Escape Velocity

It is the minimum speed an object needs to reach in order to break free from a planet's or celestial body's gravitational pull without any additional force or propulsion. For Earth, this speed is about 11.2 km/s. It depends on the mass and size (radius) of the celestial body and is calculated using principles of energy conservation. Escape velocity is a key concept in physics, especially when studying motion, gravity, and space exploration. 

1.0Gravitational Potential Energy

Gravitational potential energy is the work needed to move a particle from infinity to a point in a gravitational field without altering its kinetic energy.

$U=-G\frac{m_1{m}_2}{r}$, negative sign shows the boundedness of the two bodies

• It is a scalar quantity.
•  Its SI unit is joule and Dimensions are [M¹L²T^–2] 

• The gravitational potential energy of a particle with mass 'm' on Earth's surface (mass 'M', radius 'R') is given by:

$U=-G\frac{m_1{m}_2}{R}$

2.0Definition of Escape Velocity

It is the minimum velocity required to launch a body vertically upward so that it can just overcome Earth's gravitational field and escape into space.

3.0Derivation of Escape Velocity

• Consider Earth as a sphere of mass M and radius R, with its center at O. A body of mass m is located at point P, a distance x from the center.
• The gravitational force of attraction on the body P is $F=G\frac{Mm}{x^2}$
• The small work done in moving the body through small distance PQ=dx against the gravitational force is given by, $dW=Fdx=G\frac{Mm}{x^2}dx$
• The total work done in moving the body from Earth's surface (x = R) to a point beyond Earth's gravitational influence (x = ∞) is:

$W=dW={\int }_R^{\infty }G\frac{Mm}{x^2}dx$

$W=GMm{\int }_R^{\infty }{x}^{-2}dx=GMm[-\frac{1}{x}{]}_R^{\infty }$

$W=GMm[-\frac{1}{\infty }+\frac{1}{R}]=\frac{GMm}{R}$

If $v_e$ is the escape velocity of the body, then the kinetic energy $\frac{1}{2}m{v}_e^2$, The work imparted to the body at Earth's surface will be enough to perform:

$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}\Rightarrow v_e^2=\frac{2GM}{R}$

$v_e=\sqrt{\frac{2GM}{R}}$

$g=\frac{GM}{R^2}$ Or $GM=g{R}^2$

$v_e=\sqrt{\frac{2g{R}^2}{R}}=v_e=\sqrt{2gR}$

If $\rho$ is the density of the earth ,than  $M=\frac{4}{3}\pi R^3\rho$

$v_e=\sqrt{\frac{2G}{R}\times \frac{4}{3}\pi R^3\rho }=\sqrt{\frac{8\pi \rho G{R}^2}{3}}$

Note: Escape velocity does not depend on the mass of the body projected.

Alternative Aspect:

• Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical body or system), of radius R and mass M with escape speed $v_e$.
• When the projectile just escapes to infinity, it has neither kinetic energy nor potential energy.
• From conservation of mechanical energy

$\frac{1}{2}m{v}_e^2+(-\frac{GMm}{R})=0+0\Rightarrow v_e=\sqrt{\frac{2GM}{R}}$

The escape velocity of a body from a location which is at height 'h' above the surface of planet, we can use :-

$v_e=\sqrt{\frac{2GM}{r}}=v_e=\sqrt{\frac{2GM}{R+h}}(\because r=R+h)$

Where, r = Distance from the centre of the planet, h = Height above the surface of the planet

Escape speed depends on :

• Mass (M) and radius (R) of the planet
• Position from where the particle is projected.

Escape speed does not depend on :

• Mass (m) of the body which is projected
• Angle of projection.

If a body is thrown from Earth's surface with escape speed, it will break free from Earth's gravitational field and never return.

For Earth

$v_e=\sqrt{2gR}=\sqrt{2\times 9.8\times 6.4\times 10^6}=11.2\times 10^{3}m/s=11.2km/s$

4.0Escape Velocity From A Point Other Than Surface

1. Total energy is zero at any point when particles start moving with escape velocity, TE=0
2. If given point is at distance (r >R) from center of Earth, $P{E}_{out}=-\frac{G{M}_{e}m}{r}$
3. If given point is at distance (r <R) from the center of Earth, $P{E}_{In}=-\frac{GM(3R^2-r^2)m}{2R^3}$

5.0Satellite Trajectory

A trajectory is the path a satellite follows under the influence of gravity and momentum. It depends on the satellite's speed, altitude, and launch angle. Trajectories can be:

1. Circular
2. Elliptical
3. Parabolic
4. Hyperbolic

• Below Escape Velocity: The object can only orbit or fall back.
• At Escape Velocity: The trajectory becomes parabolic—barely escapes Earth's gravity.
• Above Escape Velocity: The object moves in a hyperbolic trajectory—leaving Earth with excess energy, possibly reaching other planets or stars.

6.0Trajectories and Velocity Thresholds

7.0Escape Energy-Binding Energy

The minimum kinetic energy required for a particle to just escape Earth's gravitational field.

Magnitude of escape energy= $\frac{GMm}{R}$

(–ve of PE on the Earth's surface)

Escape energy = Kinetic Energy Analogous to the escape velocity

$\frac{GMm}{R}=\frac{1}{2}m{v}_e^2$

Binding energy

Total energy of a particle near Earth.

$BE<0\Rightarrow$  Particle cannot escape the gravitational field of Earth

$BE\ge 0\Rightarrow$Particle can escape the gravitational field of Earth

Illustration-1. An unknown planet is of twice the size and half the mass of Earth. If escape velocity from Earth surface is V0What would be the escape velocity of a particle from an unknown planet.

Solution:

$V_0=\sqrt{\frac{2GM}{r}}\Rightarrow V_0\propto \sqrt{\frac{M}{R}}$

$\frac{V}{V_0}=\sqrt{\frac{M/2}{M}\times \frac{R}{2R}}\Rightarrow \frac{V}{V_0}=\sqrt{\frac{1}{4}}\Rightarrow V=\frac{V_0}{2}$

Illustration-2. If M is the mass of a planet, then in order to become black hole, what should be the radius of the planet?

Solution: For an object to become a black body, even light must be unable to escape its gravitational pull.

$v_e\ge c$

$\sqrt{\frac{2GM}{r}}\ge c$  c = speed of light

$\frac{2GM}{c^2}\ge r\Rightarrow r\le \frac{2GM}{c^2}$

Illustration-3. A narrow tunnel is dug along Earth's diameter (radius RR), with a particle of mass mm placed at a distance R/2 from the center. Find the escape speed of the particle from that position.

Solution: Suppose we project the particle with speed Ve. So that it just reaches infinity (r → ∞ )

By conservation of mechanical energy

$K_i+U_i=K_f+U_f$

$\frac{1}{2}m{V}_e^2+m[-\frac{G{M}_e}{2R^3}(3R^2-(\frac{R}{2}{)}^2)]=0$

$V_e=\sqrt{\frac{11G{M}_e}{4R}}$