Physics and Mathematics
Escape Velocity
1. Concept Overview
The escape velocity is the minimum velocity with which a body must be projected vertically upward from the surface of a planet (like Earth) so that it can overcome the planet’s gravitational pull and move to an infinite distance without any further propulsion.
In simpler words, it’s the speed needed to “escape” gravity.
For Earth, the escape velocity is approximately 11.2 km/s.
2. Explanation and Derivation
Let
- Mass of planet = [M]
- Radius of planet = [R]
- Mass of body = [m]
- Gravitational constant = [G]
- Escape velocity = [v_e]
When the body is at the planet’s surface, its potential energy is:
[
U = -\dfrac{G M m}{R}
]
and its kinetic energy is:
[
K = \dfrac{1}{2} m v_e^2
]
At the point of escape (infinity):
- Velocity = 0 (the body just escapes)
- Potential energy = 0
By conservation of energy:
[
K + U = 0
]
[
\dfrac{1}{2} m v_e^2 – \dfrac{G M m}{R} = 0
]
[
\Rightarrow v_e = \sqrt{\dfrac{2 G M}{R}}
]
Hence,
[
\boxed{v_e = \sqrt{\dfrac{2 G M}{R}}}
]
Relation with Acceleration due to Gravity
We know that ([g = \dfrac{G M}{R^2}]), hence:
[
v_e = \sqrt{2 g R}
]
For Earth:
[v_e] [= \sqrt{2 × 9.8 × 6.4 × 10^6}] [≈ 11.2 × 10^3 \text{ m/s}]
[
\boxed{v_e = 11.2 \text{ km/s}}
]
3. Dimensions and Units
| Quantity | Symbol | Dimensions | SI Unit |
|---|---|---|---|
| Escape velocity | [v_e] | [L T⁻¹] | [m/s] |
| Gravitational constant | [G] | [M⁻¹ L³ T⁻²] | [N·m²·kg⁻²] |
| Radius of planet | [R] | [L] | [m] |
| Acceleration due to gravity | [g] | [L T⁻²] | [m/s²] |
4. Key Features
- Escape velocity does not depend on the mass of the body.
- It depends only on the mass and radius of the planet.
- It is independent of the direction of projection (if air resistance is ignored).
- No external energy supply is needed once it’s launched at escape velocity.
- The greater the mass or smaller the radius of the planet, the higher the escape velocity.
5. Important Formulas to Remember
| Formula | Description |
|---|---|
| [v_e = \sqrt{\dfrac{2 G M}{R}}] | Escape velocity from the planet |
| [v_e = \sqrt{2 g R}] | Using acceleration due to gravity |
| [E = K + U = 0] | Condition for escape |
| [v_e ∝ \sqrt{\dfrac{M}{R}}] | Dependency relation |
| [v_e(\text{Earth}) = 11.2 \text{ km/s}] | Numerical value for Earth |
6. Conceptual Questions with Solutions
1. What is escape velocity?
It is the minimum velocity needed by a body to escape a planet’s gravitational field without further propulsion.
2. Does escape velocity depend on the mass of the body?
No. It depends only on the planet’s mass and radius.
3. Why is the escape velocity the same in all directions?
Because the gravitational field is central and isotropic.
4. What happens if a body is projected with velocity less than escape velocity?
It will not escape; it will return to the planet’s surface after reaching a certain height.
5. What is the escape velocity from the Moon’s surface?
Using [v_e = \sqrt{2 g R}], with [g = 1.62 \text{ m/s}^2] and [R = 1.74 × 10^6 \text{ m}], [v_e ≈ 2.38 \text{ km/s}].
6. What is the escape velocity for Mars?
About [5.03 \text{ km/s}].
7. If Earth’s radius doubles and mass remains the same, what happens to escape velocity?
[v_e ∝ \dfrac{1}{\sqrt{R}}], so escape velocity decreases by a factor of [\sqrt{2}].
8. What happens to escape velocity if the planet’s mass doubles?
[v_e ∝ \sqrt{M}], so escape velocity increases by [\sqrt{2}].
9. Why do rockets not actually reach escape velocity instantly?
Because they gain speed gradually due to continuous thrust overcoming gravity and air resistance.
10. Why do planets with small mass have lower escape velocities?
Because gravitational attraction is weaker.
11. What is the significance of total energy being zero at escape?
It means the body just escapes Earth’s gravity, reaching infinity with zero kinetic energy.
12. What is the relation between escape velocity and orbital velocity?
[v_e = \sqrt{2} v_o], where [v_o] is the orbital velocity.
13. How does escape velocity vary with height above Earth?
[v_e = \sqrt{\dfrac{2GM}{R + h}}]; it decreases with altitude.
14. Why can’t we achieve escape velocity on Earth easily?
Because [11.2 \text{ km/s}] is extremely high — requiring huge energy and fuel.
15. What is the escape velocity from the Sun?
Approximately [618 \text{ km/s}].
7. FAQ / Common Misconceptions
1. Escape velocity is the velocity at which a body leaves the atmosphere.
❌ False. It’s the velocity needed to overcome the planet’s gravity completely, not just the atmosphere.
2. Escape velocity depends on the mass of the object.
❌ False. It depends only on the planet’s properties.
3. A rocket must instantly reach escape velocity to escape Earth.
❌ False. It can gradually achieve escape using continuous propulsion.
4. Escape velocity changes with object shape or material.
❌ False. It is purely a function of [M] and [R].
5. At escape velocity, body stops at infinity.
✅ True. It reaches infinity with zero kinetic energy.
6. Escape velocity and orbital velocity are the same.
❌ False. [v_e = \sqrt{2} v_o].
7. Escape velocity depends on altitude.
✅ True. It decreases with increase in altitude.
8. Escape velocity can be achieved by horizontal motion only.
❌ False. It requires velocity in any direction that overcomes gravitational potential energy.
9. Escape velocity is the same for all planets.
❌ False. It varies according to [M] and [R].
10. Satellites move at escape velocity.
❌ False. Satellites move at orbital velocity, which is less than escape velocity.
8. Practice Questions (With Step-by-Step Solutions)
Q1. Derive the expression for escape velocity.
Solution:
By conservation of energy:
[\dfrac{1}{2} m v_e^2 – \dfrac{G M m}{R}] [= 0]
[
v_e = \sqrt{\dfrac{2 G M}{R}}
]
Q2. Calculate the escape velocity from Earth’s surface.
Given: [G = 6.67 × 10^{-11}], [M = 6 × 10^{24}], [R = 6.4 × 10^6].
[v_e] [= \sqrt{\dfrac{2 × 6.67 × 10^{-11} × 6 × 10^{24}}{6.4 × 10^6}}] [= 11.2 × 10^3 \text{ m/s}]
Q3. Find the escape velocity from the Moon.
Given: [g = 1.62 \text{ m/s}^2], [R = 1.74 × 10^6 \text{ m}].
[v_e] [= \sqrt{2 g R}] [= \sqrt{2 × 1.62 × 1.74 × 10^6}] [≈ 2.38 × 10^3 \text{ m/s}]
Q4. What is the ratio of escape velocity on Earth to that on Moon?
[\dfrac{v_{e, \text{Earth}}}{v_{e, \text{Moon}}}] [= \sqrt{\dfrac{g_E R_E}{g_M R_M}}] [= \sqrt{\dfrac{9.8 × 6.4 × 10^6}{1.62 × 1.74 × 10^6}}] [≈ 4.7]
Q5. A planet has twice the radius and eight times the mass of Earth.
Find its escape velocity relative to Earth’s.
[v_e’] [= \sqrt{\dfrac{2 G (8M)}{2R}}] [= \sqrt{4} v_e] [= 2 v_e]
Thus, the escape velocity is twice that of Earth.
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