Escape Velocity Formula and Detailed Derivation (With Examples)
Escape velocity is a fundamental concept in gravitation. It refers to the minimum velocity an object must attain to completely break free from the gravitational pull of a planet—without any further propulsion—so it can move infinitely far away.
This concept is crucial for understanding how satellites, probes, and even natural particles like atmospheric gases can leave a planet’s surface. The escape velocity depends on two main factors for any planet: its mass and its radius.
Detailed Step-by-Step Derivation
-
Understanding the Problem
Let:
- m = mass of the escaping body
- Mp = mass of the planet
- Rp = radius of the planet
- ve = escape velocity
-
Applying the Principle of Conservation of Energy
The total energy (kinetic + potential) at the planet’s surface must equal the total energy at infinity.
At the surface:- Kinetic Energy, KEi = (1/2) m ve²
- Gravitational Potential Energy, PEi = -GMp m / Rp
Ei = (1/2) m ve² – GMp m / Rp -
Energy at Infinity
At infinity, both kinetic and potential energy become zero:
Ef = 0 + 0 = 0 -
Setting Up the Energy Equation
By conservation of energy:
Ei = Ef
(1/2) m ve² – GMp m / Rp = 0 -
Solving for Escape Velocity
Rearranging:
(1/2) m ve² = GMp m / Rp
Divide both sides by m (m ≠ 0):
(1/2) ve² = GMp / Rp
Multiply both sides by 2:
ve² = 2GMp / Rp
Take the square root:
ve = √(2GMp / Rp)
This is the standard formula for escape velocity from a planet of mass Mp and radius Rp. Notice that the escape velocity does not depend on the mass of the escaping object (m), only on the planet’s properties.
Key Formula and Components
| Formula | Meaning of Terms | SI Units |
|---|---|---|
| ve = √(2GMp / Rp) |
G: Universal Gravitational Constant (6.67 × 10-11 N·m²/kg² ) Mp: Mass of planet (kg) Rp: Radius of planet (m) |
m/s |
Worked Example: Calculating Escape Velocity for Earth
Let’s calculate the escape velocity from Earth’s surface using the values:
G = 6.67 × 10-11 N·m²/kg²
Mp (Earth) = 5.97 × 1024 kg
Rp (Earth) = 6.37 × 106 m
| Step | Operation | Result |
|---|---|---|
| 1 | Substitute values in formula: ve = √(2GMp / Rp) |
ve = √[2 × 6.67 × 10-11 × 5.97 × 1024 / 6.37 × 106] |
| 2 | Calculate numerator: | 2 × 6.67 × 10-11 × 5.97 × 1024 ≈ 7.97 × 1014 |
| 3 | Divide numerator by denominator | 7.97 × 1014 / 6.37 × 106 ≈ 1.25 × 108 |
| 4 | Take the square root | ve ≈ 1.12 × 104 m/s = 11,200 m/s = 11.2 km/s |
Therefore, the escape velocity from Earth’s surface is approximately 11.2 km/s.
Key Applications and Comparative Table
| Context | Escape Velocity | Formula/Value |
|---|---|---|
| Earth (surface) | Approximate | 11.2 km/s |
| Moon (surface) | Lower due to smaller mass and radius | 2.4 km/s |
| Depends on | Only on mass and radius of planet | Independent of the object’s mass |
Practice Question
If the radius of a planet doubles but its mass stays the same, what happens to the escape velocity?
| Initial Formula | After Doubling Radius | Ratio |
|---|---|---|
| ve = √(2GMp / Rp) | ve (new) = √(2GMp / 2Rp) = ve/√2 | Reduces by factor 1/√2 (~0.707) |
This shows that increasing the planet's radius (keeping mass constant) will decrease its escape velocity.
Further Reading and Related Vedantu Resources
- Escape Velocity Formula: Concepts and Examples
- Gravitational Force and Escape Velocity
- Escape Velocity and Orbital Velocity: Key Differences
Use the above resources to deepen your conceptual understanding and to practice more questions on escape velocity, planetary motion, and gravitation.
FAQs on Derivation of Escape Velocity: Step-by-Step Guide for Students
1. What is escape velocity?
Escape velocity is the minimum speed required by an object to break free from the gravitational influence of a planet or celestial body, without any further propulsion. For Earth, this value is approximately 11.2 km/s.
• It depends on the mass and radius of the planet.
• If a body is projected with this speed, it will not return due to the planet’s gravity.
2. State the formula for escape velocity from a planet.
The escape velocity formula is:
vesc = √(2GM/R)
where:
• G = Universal Gravitational Constant = 6.674 × 10⁻¹¹ N·m²/kg²
• M = Mass of the planet
• R = Radius of the planet
3. Derive an expression for the escape velocity of a body from any planet.
To derive the escape velocity (vesc) of a body from any planet:
1. Use conservation of energy:
• Initial total energy at the surface: Ei = (1/2) m vesc² – GMm/R
• Final energy at infinity: 0 (both potential and kinetic)
2. Set Ei = 0:
(1/2) m vesc² – GMm/R = 0
3. Rearranging:
(1/2) vesc² = GM/R
vesc² = 2GM/R
vesc = √(2GM/R)
4. On what factors does escape velocity depend?
Escape velocity depends on:
• Mass of the planet/celestial body (M): Higher mass means higher escape velocity.
• Radius of the planet (R): Larger radius reduces escape velocity.
• It does NOT depend on the mass of the escaping object.
5. What is the difference between escape velocity and orbital velocity?
Key differences:
• Escape velocity is the minimum speed to leave a planet’s gravitational field.
• Orbital velocity is the speed needed to stay in a stable orbit close to the planetary surface.
• Escape velocity formula: √(2GM/R)
• Orbital velocity formula: √(GM/R)
• Escape velocity = √2 × orbital velocity at any radius.
6. Is escape velocity dependent on the direction of projection?
No, escape velocity is independent of the direction of projection. It is a speed (not a velocity), so it does not depend on the launch angle or direction—only its magnitude matters.
7. How does escape velocity change if the radius of a planet is doubled but its mass remains the same?
If the radius doubles (R → 2R) and mass is constant:
• New escape velocity: v'esc = √(2GM / 2R) = vesc/√2
• It decreases by a factor of 1/√2.
8. Why is the escape velocity on the Moon much less than on Earth?
The Moon has a much smaller mass and radius compared to Earth.
• Lower mass means a weaker gravitational pull.
• Thus, the escape velocity is much less than that of Earth (about 2.4 km/s for the Moon compared to 11.2 km/s for Earth).
9. What happens to a projectile if it is given exactly escape velocity from Earth's surface?
If a body is projected with exactly escape velocity, it will just reach infinity with zero velocity, overcoming Earth's gravitational field, and will not return.
10. Is escape velocity higher at the poles or the equator of the Earth?
The escape velocity is slightly higher at the poles than at the equator due to Earth's rotation and its oblate shape (radius at poles is less than at equator, increasing vesc).
11. Can escape velocity be achieved in steps, or must it be given instantly?
Escape velocity is the minimum speed required at launch (instantly), assuming no further propulsion. In reality, rockets provide continuous acceleration, but the total energy imparted must at least match that given by the escape velocity formula.
12. What misconceptions do students commonly have about escape velocity?
Common misconceptions:
• Believing escape velocity depends on the body's mass (it does not).
• Confusing escape velocity with orbital velocity.
• Thinking direction matters (only speed matters).
• Forgetting to use correct units or planetary data in calculations.
