Notes : Derivation For Escape Velocity ( Speed ) of a Body - Class 11 physics - Gravitation

Notes : Derivation For Escape Velocity ( Speed ) of a Body - Class 11 physics - Gravitation 

Definition

Escape velocity is the minimum speed with which a body must be projected from the surface of the Earth so that it can escape the Earth's gravitational field.

Consider a Body Projected Upward

Let a body of mass m be projected vertically upward from the Earth's surface with speed \(v_e\).

Initial Energies

Kinetic Energy

\[ K_i=\frac{1}{2}mv_e^2 \]

Potential Energy

\[ U_i=-\frac{GMm}{R} \]

where:

• \(G\) = Universal Gravitational Constant
• \(M\) = Mass of Earth
• \(R\) = Radius of Earth

Total Initial Energy

\[ E_i=\frac{1}{2}mv_e^2-\frac{GMm}{R} \tag{1} \]

At Height \(h\)

Suppose the body reaches a height \(h\) above the Earth's surface with velocity \(v_f\).

Kinetic Energy

\[ K_f=\frac{1}{2}mv_f^2 \]

Potential Energy

\[ U_f=-\frac{GMm}{R+h} \]

Total Final Energy

\[ E_f=\frac{1}{2}mv_f^2-\frac{GMm}{R+h} \tag{2} \]

Applying Conservation of Energy

According to the law of conservation of mechanical energy,

\[ E_i=E_f \]

Substituting equations (1) and (2),

\[ \frac{1}{2}mv_e^2-\frac{GMm}{R} = \frac{1}{2}mv_f^2-\frac{GMm}{R+h} \]

Rearranging,

\[ \frac{1}{2}mv_f^2 = \left(\frac{1}{2}mv_e^2-\frac{GMm}{R}\right) + \frac{GMm}{R+h} \tag{3} \]

Condition for Escape

Since,

\[ \frac{1}{2}mv_f^2 \ge 0 \]

and

\[ \frac{GMm}{R+h}>0 \]

therefore, for escape,

\[ \frac{1}{2}mv_e^2-\frac{GMm}{R}\ge 0 \]

\[ \frac{1}{2}mv_e^2\ge\frac{GMm}{R} \]

\[ v_e^2\ge\frac{2GM}{R} \]

\[ v_e\ge\sqrt{\frac{2GM}{R}} \]

Minimum Escape Speed

Thus , the minimum speed needed to project a body upward from the surface of the earth so that it never returns to surface of the earth is given by : 

\[ v_e=\sqrt{\frac{2GM}{R}} \tag{4} \]

Escape Velocity in Terms of \(g\)

Since

\[ g=\frac{GM}{R^2} \]

\[ GM=gR^2 \]

Substituting in equation (4),

\[ v_e=\sqrt{\frac{2gR^2}{R}} \]

\[ v_e=\sqrt{2gR} \]

\[ \boxed{v_e=\sqrt{2gR}} \tag{5} \]

Final Expression

\[ \boxed{v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}} \]

Important Points

• Escape velocity is independent of the mass of the body.
• It depends only on the mass and radius of the Earth.
• Escape velocity is the minimum speed required to overcome Earth's gravitational attraction.
• The escape velocity from Earth is approximately \(11.2\,\text{km s}^{-1}\).

Frequently Asked Questions (FAQ)

1. What is escape velocity?

Escape velocity is the minimum speed required for a body to escape the gravitational field of a planet.

2. Does escape velocity depend on the mass of the body?

No, escape velocity is independent of the mass of the body.

3. What is the escape velocity from Earth?

\[ 11.2\,\text{km s}^{-1} \]

4. Why is gravitational potential energy negative?

Because gravitational force is attractive and gravitational potential energy is taken as zero at infinity.

Quiz

1. Escape velocity from Earth is:

A) 7.9 km/s
B) 9.8 km/s
C) 11.2 km/s
D) 15 km/s

Answer: C

2. Escape velocity depends on:

A) Mass of body only
B) Radius only
C) Mass and radius of Earth
D) Shape of Earth

Answer: C

3. The SI unit of escape velocity is:

A) m
B) m/s
C) N
D) J

Answer: B

4. Escape velocity is obtained using:

A) Conservation of Momentum
B) Newton's Third Law
C) Conservation of Energy
D) Bernoulli's Principle

Answer: C

5. The formula for escape velocity is:

A) \(\sqrt{gR}\)
B) \(\sqrt{2gR}\)
C) \(2gR\)
D) \(gR\)

Answer: B