Escape Velocity

Escape velocity is the minimum speed an object must reach to break free from the gravitational pull of a celestial body without further propulsion. It's a fundamental concept in astrophysics, representing the threshold where the kinetic energy of an object equals the gravitational energy pulling it back. This concept is not just confined to space exploration; it can be analogously related to real-life situations. For instance, in personal development, "escape velocity" could symbolize the point at which one's efforts, skills, and determination overcome the barriers of fear, doubt, and external obstacles. Just as a spacecraft must reach a certain speed to leave Earth's orbit, individuals must muster enough momentum—through hard work and perseverance—to break free from limiting circumstances and achieve their goals.

Definition of Escape Velocity

Escape velocity is defined as the minimum velocity an object needs to achieve in order to escape the gravitational influence of a celestial body, such as a planet or a star, without further propulsion.

Escape Velocity ( In Terms of the Radius of the Earth)

To escape a body from the earth's surface means to displace it from the surface of the earth to infinity.

The work done to displace a body from the surface of the earth $(r=R)$ to infinity $(r=\infty)$ is

$
W=\int_R^{\infty} \frac{G M m}{x^2} d x=\frac{G M m}{R}
$
So if we provide kinetic energy equal to W to the body at the surface of the earth then it will be able to escape from the earth's gravitational pull.

$
\text { So } K E=\frac{G M m}{R}
$
And Kinetic energy can be written as

$
K E=\frac{1}{2} m V_e^2
$
Where $V_e$ is the required escape velocity.

By comparing we get
$
\begin{aligned}
& \frac{1}{2} m V_e^2=\frac{G M m}{R} \\
\Rightarrow & V_e=\sqrt{\frac{2 G M}{R}}
\end{aligned}
$
Using $G M=g R^2$
We get $V_e=\sqrt{2 g R}$
$V_e \rightarrow$ Escape velocity
$R \rightarrow$ Radius of earth
And using $g=\frac{4}{3} \pi \rho G R$

$
V_e=R \sqrt{\frac{8}{3} \pi G \rho}
$
For the earth

$
V_e=11.2 \mathrm{Km} / \mathrm{s}
$

More About Escape Velocity

• Escape velocity is independent of the mass of the body.

• Escape velocity is independent of the direction of projection of the body.

• Escape velocity depends on the mass and radius of the earth/planet. i.e. Greater the value of $\frac{M}{R}$ or $(g R)$ of the planet, the greater the escape velocity

• If the body projected with velocity less than escape velocity $\left(V<V_e\right)$

In this case, the first body will reach a certain maximum height $\left(H_{\max }\right)$

After that, it may either move in an orbit around the earth/planet or may fall back down towards the earth/planet.

Let's find the Maximum height attained by the body

At maximum height, the velocity of the particle is zero

So at $\mathrm{h}=H_{\max }$ it's Kinetic energy=0

By the law of conservation of energy

Total energy at the surface = Total energy at the height $H_{\max }$
$
\begin{aligned}
& \frac{-G M m}{R}+\frac{1}{2} m V^2=\frac{-G M m}{H_{\max }}+0 \\
& \text { And using } V_e=\sqrt{\frac{2 G M}{R}}
\end{aligned}
$

We get

$
H_{\max }=R\left[\frac{V^2}{V_e^2-V^2}\right]
$

$V_e \rightarrow$ escape velocity
$V \rightarrow$ Projection velocity of the body
$R \rightarrow$ Radius of planet
If a body is projected with a velocity greater than escape velocity ( $V>V_e$ )
Then By the law of conservation of energy
Total energy at surface $=$ Total energy at infinity

$
\frac{-G M m}{R}+\frac{1}{2} m V^2=0+\frac{1}{2} m\left(V^{\prime}\right)^2
$
And using

$
V_e=\sqrt{\frac{2 G M}{R}}
$

We get

$\begin{aligned} & V^{\prime}=\sqrt{V^2-V_e^2} \\ & \text { new velocity of the body at infinity }=V^{\prime} \\ & V \rightarrow \text { projection velocity } \\ & \quad V_e \rightarrow \text { Escape velocity }\end{aligned}$

Escape Energy

Escape energy is the amount of kinetic energy required for an object to escape the gravitational pull of a celestial body, such as a planet or star, without any further propulsion. This energy is what enables the object to overcome the gravitational potential energy binding it to the celestial body.

Energy to be given to an object on the surface of the earth so that its total energy is 0

$\begin{aligned} & \frac{G M m}{R}=\text { Escape Energy } \\ & M \rightarrow \text { Mass of planet } \\ & m \rightarrow \text { mass of the body } \\ & G \rightarrow \text { Gravitational constant }\end{aligned}$

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Solved Examples Based on Escape Velocity

Example 1: A planet in a distant solar system is 10 times more massive than the Earth and its radius is 10 times smaller Given that the escape velocity from the Earth is 11 km s^-1, the escape velocity ( in km s^-1) from the surface of the planet would be :

1) 110

2) 1.1

3) 11

4) 0.11

Solution:

Escape velocity $V_e=\sqrt{\frac{2 G M}{R}}$ for the earth

$
V_e=11 \mathrm{kms}^{-1}
$
Hence the correct mass of the planet $=10 \mathrm{M}_{\mathrm{e}}$
The radius of the planet $\frac{R}{10}$

$
\begin{aligned}
& V_e=\sqrt{\frac{2 G M \times 10}{\frac{R}{10}}} \\
& =10 \times 11=110 \mathrm{kms}^{-1}
\end{aligned}
$

Hence, the answer is the option (1).

Example 2: Two stars of masses $3 \times 10^{31} \mathrm{~kg}$ each, and at a distance $2 \times 10^{11} \mathrm{~m}$ each, rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that a meteorite should have at O is :

(Take Gravitational constant G= 6.67 X 10^-11 Nm² kg ^-2)

1) $3.8 \times 10^4 \mathrm{~m} / \mathrm{s}$
2) $1.4 \times 10^5 \mathrm{~m} / \mathrm{s}$
3) $2.8 \times 10^5 \mathrm{~m} / \mathrm{s}$
4) $2.4 \times 10^4 \mathrm{~m} / \mathrm{s}$

Solution:

By energy conservation
$
-\frac{G M m}{r}-\frac{G M m}{r}+\frac{1}{2} m V^2=0+0
$
M - the mass of the star
m - mass of meteorite

$
V=\sqrt{\frac{4 G m}{r}}=2.8 \times 10^5 \mathrm{~m} / \mathrm{s}
$

Hence, the answer is the option (3).

Example 3: The kinetic energy needed to project a body of mass m from the earth's surface (radius R) to infinity is :

1) $m g R / 2$
2) $2 m g R$
3) $m g R$
4) $\mathrm{mgR} / 4$

Solution:

Escape velocity ( in terms of the radius of the planet)
$
\begin{aligned}
V_c & =\sqrt{\frac{2 G M}{R}} \\
V_c & =\sqrt{2 g R}
\end{aligned}
$

$V_c \rightarrow$ Escape velocity
$R \rightarrow$ Radius of earth
wherein
- depends on the reference body
- greater the value of $\frac{M}{R}$ or $(g R)$ greater will be the escape velocity $V_e=11.2 \mathrm{Km} / \mathrm{s}$ For earth

$
\begin{gathered}
v_e=\sqrt{2 g R} \\
\therefore \text { Kinetic Energy }=\frac{1}{2} m v_e^2=\frac{1}{2} m(2 g R)=m g R
\end{gathered}
$
Hence, the answer is the option (3).

Example 4: The escape velocity of a body depends upon mass:

1) $m^0$
2) $m^1$
3) $\mathrm{m}^2$
4) $m^3$

Solution:

Escape velocity ( in terms of the radius of the planet)
$
\begin{aligned}
V_c & =\sqrt{\frac{2 G M}{R}} \\
V_c & =\sqrt{2 g R}
\end{aligned}
$

$V_c \rightarrow$ Escape velocity
$R \rightarrow$ Radius of earth
wherein
- depends on the reference body
- greater the value of $\frac{M}{R}$ or $(g R)$ greater will be the escape velocity $V_e=11.2 \mathrm{Km} / \mathrm{s}$ For earth Escape velocity $=\sqrt{2 g R}=\sqrt{\frac{2 G M_e}{R}}$

$
V_e \propto m^o
$

Hence, the answer is the option (1).

Example 5: A satellite is revolving in a circular orbit at a height h from the earth's surface, such that h<<R where R is the radius of the earth. Assuming that the effect of the earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of the earth is:

1) $\sqrt{g R}$
2) $\sqrt{\frac{g R}{2}}$
3) $\sqrt{2 g R}$
4) $\sqrt{g R}(\sqrt{2}-1)$

Solution:
$
\begin{aligned}
V_c & =\sqrt{\frac{2 G M}{R}} \\
V_c & =\sqrt{2 g R} \\
V_c & \rightarrow \text { Escape velocity } \\
R & \rightarrow \text { Radius of earth }
\end{aligned}
$
It depends on the reference body
The greater the value of $\frac{M}{R}$ or $(g R)$, the greater will be the escape velocity $V_e=11.2 \mathrm{Km} / \mathrm{s}$
For earth

$
\begin{aligned}
& V_o=\sqrt{g(R+h)} \\
& \approx \sqrt{g R} \\
& V_e=\sqrt{2 g(R+h)} \approx \sqrt{2 g R} \\
& \Delta V=V_e-V_o=(\sqrt{2}-1) \sqrt{g R}
\end{aligned}
$

Hence, the answer is the option (4).

Summary

Escape velocity, as used in astronomy and space exploration, is the speed at which a body must leave a gravitational centre of attraction without accelerating further. The velocity needed to keep a circular orbit at the same height is equal to the square root of two, or around 1.414, times the velocity needed to break free. This velocity decreases with altitude. At the Earth's surface, the escape velocity would be about 11.2 kilometres per second, or 6.96 miles per second, if air resistance was disregarded. The less massive Moon can be escaped from its surface at a speed of about 2.4 km/s. If a planet's escape velocity is near the average velocity of the gas molecules that comprise the atmosphere, the planet (or satellite) will not be able to support an atmosphere for very long.