Escape Velocity Definition
When an object reaches the speed that it is capable of breaking the Earth’s gravitational barrier then we say that it is travelling at escape velocity. i.e. Escape velocity is the minimum velocity that when an object travels at that speed can break the gravitational barrier of that celestial body.
The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational pull, is known as escape velocity.
Escape Velocity Formula
The escape velocity equation is obtained by equating the kinetic energy of an object with mass m and travelling with a velocity of v and the gravitational potential energy of the same object. When the Kinetic energy of the object equals the gravitational energy of the celestial object the equation so obtained is called the Escape Velocity Equation.
For Earth let the escape velocity is ve then the kinetic energy is 1/2 m(ve)2 if m is the mass of the object and now the gravitational potential energy is equal to the kinetic energy and the gravitational potential energy of the object is GMm/r where G is the universal gravitational constant and M is the mass of the Earth then,
1/2 m(ve)2 = GMm/r
Thus, the equation of Escape Velocity is given by
ve = √(2GM/r)
Where,
- ve is the Escape Velocity of the object,
- G is the Universal Gravitational Constant,
- M is the Mass of the Celestial Object whose gravitational force is to break, and
- r is the distance from the object to the centre of mass of the body which has to be escaped.
Escape Velocity Derivation
The minimum amount of velocity for which the particle escapes the gravitational sphere of influence of a planet is known as escape velocity (ve). When escape velocity is provided to a body, it goes to infinity theoretically.
As the gravitational force is a conservative force, the law of conservation of energy holds good. Applying the law of conservation of energy for the particle which is provided with a required minimum velocity to go to infinity,
\(\begin{array}{l}{{U}_{i}}+{{K}_{i}}={{U}_{f}}+{{K}_{f}}\end{array} \).
At infinity, the particles experience no interaction so the final potential energy, and we know from motion in the 1D chapter, the final velocity of a body becomes zero after reaching its maximum height. So from this, we can deduce the final kinetic energy of the particle.
Then,
\(\begin{array}{l}{{U}_{i}}+{{K}_{i}}=0~\text{and }\!\!~\!\!\text{ we }\!\!~\!\!\text{ know }\!\!~\!\!\text{ that},{{U}_{i}}=\frac{-GMm}{R}~,{{K}_{i}}=~\frac{1}{2}m{{v}_{e}}^{2}\end{array} \)
We get,
\(\begin{array}{l}\frac{1}{2}m{{v}_{e}}^{2}+\left( \frac{-GMm}{R} \right)=0\Rightarrow \frac{1}{2}m{{v}_{e}}^{2}=\frac{GMm}{R}\end{array} \)
That implies,
\(\begin{array}{l}{{v}_{e}}=\sqrt{\frac{2GM}{R}}\end{array} \)
From the above formula, it is clear that escape velocity does not depend upon the test mass (m).
If the source mass is earth, then the escape velocity has a value of 11.2 km/s.
If v = ve, the body escapes the planet’s gravitational sphere of influence; if 0 ≤ v < ve, the body either falls back onto the earth or continues to orbit around the planet within the sphere of influence of the planet.
Escape Velocity of Earth
The velocity required by an object to escape from Earth’s Gravitational Force is called the Escape Velocity of Earth. Now,
Let the escape velocity of the Earth be ve then,
ve = √(2GM/r)
**Note : Escape velocity is independent of the mass of the object (m) being launched; it depends only on the mass (M) and radius of the planet (r).
We know that,
Put GM/r2 = g (where g is Earth’s gravity) in the above equation. Now the equation of escape velocity becomes
ve = √(2gr)
We know, Radius of Earth, r = 6.4 × 106 m, and g = 9.8 m/s2
Now the escape velocity of the Earth,
ve = √(2 × 9.8 × 6.4 × 106)
ve = 11.2 × 103 m/s
ve = 11.2 km/s
Thus, the escape velocity of the Earth is 11.2 km/s.
Unit and Dimension of Escape Velocity
Escape velocity is the velocity that is required to break the gravitational barrier of any celestial object so the unit of escape velocity is similar to the unit of velocity.
The Escape Velocity is measured in Km/sec, m/sec, or in the other units of velocity and the dimensional formula of Escape Velocity is [LT-2].
Relation Between Escape Velocity and Density of Planet
The formula for escape velocity is
ve = √(2gR)
Since, g = 4/3 [πρRG], put the value of g in above equation, we get
ve = √(2 x 4/3 [πρRG]R)
ve = R √(8/3 πρG)
** Note : If density of planet is given in the problem, the
ve ∝ R
Escape Velocity on Different Celestial Objects
Different Celestial Objects in our solar system have different escape velocities due to different gravitation forces of attraction (as they have different masses and different radii). The Escape velocity of different objects in our solar system is discussed in the table below,
| Celestial Body | Escape Velocity (in km/s) |
|---|---|
| Sun | 618 km/s |
| Moon | 2.38 km/s |
| Earth | 11.2 km/s |
| Jupiter | 59.5 km/s |
| Mercury | 4.25 km/s |
| Mars | 5.03 km/s |
| Venus | 10.36 km/s |
Solved Examples on Escape Velocity
Example 1: Find the escape velocity for a planet whose mass is 7.35 × 1022 Kg and radius is 1.5 × 106m.
Solution:
The formula for escape velocity is given by,
ve = √(2GM/R)
Given,
M = 7.35 × 1022 Kg
R = 1.5 × 106m
G = 6.6 × 10-11 N m2/kg2
plugging the values into the equation, we get
ve = 7.6 × 105 m/s
Example 2: Find the escape velocity for a planet whose mass is 14.7 × 1022 Kg and radius is 3 × 106m.
Solution:
The formula for escape velocity is given by,
ve = √(2GM/R)
Given,
M = 14.7 × 1022 Kg
R = 3 × 106 m
G = 6.6 × 10-11 N m2/kg2
Putting the values in equation
ve = 3.04 × 103 m/s
Escape Velocity of the planet is 3.04 × 103 m/s
Example 3: Find the mass of a planet whose escape velocity is 2 × 104 m/s and radius is 2 × 105m.
Solution:
The formula for escape velocity is given by,
ve = √(2GM/R)
Given,
R = 2 × 105m
G = 6.6 × 10-11 N m2/kg2
ve = 2 × 104 m/s
Putting the values in equation
M = 6 × 1023 Kg
The mass of the planet is 6 × 1023 Kg
Example 4: Find the escape velocity if the radius of the Earth is increased by 4 times.
Solution:
The formula for escape velocity In Earth’s case is given by,
ve = √(2gr)
If the radius is made four times, this means that the velocity must be doubled.
Escape Velocity of Earth = 11.2 Km/s.
Escape Velocity of Earth with with 4 times radius = 11.2 × 2 = 22.4 Km/s.
Escape Velocity – Important Question-Answers in Physics
Q1: What is Escape Velocity?
Answer: The minimum velocity with which a body must be projected upwards to overcome the gravitational pull of the Earth is called escape velocity.
Q2: What is the formula for escape velocity?
Answer: The escape velocity is given by
ve = √(2GM/r)
Where,
- ve is the Escape Velocity of the object,
- G is the Universal Gravitational Constant,
- M is the Mass of the Celestial Object whose gravitational force is to break, and
- r is the distance from the object to the centre of mass of the body which has to be escaped.
Q3: What is the value of escape velocity on Earth?
Answer: The formula of escape velocity for case of earth is :
ve = √(2gr)
We know, Radius of Earth, r = 6.4 × 106 m, and g = 9.8 m/s2
Now the escape velocity of the Earth,
ve = √(2 × 9.8 × 6.4 × 106)
ve = 11.2 × 103 m/s
ve = 11.2 km/s
Q4: Does escape velocity depend on the mass of the object?
Answer: No, escape velocity is independent of the mass of the object being launched; it depends only on the mass and radius of the planet.
Q5: How does escape velocity change for other planets?
Answer: Escape velocity varies for different planets because it depends on their mass and radius. For example:
- Moon: 2.4 km/s
- Mars: 5.0 km/s
- Jupiter: 59.5 km/s
MCQs on Escape Velocity
Q1: Escape velocity depends on which of the following?
(a) Mass of the object
(b) Radius of the planet
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Answer: (b) Radius of the planet
Explanation: Escape velocity is given by, ve = √(2GM/r)
**Note : Escape velocity is independent of the mass of the object (m) being launched; it depends only on the mass (M) and radius of the planet (r)
Q2: If the radius of a planet doubles but mass remains the same, how does the escape velocity change?
(a) Increases by 2 times
(b) Decreases by √2 times
(c) Remains the same
(d) Decreases by 2 times
Answer: (b) Decreases by √2 times
Explanation: Since ve = √(2GM/r), if the radius r is doubled decreases proportionally by √2.
Q3: What will be the escape velocity of a body on a planet with twice the Earth’s mass and the same radius?
(a) 11.2 km/s
(b) 15.8 km/s
(c) 22.4 km/s
(d) 7.9 km/s
Answer: (b) 15.8 km/s
Explanation: ve = √(2GM/r). Escape velocity is proportional to the square root of mass. If mass doubles, escape velocity increases by √2 times, i.e., 11.2×√2 ≈ 15.811.2 approx 15.8 km/s.
Frequently Asked Questions (FAQs)
Q1: Why is escape velocity important?
Answer: Escape velocity is essential in space missions, as it determines the minimum speed required to send spacecraft beyond Earth’s gravitational influence.
Q2: Can an object escape Earth’s gravity at a lower velocity?
Answer: No, unless an external force (such as continuous propulsion) is applied, the object needs at least the escape velocity to break free from Earth’s gravity.
Q3: How does atmospheric resistance affect escape velocity?
Answer: Atmospheric drag slows down objects, requiring additional thrust to achieve escape velocity in real-world scenarios.
Q4: Does the Moon have an escape velocity?
Answer: Yes, the Moon has an escape velocity of approximately 2.4 km/s, which is much lower than Earth’s due to its smaller mass and radius.
Q5: How does escape velocity relate to black holes?
Answer: A black hole’s escape velocity is greater than the speed of light, meaning nothing, not even light, can escape its gravitational pull.
Q6: What is the difference between orbital velocity and escape velocity?
Answer: Orbital Speed: The speed required to reach the orbit of a planet or any celestial object is called as the orbital speed of that object. Orbital speed of the Earth is 7.34 km/sec.
Escape Velocity: The velocity required to break the gravitational barrier of the celestial object is called the escape velocity of that object. Escape velocity of Earth is 11.2 km/sec.
Q7: Which planet has the highest escape velocity in our solar system?
Answer: Jupiter has the highest escape velocity among all the planets in our solar system. The escape velocity of Jupiter is 59.5 km/sec.
Q8: Which planet has the lowest escape velocity in our solar system?
Answer: Mercury has the lowest escape velocity among all the planets in our solar system. The escape velocity of Mercury is 4.2.5 km/sec.
Q9: What is the escape velocity of the Sun?
Answer: The escape velocity of the Sun is equal to 618 km/s.
Q10: What is the escape velocity of the Moon?
Answer: The escape velocity of the Moon is equal to 2.38 km/s.
Q11: What is the escape velocity of Jupiter?
Answer: The escape velocity of Jupiter is equal to 59.5 km/s.
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