(v_e\right)_{\min }=\sqrt{\frac{2 G M}{R}}
\)
Using the relation \(g=G M / R^2\), we get
\(
\left(v_e\right)_{\min }=\sqrt{2 g R} \dots(6)
\)

Example 1: Calculate the escape velocity from the moon. The mass of the moon \(=7.4 \times 10^{22} \mathrm{~kg}\) and radius of the moon \(=1740 \mathrm{~km}\).

Solution: The escape velocity is
\(v=\sqrt{\frac{2 G M}{R}}\)
\(=\sqrt{\frac{2 \times 6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2 \times 7 \cdot 4 \times 10^{22} \mathrm{~kg}}{1740 \times 10^3 \mathrm{~m}}}\)
\(=2 \cdot 4 \mathrm{~km} \mathrm{~s}^{-1}\)

Example 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?

Solution: 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 \(=5.98 \times 10^{24} \mathrm{~kg}\)
The Radius of the Earth \(=6.38 \times 10^6 \mathrm{~m}\)
Newtons Gravitational Constant, \(G=6.673 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}\)
The Mass of the planet \(=8 \times 5.98 \times 10^{24} \mathrm{~kg}\)
The Radius of the planet \(=2 \times 6.38 \times 10^6 \mathrm{~m}\)
Now using the escape velocity formula,
Now we have: \(v_e=\sqrt{\frac{2 G M}{R}}\)
\(
\begin{aligned}
&\Rightarrow v_e=\sqrt{\frac{2 \times\left(6.673 \times 10^{11}\right)\left(8 \times 5.98 \times 10^{20}\right)}{2 \times 6.38 \times 10^6}} \\
&\Rightarrow v_e=22368.96 \mathrm{~m} \mathrm{~s}^{-1} \\
&\Rightarrow v_e=22.36 \mathrm{~km} \mathrm{~s}^{-1}
\end{aligned}
\)
The escape velocity of the planet is \(22.36 \mathrm{~km} \mathrm{~s}^{-1}\).