Class 11 Physics Chapter 7: Gravitation Notes & Formulas

Class 11 Physics Chapter 7: Gravitation

Gravitation is an important chapter of Class 11 Physics. It explains the force of attraction between masses and helps us understand the motion of planets, satellites and other celestial bodies.

Chapter Focus: Kepler's laws, universal law of gravitation, acceleration due to gravity, gravitational potential and energy, escape velocity, orbital velocity and satellites.

In this chapter:

• Kepler's Laws
• Universal Law of Gravitation
• Acceleration Due to Gravity
• Variation of g
• Gravitational Potential
• Gravitational Potential Energy
• Escape Velocity
• Satellites
• Orbital Velocity
• Energy of a Satellite
• Geostationary Satellites

1. Kepler's Laws of Planetary Motion

Johannes Kepler proposed three laws to describe the motion of planets around the Sun. These laws provided an important foundation for understanding planetary motion.

Kepler's First Law

Every planet moves around the Sun in an elliptical orbit with the Sun at one of the two foci of the ellipse.

Kepler's Second Law

The line joining a planet and the Sun sweeps out equal areas in equal intervals of time.

Meaning: A planet moves faster when it is closer to the Sun and slower when it is farther away.

Kepler's Third Law

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

T² ∝ r³

For planets orbiting the same central body:

T²/r³ = Constant

2. Universal Law of Gravitation

Newton proposed that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

F = Gm₁m₂/r²

Here:

• F = gravitational force
• m₁, m₂ = masses of the two bodies
• r = distance between their centres
• G = universal gravitational constant

G = 6.67 × 10⁻¹¹ N m²/kg²

Important: Gravitational force is always attractive and acts along the line joining the centres of the two bodies.

3. Acceleration Due to Gravity

The acceleration produced in a body due to the gravitational attraction of the Earth is called acceleration due to gravity. It is represented by g.

For a body of mass m near the surface of Earth:

F = GMm/R²

Using Newton's second law:

F = mg

Therefore:

g = GM/R²

The approximate value of g near the Earth's surface is:

g ≈ 9.8 m/s²

4. Variation of g with Height

The value of acceleration due to gravity decreases as we move away from the Earth's surface.

At a height h above the Earth's surface:

g_h = GM/(R+h)²

Therefore:

g_h = g [R/(R+h)]²

For small height compared with Earth's radius:

g_h ≈ g(1 - 2h/R)

Remember: As height increases, the value of g decreases.

5. Variation of g with Depth

The value of acceleration due to gravity also decreases as we move below the Earth's surface.

At a depth d below the surface, assuming Earth has uniform density:

g_d = g(1 - d/R)

At the centre of Earth:

d = R   ⇒   g = 0

6. Gravitational Potential

Gravitational potential at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point without acceleration.

The gravitational potential at a distance r from a mass M is:

V = -GM/r

The negative sign indicates that gravitational potential is taken as zero at infinity and becomes negative closer to the attracting mass.

The SI unit of gravitational potential is J/kg.

7. Gravitational Potential Energy

The gravitational potential energy of a mass m at a distance r from the centre of a spherical mass M is:

U = -GMm/r

Near the Earth's surface, for small heights:

U = mgh

Important: The formula mgh is useful near Earth's surface, while -GMm/r is the general expression when distance from Earth's centre is considered.

8. Escape Velocity

Escape velocity is the minimum speed with which an object must be projected from the Earth's surface so that it can escape the Earth's gravitational field without further propulsion.

v_e = √(2GM/R)

Using g = GM/R²:

v_e = √(2gR)

For Earth, the escape velocity is approximately:

v_e ≈ 11.2 km/s

Key Point: Escape velocity does not depend on the mass of the object being projected.

9. Artificial Satellites

An artificial satellite is a human-made object placed in orbit around Earth or another celestial body.

A satellite remains in orbit because the gravitational force provides the required centripetal force.

Gravitational force = Centripetal force

GMm/r² = mv²/r

10. Orbital Velocity

Orbital velocity is the minimum horizontal velocity required by a satellite to remain in a circular orbit around Earth at a given height.

v_o = √(GM/r)

For a satellite close to Earth's surface:

v_o = √(GM/R)

Using g = GM/R²:

v_o = √(gR)

Therefore:

v_e = √2 v_o

11. Time Period of a Satellite

The time period of a satellite is the time taken to complete one revolution around Earth.

T = 2π√(r³/GM)

Therefore:

T² ∝ r³

This relation is also consistent with Kepler's third law.

12. Energy of an Orbiting Satellite

The total mechanical energy of a satellite is the sum of its kinetic and gravitational potential energies.

For a satellite of mass m in a circular orbit of radius r:

K = GMm/2r

U = -GMm/r

Therefore total energy is:

E = -GMm/2r

Important: The total energy of a bound satellite is negative, showing that the satellite is gravitationally bound to Earth.

13. Geostationary Satellites

A geostationary satellite appears stationary relative to an observer on Earth. It revolves from west to east in the equatorial plane with the same angular velocity as Earth.

Important Conditions

• Orbit must be circular.
• Orbit must lie in the equatorial plane.
• Satellite must move from west to east.
• Its time period must equal Earth's rotational period.

A geostationary satellite has an orbital period of approximately 24 hours.

Quick Revision

Concept	Formula
Gravitational force	F = Gm₁m₂/r²
Acceleration due to gravity	g = GM/R²
Gravity at height h	gh = GM/(R+h)²
Gravity at depth d	gd = g(1-d/R)
Gravitational potential	V = -GM/r
Potential energy	U = -GMm/r
Escape velocity	ve = √(2GM/R)
Orbital velocity	vo = √(GM/r)
Satellite time period	T = 2π√(r³/GM)
Satellite total energy	E = -GMm/2r

Important Questions – Gravitation

Very Short Answer Questions

1. State Newton's universal law of gravitation.
2. What is the value of the universal gravitational constant G?
3. Define acceleration due to gravity.
4. Write the relation between G and g.
5. What happens to g as we move above the Earth's surface?
6. What happens to g at the centre of Earth?
7. State Kepler's first law.
8. State Kepler's second law.
9. State Kepler's third law.
10. Define gravitational potential.
11. What is escape velocity?
12. What is orbital velocity?
13. What is a geostationary satellite?
14. Why is gravitational potential energy negative?

Short Answer Questions

1. Explain Newton's universal law of gravitation.
2. Derive the expression g = GM/R².
3. Why does the value of g decrease with altitude?
4. Why does g become zero at the centre of Earth?
5. Explain Kepler's three laws of planetary motion.
6. Differentiate between gravitational potential and gravitational potential energy.
7. Why is escape velocity independent of the mass of the projectile?
8. Why does a satellite not fall directly onto Earth?
9. Differentiate between orbital velocity and escape velocity.
10. What conditions are necessary for a satellite to be geostationary?

Long Answer Questions

1. Derive the expression for acceleration due to gravity at the Earth's surface.
2. Derive the expression for escape velocity from Earth's surface.
3. Derive the expression for orbital velocity of a satellite.
4. Derive the expression for the time period of a satellite.
5. Derive the total energy of a satellite in a circular orbit.
6. Explain the variation of acceleration due to gravity with height and depth.
7. Explain Kepler's laws and their significance in planetary motion.

Important MCQs

1. The gravitational force between two masses is proportional to:
   (a) sum of masses   (b) product of masses   (c) difference of masses   (d) square of distance
   Answer: (b) product of masses
2. The value of g at the centre of Earth is:
   (a) 9.8 m/s²   (b) 4.9 m/s²   (c) zero   (d) infinite
   Answer: (c) zero
3. The gravitational force varies with distance r as:
   (a) r   (b) r²   (c) 1/r   (d) 1/r²
   Answer: (d) 1/r²
4. According to Kepler's third law:
   (a) T ∝ r
   (b) T² ∝ r³
   (c) T³ ∝ r²
   (d) T ∝ r²
   Answer: (b) T² ∝ r³
5. The escape velocity from Earth is approximately:
   (a) 5.6 km/s   (b) 9.8 km/s   (c) 11.2 km/s   (d) 15 km/s
   Answer: (c) 11.2 km/s
6. The orbital velocity of a satellite close to Earth's surface is:
   (a) √gR   (b) √2gR   (c) gR   (d) 2gR
   Answer: (a) √gR
7. The gravitational potential at infinity is generally taken as:
   (a) 1   (b) -1   (c) zero   (d) infinite
   Answer: (c) zero
8. A geostationary satellite has a time period of approximately:
   (a) 12 hours   (b) 24 hours   (c) 48 hours   (d) 6 hours
   Answer: (b) 24 hours

Numerical Problems

Numerical 1: Gravitational Force

Two bodies of masses 10 kg and 20 kg are separated by a distance of 2 m. Calculate the gravitational force between them. Take G = 6.67 × 10⁻¹¹ N m²/kg².

F = Gm₁m₂/r²
F = (6.67 × 10⁻¹¹ × 10 × 20)/4
F = 3.335 × 10⁻⁹ N

Answer: 3.335 × 10⁻⁹ N

Numerical 2: Weight of a Body

Find the weight of a 50 kg student on Earth. Take g = 9.8 m/s².

W = mg
W = 50 × 9.8
W = 490 N

Answer: 490 N

Numerical 3: Gravitational Potential

Find the gravitational potential at a point 2R from the centre of a planet of mass M.

V = -GM/r
V = -GM/2R

Answer: -GM/2R

Numerical 4: Escape Velocity

Calculate the escape velocity from a planet having g = 10 m/s² and radius R = 6.4 × 10⁶ m.

v_e = √(2gR)
= √(2 × 10 × 6.4 × 10⁶)
≈ 11.3 × 10³ m/s

Answer: Approximately 11.3 km/s

Numerical 5: Orbital Velocity

A satellite is orbiting Earth at a distance r from its centre. Write its orbital velocity in terms of G, M and r.

v_o = √(GM/r)

Answer: √(GM/r)

Numerical 6: Satellite Energy

Find the total energy of a satellite of mass m moving in a circular orbit of radius r around a planet of mass M.

E = -GMm/2r

Answer: -GMm/2r

Assertion and Reason Questions

1. Assertion: The value of g decreases as height above Earth's surface increases.
   Reason: The distance from the centre of Earth increases with height.
   Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
2. Assertion: The value of g is zero at the centre of Earth.
   Reason: The gravitational forces due to all directions cancel at the centre.
   Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
3. Assertion: Escape velocity is independent of the mass of the projectile.
   Reason: The mass of the projectile cancels out while deriving escape velocity.
   Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.
4. Assertion: A satellite in a circular orbit is continuously accelerating.
   Reason: Its velocity continuously changes direction due to centripetal acceleration.
   Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Exam Tips

• Memorise the three Kepler's laws and understand their meaning.
• Practice the derivations of g, orbital velocity and escape velocity.
• Remember that gravitational force follows the inverse-square law.
• Do not confuse gravitational potential with gravitational potential energy.
• Remember that orbital velocity is less than escape velocity.
• Practice satellite energy and time-period numericals.
• Learn the conditions for geostationary satellites.
• In numerical problems, check whether the given distance is from Earth's surface or from its centre.

Class 11 Physics Chapter 7 FAQs

1. What is Chapter 7 of Class 11 Physics?

Chapter 7 is Gravitation. It covers Kepler's laws, Newton's law of gravitation, acceleration due to gravity, gravitational potential, potential energy, escape velocity and satellite motion.

2. What is Newton's universal law of gravitation?

Every two masses attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

3. What is the formula for gravitational force?

The formula is F = Gm₁m₂/r².

4. What is the formula for acceleration due to gravity?

At Earth's surface, the acceleration due to gravity is g = GM/R².

5. What happens to g at the centre of Earth?

The value of g becomes zero at the centre of Earth because the gravitational effects from all directions balance each other.

6. What is escape velocity?

Escape velocity is the minimum speed required for an object to escape the gravitational field of a celestial body without further propulsion.

7. What is the escape velocity from Earth?

The escape velocity from Earth's surface is approximately 11.2 km/s.

8. What is orbital velocity?

Orbital velocity is the velocity required by a satellite to maintain a particular circular orbit around a celestial body. It is given by v_o = √(GM/r).

9. What is a geostationary satellite?

A geostationary satellite is a satellite that appears stationary relative to Earth because it moves in the equatorial plane from west to east with a period equal to Earth's rotation period.

10. Why is gravitational potential energy negative?

When gravitational potential energy is taken as zero at infinity, a bound object has negative gravitational potential energy because work must be supplied to take it from its position to infinity.

Chapter 7 Formula Sheet

Topic	Formula
Gravitational force	F = Gm₁m₂/r²
Acceleration due to gravity	g = GM/R²
Gravity at height	gh = g[R/(R+h)]²
Gravity at depth	gd = g(1-d/R)
Gravitational potential	V = -GM/r
Gravitational potential energy	U = -GMm/r
Near-Earth potential energy	U = mgh
Escape velocity	ve = √(2GM/R)
Escape velocity near Earth	ve = √(2gR)
Orbital velocity	vo = √(GM/r)
Relation	ve = √2 vo
Satellite time period	T = 2π√(r³/GM)
Satellite kinetic energy	K = GMm/2r
Satellite total energy	E = -GMm/2r

One-Minute Revision

• Gravitational force: F = Gm₁m₂/r².
• Acceleration due to gravity: g = GM/R².
• Kepler's third law: T² ∝ r³.
• Gravitational potential: V = -GM/r.
• Potential energy: U = -GMm/r.
• Escape velocity: v_e = √(2GM/R).
• Orbital velocity: v_o = √(GM/r).
• Relation: v_e = √2v_o.
• Satellite period: T = 2π√(r³/GM).
• Satellite energy: E = -GMm/2r.
• At Earth's centre: g = 0.

Conclusion

Class 11 Physics Chapter 7 – Gravitation connects Newton's laws with the motion of planets and satellites. For CBSE preparation, students should focus especially on Kepler's laws, variation of g, gravitational potential and energy, escape velocity, orbital velocity and satellite numericals.

A strong understanding of the formulas along with regular numerical practice will make this chapter much easier to revise before examinations.