What would happen if an object is projected exactly at escape velocity?

If projected vertically at exactly escape velocity, the object will continue to move upward, slowing down due to gravity, and reach an infinite distance with zero final velocity. It will just escape the gravitational pull without falling back or needing further propulsion.

Why is escape velocity considered a "threshold" velocity?

Escape velocity is the minimum speed needed to just overcome a celestial body's gravitational attraction. Below this threshold, an object will eventually fall back. At or above this speed (assuming no energy losses), the object can escape the gravitational field entirely and move into space.

Can escape velocity ever be zero? Under what conditions?

Escape velocity would be zero only if the celestial body had no mass, which is physically impossible. Practically, as you move infinitely far from a mass, the local escape velocity approaches zero because gravity becomes negligible.

Why is escape velocity treated as a scalar quantity?

Escape velocity is the minimum required speed, regardless of direction, and is derived from energy (which is scalar). It does not provide directional information, so it's treated as a scalar.

How does the concept of escape velocity apply to satellites and space missions?

Escape velocity determines whether a satellite or probe can leave Earth's orbit and travel to another planet or into deep space. Mission planners calculate this to determine fuel requirements, engine thrust, and optimal trajectories.

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JEE Physics

Escape Velocity

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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 $F = G \frac{M m}{x ^{2}}$
The small work done in moving the body through small distance PQ=dx against the gravitational force is given by, $d W = F d x = G \frac{M m}{x ^{2}} d x$
The total work done in moving the body from Earth's surface (x = R) to a point beyond Earth's gravitational influence (x = ∞) is:

$W = d W = \int_{R}^{\infty} G \frac{M m}{x ^{2}} d x$

$W = GM m \int_{R}^{\infty} x^{- 2} d x = GM m [- \frac{1}{x}]_{R}^{\infty}$

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

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:

$\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R} \Rightarrow v_{e}^{2} = \frac{2 GM}{R}$

$v_{e} = \frac{2 GM}{R}$

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

$v_{e} = \frac{2 g R ^{2}}{R} = v_{e} = 2 g R$

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

$v_{e} = \frac{2 G}{R} \times \frac{4}{3} π R^{3} ρ = \frac{8 π ρG R ^{2}}{3}$

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

$\frac{1}{2} m v_{e}^{2} + (- \frac{GM m}{R}) = 0 + 0 \Rightarrow v_{e} = \frac{2 GM}{R}$

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

$v_{e} = \frac{2 GM}{r} = v_{e} = \frac{2 GM}{R + h} (∵ r = R + h)$

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

$v_{e} = 2 g R = 2 \times 9.8 \times 6.4 \times 1 0^{6} = 11.2 \times 1 0^{3} m / s = 11.2 km / s$

4.0Escape Velocity From A Point Other Than Surface

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

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:

Circular
Elliptical
Parabolic
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

Speed	Trajectory Type	Result
< 7.9 km/s	Sub-orbital	Falls back to Earth
= 7.9 km/s	Circular orbit	Low Earth Orbit (LEO)
7.9–11.2 km/s	Elliptical orbit	Earth orbit but not circular
= 11.2 km/s	Parabolic escape	Just escapes Earth's gravity
> 11.2 km/s	Hyperbolic trajectory	Leaves Earth with residual speed

7.0Escape Energy-Binding Energy

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

Escape energy or binding energy of the earth

Magnitude of escape energy= $\frac{GM m}{R}$

(–ve of PE on the Earth's surface)

Escape energy = Kinetic Energy Analogous to the escape velocity

$\frac{GM m}{R} = \frac{1}{2} m v_{e}^{2}$

KE < Escape Energy	$v_{0} < v_{e}$	Body returns to Earth surface
KE = Escape Energy	$v_{0} = v_{e}$	Body comes to rest at infinity
KE > Escape Energy	$v_{0} > v_{e}$	Body has residual velocity at infinity

Binding energy

Total energy of a particle near Earth.

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

$BE \geq 0 \Rightarrow$ 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:

$V_{0} = \frac{2 GM}{r} \Rightarrow V_{0} \propto \frac{M}{R}$

$\frac{V}{V _{0}} = \frac{M /2}{M} \times \frac{R}{2 R} \Rightarrow \frac{V}{V _{0}} = \frac{1}{4} \Rightarrow V = \frac{V _{0}}{2}$

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.

$v_{e} \geq c$

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

$\frac{2 GM}{c ^{2}} \geq r \Rightarrow r \leq \frac{2 GM}{c ^{2}}$

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

$K_{i} + U_{i} = K_{f} + U_{f}$

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

$V_{e} = \frac{11 G M _{e}}{4 R}$

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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 $F = G \frac{M m}{x ^{2}}$
The small work done in moving the body through small distance PQ=dx against the gravitational force is given by, $d W = F d x = G \frac{M m}{x ^{2}} d x$
The total work done in moving the body from Earth's surface (x = R) to a point beyond Earth's gravitational influence (x = ∞) is:

$W = d W = \int_{R}^{\infty} G \frac{M m}{x ^{2}} d x$

$W = GM m \int_{R}^{\infty} x^{- 2} d x = GM m [- \frac{1}{x}]_{R}^{\infty}$

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

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:

$\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R} \Rightarrow v_{e}^{2} = \frac{2 GM}{R}$

$v_{e} = \frac{2 GM}{R}$

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

$v_{e} = \frac{2 g R ^{2}}{R} = v_{e} = 2 g R$

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

$v_{e} = \frac{2 G}{R} \times \frac{4}{3} π R^{3} ρ = \frac{8 π ρG R ^{2}}{3}$

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

$\frac{1}{2} m v_{e}^{2} + (- \frac{GM m}{R}) = 0 + 0 \Rightarrow v_{e} = \frac{2 GM}{R}$

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

$v_{e} = \frac{2 GM}{r} = v_{e} = \frac{2 GM}{R + h} (∵ r = R + h)$

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

$v_{e} = 2 g R = 2 \times 9.8 \times 6.4 \times 1 0^{6} = 11.2 \times 1 0^{3} m / s = 11.2 km / s$

4.0Escape Velocity From A Point Other Than Surface

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

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:

Circular
Elliptical
Parabolic
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

Speed	Trajectory Type	Result
< 7.9 km/s	Sub-orbital	Falls back to Earth
= 7.9 km/s	Circular orbit	Low Earth Orbit (LEO)
7.9–11.2 km/s	Elliptical orbit	Earth orbit but not circular
= 11.2 km/s	Parabolic escape	Just escapes Earth's gravity
> 11.2 km/s	Hyperbolic trajectory	Leaves Earth with residual speed

7.0Escape Energy-Binding Energy

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

Magnitude of escape energy= $\frac{GM m}{R}$

(–ve of PE on the Earth's surface)

Escape energy = Kinetic Energy Analogous to the escape velocity

$\frac{GM m}{R} = \frac{1}{2} m v_{e}^{2}$

KE < Escape Energy	$v_{0} < v_{e}$	Body returns to Earth surface
KE = Escape Energy	$v_{0} = v_{e}$	Body comes to rest at infinity
KE > Escape Energy	$v_{0} > v_{e}$	Body has residual velocity at infinity

Binding energy

Total energy of a particle near Earth.

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

$BE \geq 0 \Rightarrow$ 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:

$V_{0} = \frac{2 GM}{r} \Rightarrow V_{0} \propto \frac{M}{R}$

$\frac{V}{V _{0}} = \frac{M /2}{M} \times \frac{R}{2 R} \Rightarrow \frac{V}{V _{0}} = \frac{1}{4} \Rightarrow V = \frac{V _{0}}{2}$

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.

$v_{e} \geq c$

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

$\frac{2 GM}{c ^{2}} \geq r \Rightarrow r \leq \frac{2 GM}{c ^{2}}$

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

$K_{i} + U_{i} = K_{f} + U_{f}$

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

$V_{e} = \frac{11 G M _{e}}{4 R}$

Frequently Asked Questions

What would happen if an object is projected exactly at escape velocity?

Why is escape velocity considered a "threshold" velocity?

Can escape velocity ever be zero? Under what conditions?

Why is escape velocity treated as a scalar quantity?

How does the concept of escape velocity apply to satellites and space missions?

Join ALLEN!

Escape Velocity

1.0Gravitational Potential Energy

2.0Definition of Escape Velocity

3.0Derivation of Escape Velocity

4.0Escape Velocity From A Point Other Than Surface

5.0Satellite Trajectory

6.0Trajectories and Velocity Thresholds

7.0Escape Energy-Binding Energy

Table of Contents

Frequently Asked Questions

What would happen if an object is projected exactly at escape velocity?

Why is escape velocity considered a "threshold" velocity?

Can escape velocity ever be zero? Under what conditions?

Why is escape velocity treated as a scalar quantity?

How does the concept of escape velocity apply to satellites and space missions?

Join ALLEN!

Escape Velocity

1.0Gravitational Potential Energy

2.0Definition of Escape Velocity

3.0Derivation of Escape Velocity

4.0Escape Velocity From A Point Other Than Surface

5.0Satellite Trajectory

6.0Trajectories and Velocity Thresholds

7.0Escape Energy-Binding Energy

Table of Contents