Chapter 8: Gravitation

Introduction

Historical Background

• Nicolaus Copernicus (1473–1543): Proposed heliocentric model with planets moving in circular orbits around the Sun.

• Galileo Galilei: Supported Copernicus; faced prosecution for his beliefs.

• Tycho Brahe (1546–1601): Made precise naked-eye planetary observations.

• Johannes Kepler (1571–1640): Derived three laws of planetary motion from Brahe's data.

• Isaac Newton: Used Kepler's laws to formulate universal law of gravitation.

Solved Examples

Ex 1

• Identify the scientist who proposed elliptical orbits: Kepler.

• State the law that relates period and orbit size: Kepler's Third Law.

Practice Set

• Q1: Who proposed that planets move in elliptical orbits?
  A. Newton
  B. Copernicus
  C. Kepler
  D. Galileo

• Q2: Which scientist compiled precise planetary data with naked eye?
  A. Kepler
  B. Brahe
  C. Newton
  D. Copernicus

• Q3: Newton used data from which scientist to propose universal gravitation?
  A. Galileo
  B. Brahe
  C. Kepler
  D. Copernicus

Answer Key

• Q1: C - Kepler formulated elliptical orbits.

• Q2: B - Tycho Brahe made detailed observations.

• Q3: C - Newton used Kepler's laws for gravitation.

Memory Aids

• Copernicus: Circle model

• Kepler: Elliptical orbits + 3 laws

• Brahe: Data collector

• Newton: Gravity from Kepler's laws

Kepler's Laws

Law of Orbits

• Planets move in elliptical orbits.

• Sun is at one focus of ellipse.

Law of Areas

• Line joining planet and Sun sweeps equal areas in equal times.

• Area swept in small time \( \Delta t \): \( \Delta A = \frac{1}{2} r v \Delta t \)

• Vector form: \( \Delta \vec{A} = \frac{1}{2} (\vec{r} \times \vec{v} \Delta t) \)

• Rate of area swept: \( \frac{\Delta \vec{A}}{\Delta t} = \frac{1}{2} (\vec{r} \times \vec{v}) \)

• Using momentum \( \vec{p} = m \vec{v} \): \( \frac{\Delta \vec{A}}{\Delta t} = \frac{\vec{L}}{2m} \) where \( \vec{L} = \vec{r} \times \vec{p} \) is angular momentum.

• Angular momentum \( \vec{L} \) is constant for central force → area rate constant.

Law of Periods

• Square of orbital period \( T \) is proportional to cube of semi-major axis \( R \):

• \( T^2 \propto R^3 \)

Solved Examples

Ex 1

• Calculate area swept in 10 s if \( r = 1.5 \times 10^{11} \) m, \( v = 3 \times 10^4 \) m/s.
  \( \Delta A = \frac{1}{2} r v \Delta t = 0.5 \times 1.5 \times 10^{11} \times 3 \times 10^4 \times 10 = 2.25 \times 10^{16} \, m^2 \)

Ex 2

• If Earth's orbital radius is \( 1.5 \times 10^{11} \) m and period is 1 year, find period for a planet at \( 2.25 \times 10^{11} \) m.
  Using \( T^2 \propto R^3 \):
  \( \frac{T_2^2}{T_1^2} = \left( \frac{R_2}{R_1} \right)^3 = \left( \frac{2.25}{1.5} \right)^3 = (1.5)^3 = 3.375 \)
  \( T_2 = T_1 \sqrt{3.375} = 1 \times 1.837 = 1.837 \) years

Practice Set

• Q1: According to Kepler's second law, the area swept by the radius vector in equal time intervals is:
  A. Increasing
  B. Decreasing
  C. Constant
  D. Zero

• Q2: The angular momentum of a planet about the Sun is:
  A. Increasing
  B. Decreasing
  C. Constant
  D. Zero

• Q3: If the semi-major axis of a planet's orbit is doubled, its orbital period becomes:
  A. Double
  B. Four times
  C. \( 2\sqrt{2} \) times
  D. \( 2\sqrt{8} \) times

Answer Key

• Q1: C - Equal areas in equal times means constant area sweep rate.

• Q2: C - Angular momentum is conserved under central force.

• Q3: D - \( T^2 \propto R^3 \) so \( T = R^{3/2} \). Doubling R: \( T_2 = 2^{3/2} T_1 = 2\sqrt{2} T_1 \).

Memory Aids

• Law 1: Ellipse with Sun at focus

• Law 2: Equal areas in equal times → constant angular momentum

• Law 3: Period squared ∝ radius cubed

• Formula: \( \Delta A = \frac{1}{2} r v \Delta t \)

Universal Law of Gravitation

Newton's Law

• Every particle attracts every other particle.

• Force \( F \) is proportional to product of masses \( m_1, m_2 \).

• Force inversely proportional to square of distance \( r \) between them.

• Mathematically, \( F = \dfrac{G m_1 m_2}{r^2} \), where \( G \) is universal gravitational constant.

Characteristics of Gravitational Force

• Always attractive, acts along line joining particles.

• Equal and opposite forces on both masses (action-reaction pair): \( \vec{F}_1 = -\vec{F}_2 \).

• Independent of medium between particles.

• Valid for all distances: interatomic to interplanetary.

• Conservative force: work done depends only on initial and final positions.

• No work done in closed path motion.

• Vector form: \( \vec{F} = -G \dfrac{m_1 m_2}{r^2} \hat{r} = -G \dfrac{m_1 m_2}{|\vec{r}|^3} \vec{r} \).

Solved Examples

Ex 1

• Find force between two masses \( m_1 = 5\,kg \) and \( m_2 = 10\,kg \) separated by \( r = 2\,m \).

• Use \( G = 6.67 \times 10^{-11} \; N m^2/kg^2 \).

• \( F = \dfrac{6.67 \times 10^{-11} \times 5 \times 10}{2^2} = 8.34 \times 10^{-10} \; N \).

Practice Set

Q1

What is the direction of gravitational force between two masses?

• A) Repulsive along line joining masses

• B) Attractive along line joining masses

• C) Perpendicular to line joining masses

• D) Zero

Q2

Two masses \( m_1 = 3\,kg \) and \( m_2 = 6\,kg \) are 4 m apart. Calculate gravitational force. (Use \( G = 6.67 \times 10^{-11} \))

• A) \(7.5 \times 10^{-11} N\)

• B) \(3.0 \times 10^{-10} N\)

• C) \(7.5 \times 10^{-10} N\)

• D) \(1.0 \times 10^{-9} N\)

Q3

If distance between two masses is halved, gravitational force becomes?

• A) Halved

• B) Doubled

• C) Quadrupled

• D) Quartered

Answer Key

• Q1: B - Gravitational force is always attractive along the line joining the masses.

• Q2: C - \( F = \dfrac{6.67 \times 10^{-11} \times 3 \times 6}{4^2} = 7.5 \times 10^{-11} \times 3 = 7.5 \times 10^{-10} N \).

• Q3: C - Force varies inversely as square of distance, halving distance increases force by 4 times.

Memory Aids

• Force \( \propto m_1 m_2 / r^2 \).

• Always attractive, acts along line joining masses.

• Action = Reaction (equal & opposite forces).

• Halving distance quadruples force.

Gravitational Force Vector Sum

Force on Multiple Masses

• Force on a mass due to several masses is vector sum of individual forces.

• For mass \( m_1 \), total force \( \vec{F}_1 = \sum \dfrac{G m_i m_1}{r_{i1}^2} \hat{r}_{i1} \).

• Unit vector \( \hat{r}_{i1} \) points from \( m_1 \) to \( m_i \).

• Direction of force is attractive towards other masses.

Solved Examples

Ex 1

• Calculate net gravitational force on \( m_1 \) due to \( m_2, m_3, m_4 \) given their masses and distances.

• Calculate each force using \( F = \dfrac{G m_i m_1}{r_{i1}^2} \) and add vectorially.

Practice Set

Q1

Force on \( m_1 \) due to \( m_2 \) is \( F_{12} \). What is force on \( m_2 \) due to \( m_1 \)?

• A) \( F_{12} \)

• B) \( -F_{12} \)

• C) Zero

• D) Depends on distance

Q2

Net force on \( m_1 \) due to \( m_2 \) and \( m_3 \) is vector sum of individual forces. This is an example of?

• A) Scalar addition

• B) Vector addition

• C) No addition

• D) None

Q3

If \( m_2 = m_3 = m \), distances \( r_{21} = r_{31} = r \), and forces \( \vec{F}_{12} \) and \( \vec{F}_{13} \) are perpendicular, magnitude of net force on \( m_1 \) is?

• A) \( 2 \dfrac{G m m_1}{r^2} \)

• B) \( \sqrt{2} \dfrac{G m m_1}{r^2} \)

• C) \( \dfrac{G m m_1}{r^2} \)

• D) Zero

Answer Key

• Q1: B - Forces are equal in magnitude and opposite in direction (Newton's third law).

• Q2: B - Gravitational forces add vectorially.

• Q3: B - Net force is vector sum: \( \sqrt{F^2 + F^2} = \sqrt{2} F \).

Memory Aids

• Net force = vector sum of individual gravitational forces.

• Action-reaction forces are equal and opposite.

• Use Pythagoras for perpendicular force vectors.

Gravitational Constant

Definition and Discovery

• Gravitational constant \( G \) quantifies the strength of gravity between two masses.

• First measured by Henry Cavendish in 1798 using a torsion balance experiment.

Cavendish Experiment Setup

• Two small masses \( m \) fixed at ends A and B of a light rod, rotating about a vertical axis at the center.

• Two large spheres \( S_1 \) and \( S_2 \) placed near the small masses exert gravitational force.

• When large spheres are moved to opposite sides (positions \( S'_1 \) and \( S'_2 \)), the rod rotates due to torque change.

• Angle of rotation \( \theta \) is measured to calculate \( G \).

Equilibrium Condition and Formula

• At equilibrium, gravitational torque equals restoring torque from the wire.

• Torque due to gravity: \( \tau_g = G \frac{Mm}{r^2} L \)

• Restoring torque: \( \tau_r = k \theta \), where \( k \) is torsion constant, \( \theta \) is angle of twist.

• Equating: \( G \frac{Mm}{r^2} L = k \theta \)

• Rearranged formula for \( G \): \( G = \frac{k \theta r^2}{M m L} \)

• \( r \): distance between centers of large and small spheres; \( L \): length of the rod.

Accepted Value

• Current accepted value: \( G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)

Solved Examples

Ex 1

• Given: \( k = 2 \times 10^{-6} \text{ Nm/rad} \), \( \theta = 0.01 \text{ rad} \), \( r = 0.05 \text{ m} \), \( M = 1.0 \text{ kg} \), \( m = 0.01 \text{ kg} \), \( L = 0.4 \text{ m} \).

• Calculate \( G \): \( G = \frac{2 \times 10^{-6} \times 0.01 \times (0.05)^2}{1.0 \times 0.01 \times 0.4} = 1.25 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \).

Practice Set

Q1

In Cavendish's experiment, the torque due to gravitational force is balanced by:

• A) Magnetic torque

• B) Restoring torque of the wire

• C) Frictional torque

• D) Centrifugal torque

Q2

Which of the following is the correct formula for \( G \) in Cavendish's experiment?

• A) \( G = \frac{k \theta r^2}{M m L} \)

• B) \( G = \frac{M m L}{k \theta r^2} \)

• C) \( G = \frac{k \theta M m}{r^2 L} \)

• D) \( G = \frac{k r^2}{\theta M m L} \)

Q3

In an experiment, \( k = 3 \times 10^{-6} \text{ Nm/rad} \), \( \theta = 0.02 \text{ rad} \), \( r = 0.04 \text{ m} \), \( M = 2 \text{ kg} \), \( m = 0.02 \text{ kg} \), \( L = 0.5 \text{ m} \). Calculate \( G \).

• A) \( 4.8 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)

• B) \( 1.92 \times 10^{-10} \text{ N m}^2/\text{kg}^2 \)

• C) \( 9.6 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)

• D) \( 3.84 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)

Answer Key

• Q1: B) Restoring torque of the wire — It balances gravitational torque at equilibrium.

• Q2: A) \( G = \frac{k \theta r^2}{M m L} \) — Correct rearranged formula from torque balance.

• Q3: D) \( 3.84 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \) — Calculated using given values in formula.

Memory Aids

• Remember: \( G \) relates torque balance: gravitational torque = restoring torque.

• Formula cue: \( G = \frac{k \theta r^2}{M m L} \) — torsion constant × angle × distance squared over masses and rod length.

• Value of \( G \) is very small (~10⁻¹¹), indicating weak gravitational force.

Acceleration due to Gravity

Concepts

• Acceleration due to gravity (\( g \)) is the acceleration produced in a body falling freely under Earth's gravity.

• At Earth's surface, \( g = 9.8\, \mathrm{m/s^2} \).

• Gravitational force between Earth (mass \( M_E \)) and a body (mass \( m \)) at distance \( R_E \) (Earth's radius) is \( F = \frac{G M_E m}{R_E^2} \).

• Weight of the body: \( W = mg \).

• Equating weight and gravitational force: \( mg = \frac{G M_E m}{R_E^2} \Rightarrow g = \frac{G M_E}{R_E^2} \).

• Numerical value: \( g = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6400 \times 10^3)^2} = 9.8\, \mathrm{m/s^2} \).

Solved Examples

• Ex 1: Calculate \( g \) using \( G = 6.67 \times 10^{-11} \mathrm{Nm^2/kg^2} \), \( M_E = 6 \times 10^{24} \mathrm{kg} \), \( R_E = 6400 \times 10^3 \mathrm{m} \).

  • Use formula: \( g = \frac{G M_E}{R_E^2} \).

  • Substitute values and compute \( g = 9.8\, \mathrm{m/s^2} \).

Practice Set

• Q1: What is the acceleration due to gravity on Earth's surface?

  • A) 9.8 m/s²

  • B) 10 m/s²

  • C) 9.2 m/s²

  • D) 8.9 m/s²

• Q2: If Earth's mass doubles and radius remains the same, what happens to \( g \)?

  • A) Doubles

  • B) Halves

  • C) Remains same

  • D) Quadruples

• Q3: Calculate \( g \) on a planet with mass \( 3 \times 10^{24} \mathrm{kg} \) and radius \( 4 \times 10^6 \mathrm{m} \).

  • A) 12.5 m/s²

  • B) 10.5 m/s²

  • C) 8.3 m/s²

  • D) 15.0 m/s²

Answer Key

• Q1: A) 9.8 m/s² — Standard acceleration due to gravity on Earth.

• Q2: A) Doubles — \( g \) is directly proportional to mass \( M_E \).

• Q3: C) 8.3 m/s² — Using \( g = \frac{GM}{R^2} \), substitute values to calculate.

Memory Aids

• Formula: \( g = \frac{GM_E}{R_E^2} \).

• Remember: \( g \) depends on Earth's mass and radius.

• Standard \( g = 9.8\, \mathrm{m/s^2} \) at surface.

Relationship between g and G

Concepts

• Earth's mass: \( M_E = \frac{4}{3} \pi R_E^3 \rho \), where \( \rho \) is Earth's density.

• Substitute in \( g = \frac{G M_E}{R_E^2} \):

  • \( g = G \frac{4}{3} \pi R_E^3 \rho / R_E^2 = \frac{4}{3} \pi G R_E \rho \).

• Shows \( g \) depends on gravitational constant \( G \), Earth's radius \( R_E \), and density \( \rho \).

Solved Examples

• Ex 1: Calculate \( g \) if \( R_E = 6.4 \times 10^6 \mathrm{m} \), \( \rho = 5500 \mathrm{kg/m^3} \), \( G = 6.67 \times 10^{-11} \mathrm{Nm^2/kg^2} \).

  • Use formula: \( g = \frac{4}{3} \pi G R_E \rho \).

  • Calculate \( g \approx 9.8\, \mathrm{m/s^2} \).

Practice Set

• Q1: Which factor does NOT affect \( g \)?

  • A) Earth's radius

  • B) Earth's density

  • C) Gravitational constant \( G \)

  • D) Mass of the falling object

• Q2: If Earth's density doubles, what happens to \( g \)?

  • A) Doubles

  • B) Halves

  • C) Remains same

  • D) Quadruples

• Q3: Express \( g \) in terms of \( G, R_E, \) and \( \rho \).

  • A) \( g = G R_E \rho \)

  • B) \( g = \frac{4}{3} \pi G R_E \rho \)

  • C) \( g = \frac{G \rho}{R_E} \)

  • D) \( g = \frac{4}{3} \pi G \rho / R_E \)

Answer Key

• Q1: D) Mass of the falling object — \( g \) is independent of falling object's mass.

• Q2: A) Doubles — \( g \) is directly proportional to density \( \rho \).

• Q3: B) \( g = \frac{4}{3} \pi G R_E \rho \) — Derived formula.

Memory Aids

• Formula: \( g = \frac{4}{3} \pi G R_E \rho \).

• Think: \( g \) depends on Earth's size and density.

• Density increase → \( g \) increase.

Gravitational Field

Concepts

• Gravitational field: Space around a mass where its gravitational pull is felt.

• Field intensity (\( E \)) at a point: Force per unit mass placed at that point.

• Formula: \( E = - \frac{GM}{r^2} \), where \( M \) is mass, \( r \) is distance from center.

• Negative sign indicates attractive nature of gravity.

Solved Examples

• Ex 1: Calculate gravitational field intensity \( E \) at \( 10^7 \mathrm{m} \) from mass \( 5 \times 10^{24} \mathrm{kg} \).

  • Use \( E = - \frac{GM}{r^2} \).

  • Calculate magnitude \( E = \frac{6.67 \times 10^{-11} \times 5 \times 10^{24}}{(10^7)^2} = 33.35\, \mathrm{N/kg} \).

Practice Set

• Q1: Gravitational field intensity is defined as?

  • A) Force on unit mass

  • B) Force on unit charge

  • C) Energy per unit mass

  • D) Mass per unit volume

• Q2: Gravitational field intensity at distance \( r \) from mass \( M \) is?

  • A) \( \frac{GM}{r} \)

  • B) \( \frac{GM}{r^2} \)

  • C) \( -\frac{GM}{r^2} \)

  • D) \( -\frac{GM}{r} \)

• Q3: What does the negative sign in \( E = -\frac{GM}{r^2} \) indicate?

  • A) Repulsive force

  • B) Attractive force

  • C) Zero force

  • D) Variable force

Answer Key

• Q1: A) Force on unit mass — Definition of gravitational field intensity.

• Q2: C) \( -\frac{GM}{r^2} \) — Correct formula with negative sign.

• Q3: B) Attractive force — Negative sign shows attraction.

Memory Aids

• Formula: \( E = -\frac{GM}{r^2} \).

• Negative sign = gravity pulls inward.

• Field intensity = force per unit mass.

Acceleration due to Gravity Below and Above Earth Surface

Effect of Altitude

• Force on mass \( m \) at height \( h \): \( F(h) = \frac{GM_Em}{(R_E + h)^2} \)

• Acceleration due to gravity at height \( h \): \( g(h) = \frac{F(h)}{m} = \frac{GM_E}{(R_E + h)^2} = g \left(1 + \frac{h}{R_E}\right)^{-2} \)

• For small \( h \) (\( h \ll R_E \)), using binomial approximation: \( g(h) \approx g \left(1 - \frac{2h}{R_E}\right) \)

• Gravity decreases with altitude by factor \( (1 - 2h/R_E) \)

Effect of Depth

• Mass inside radius \( R_E - d \): \( M_s = M_E \left(\frac{R_E - d}{R_E}\right)^3 \)

• Force on mass \( m \) at depth \( d \): \( F(d) = \frac{GM_sm}{(R_E - d)^2} = \frac{GM_Em(R_E - d)}{R_E^3} \)

• Acceleration due to gravity at depth \( d \): \( g(d) = \frac{F(d)}{m} = g \left(1 - \frac{d}{R_E}\right) \)

• Gravity decreases linearly with depth by factor \( (1 - d/R_E) \)

Solved Examples

• Ex 1: Calculate gravity at 10 km above Earth surface. Given \( R_E = 6400 \text{ km} \), \( g = 9.8 \text{ m/s}^2 \).

  • Use \( g(h) \approx g (1 - 2h/R_E) \)

  • \( g(10) = 9.8 \times (1 - 2 \times 10 / 6400) = 9.8 \times 0.996875 = 9.76 \text{ m/s}^2 \)

• Ex 2: Find gravity at 5 km depth.

  • Use \( g(d) = g (1 - d/R_E) \)

  • \( g(5) = 9.8 \times (1 - 5/6400) = 9.8 \times 0.99922 = 9.79 \text{ m/s}^2 \)

Practice Set

• Q1: Gravity at 20 km above Earth surface is approximately:

  • A) 9.5 m/s²

  • B) 9.7 m/s²

  • C) 9.8 m/s²

  • D) 10 m/s²

• Q2: At 3 km depth, gravity is:

  • A) Increased by factor \( 1 + 3/R_E \)

  • B) Decreased by factor \( 1 - 3/R_E \)

  • C) Same as surface gravity

  • D) Zero

• Q3: If gravity at surface is 9.8 m/s², gravity at 100 km altitude is approximately:

  • A) 9.5 m/s²

  • B) 9.1 m/s²

  • C) 8.8 m/s²

  • D) 9.0 m/s²

Answer Key

• Q1: B - Gravity decreases slightly with altitude by \( 1 - 2h/R_E \), so 20 km gives about 9.7 m/s².

• Q2: B - Gravity decreases linearly with depth by \( 1 - d/R_E \).

• Q3: A - Using \( g(h) \approx g (1 - 2h/R_E) \), 100 km altitude gives about 9.5 m/s².

Memory Aids

• Gravity decreases with altitude as \( (1 - 2h/R_E) \)

• Gravity decreases with depth as \( (1 - d/R_E) \)

• Use binomial approximation for small heights

Variation in Acceleration due to Gravity

Factors Affecting Gravity

• Gravity decreases with height above Earth surface

• Gravity decreases with depth below surface, zero at center

• Earth's rotation reduces gravity at equator due to centrifugal force

• Gravity varies with latitude \( \lambda \): \( g_\lambda = g - R_E \omega^2 \cos^2 \lambda \)

• Minimum gravity at equator (\( \lambda=0^\circ \)), maximum at poles (\( \lambda=90^\circ \))

• Earth's shape (equatorial bulge) also reduces gravity at equator

Solved Examples

• Ex 1: Calculate difference in gravity between pole and equator.

  • \( g_p - g_e = R_E \omega^2 \)

  • Given \( R_E = 6.4 \times 10^6 \text{ m} \), \( \omega = 7.27 \times 10^{-5} \text{ rad/s} \)

  • \( g_p - g_e = 6.4 \times 10^6 \times (7.27 \times 10^{-5})^2 = 0.034 \text{ m/s}^2 \)

Practice Set

• Q1: Gravity at equator is:

  • A) Maximum

  • B) Minimum

  • C) Same as poles

  • D) Zero

• Q2: Gravity at latitude 45° is:

  • A) \( g - R_E \omega^2 \cos^2 45^\circ \)

  • B) \( g + R_E \omega^2 \cos^2 45^\circ \)

  • C) \( g \)

  • D) Zero

• Q3: Earth's equatorial radius is larger than polar radius by about:

  • A) 21 km

  • B) 100 km

  • C) 5 km

  • D) 50 km

Answer Key

• Q1: B - Gravity is minimum at equator due to rotation and bulge.

• Q2: A - Gravity reduces by \( R_E \omega^2 \cos^2 \lambda \) at latitude \( \lambda \).

• Q3: A - Earth's equatorial radius is about 21 km larger than polar radius.

Memory Aids

• Gravity minimum at equator, maximum at poles

• Gravity variation: \( g_\lambda = g - R_E \omega^2 \cos^2 \lambda \)

• Earth's bulge reduces gravity at equator

Gravitational Field and Potential of Continuous Mass Distributions

Key Relations

• Gravitational potential \( V \): work done per unit mass from infinity to point

• Relation: \( E = -\frac{dV}{dr} \)

• \( dV = - \mathbf{E} \cdot d\mathbf{r} \)

Uniform Ring (Mass \( M \), Radius \( R \))

• Field on axis at distance \( x \): \( E = - \frac{GMx}{(R^2 + x^2)^{3/2}} \)

• Potential on axis: \( V = - \frac{GM}{\sqrt{R^2 + x^2}} \)

Uniform Disc (Mass \( M \), Radius \( R \))

• Field on axis: \( E = - \frac{2GM}{R^2} \left[1 - \frac{x}{\sqrt{R^2 + x^2}}\right] \)

• Potential on axis: \( V = - \frac{2GM}{R^2} \left[x - \sqrt{R^2 + x^2}\right] \)

Thin Spherical Shell (Mass \( M \), Radius \( R \))

• Inside shell (\( r < R \)): \( E = 0 \), \( V = -\frac{GM}{R} \)

• On surface (\( r = R \)): \( E = -\frac{GM}{R^2} \), \( V = -\frac{GM}{R} \)

• Outside shell (\( r > R \)): \( E = -\frac{GM}{r^2} \), \( V = -\frac{GM}{r} \)

Solid Sphere (Mass \( M \), Radius \( R \), Uniform Density)

• Inside (\( r < R \)): \( E = -\frac{GMr}{R^3} \), \( V = -\frac{GM}{2R^3}(3R^2 - r^2) \)

• On surface (\( r = R \)): \( E = -\frac{GM}{R^2} \), \( V = -\frac{GM}{R} \)

• Outside (\( r > R \)): \( E = -\frac{GM}{r^2} \), \( V = -\frac{GM}{r} \)

• At center (\( r=0 \)): \( V = -\frac{3GM}{2R} \)

Solved Examples

• Ex 1: Find gravitational field at \( r = R/2 \) inside solid sphere.

  • \( E = -\frac{GM (R/2)}{R^3} = -\frac{GM}{2R^2} \)

• Ex 2: Potential at center of solid sphere.

  • \( V = -\frac{3GM}{2R} \)

Practice Set

• Q1: Gravitational field inside a thin spherical shell is:

  • A) \( -\frac{GM}{r^2} \)

  • B) Zero

  • C) \( -\frac{GM}{R^2} \)

  • D) \( -\frac{GM}{R^3} r \)

• Q2: Gravitational potential outside a solid sphere at distance \( r > R \) is:

  • A) \( -\frac{GM}{r} \)

  • B) \( -\frac{GM}{R} \)

  • C) Zero

  • D) \( -\frac{GM}{2R^3}(3R^2 - r^2) \)

• Q3: Gravitational field at surface of solid sphere is:

  • A) \( -\frac{GM}{R^2} \)

  • B) Zero

  • C) \( -\frac{GM}{R^3} r \)

  • D) \( -\frac{GM}{r^2} \)

Answer Key

• Q1: B - Field inside thin shell is zero.

• Q2: A - Potential outside sphere is \( -GM/r \).

• Q3: A - Field at surface is \( -GM/R^2 \).

Memory Aids

• Inside shell: zero field, constant potential

• Outside mass: field \( \propto 1/r^2 \), potential \( \propto 1/r \)

• Inside solid sphere: field \( \propto r \), potential quadratic in \( r \)

Gravitational Potential Energy

Definition and Formula

• Work done to bring a body from infinity to a point in a gravitational field.

• Between two masses M and m separated by distance r: U = -\(\frac{GMm}{r}\)

• Potential energy zero at infinite separation (r → ∞).

• Scalar quantity; Dimensional formula: [ML2T-2]; SI unit: Joule (J).

Near Earth Surface

• At height h above Earth surface: Uh = -\(\frac{GM_{e}m}{R_{e} + h}\)

• At Earth surface: Us = -\(\frac{GM_{e}m}{R_{e}}\)

• Change in potential energy moving up height h:
  \(\Delta U = U_{h} - U_{s} = GM_{e}m \left( \frac{1}{R_{e}} - \frac{1}{R_{e} + h} \right) = \frac{GM_{e}mh}{R_{e}^2 (1 + \frac{h}{R_{e}})} = \frac{mgh}{1 + \frac{h}{R_{e}}}\)

• Since \(\frac{GM_{e}}{R_{e}^2} = g\), for h << Re, \(\Delta U \approx mgh\)

Solved Examples

• Ex 1: Calculate gravitational potential energy of 2 kg mass at 10 m above Earth surface.
  Given: g = 9.8 m/s², h = 10 m
  Solution: \(\Delta U = mgh = 2 \times 9.8 \times 10 = 196\) J

• Ex 2: Find change in potential energy when 5 kg mass is moved from surface to 1000 m height.
  Given: g = 9.8 m/s², h = 1000 m, Re = 6.4 × 106 m
  Solution: \(\Delta U = \frac{mgh}{1 + \frac{h}{R_{e}}} = \frac{5 \times 9.8 \times 1000}{1 + \frac{1000}{6.4 \times 10^{6}}} \approx 49000\) J

Practice Set

• Q1: Gravitational potential energy between two masses M and m separated by distance r is:
  A) \(\frac{GMm}{r}\) B) \(-\frac{GMm}{r}\) C) \(GMmr\) D) Zero

• Q2: The change in gravitational potential energy when a 3 kg mass is raised 20 m near Earth surface (g = 9.8 m/s²) is:
  A) 588 J B) -588 J C) 60 J D) 0 J

• Q3: For a height h = 1000 m, Earth radius Re = 6.4 × 106 m, the exact change in potential energy for mass m is:
  A) \(mgh\) B) \(\frac{mgh}{1 + \frac{h}{R_{e}}}\) C) \(mgh (1 + \frac{h}{R_{e}})\) D) Zero

Answer Key

• Q1: B; Gravitational potential energy is negative and given by \(-\frac{GMm}{r}\).

• Q2: A; \(\Delta U = mgh = 3 \times 9.8 \times 20 = 588\) J (positive when raised).

• Q3: B; Exact formula includes correction factor \(1 + \frac{h}{R_{e}}\) in denominator.

Memory Aids

• Gravitational potential energy is always negative; zero at infinity.

• Near Earth, \(\Delta U \approx mgh\) for small heights.

• Use correction factor \(1 + \frac{h}{R_{e}}\) for large heights.

Escape Speed

Concept and Formula

• Minimum speed to escape a planet's gravity without returning.

• Escape speed from surface: \( v_e = \sqrt{\frac{2GM}{R}} \)

• Where, \( M \) = mass of planet, \( R \) = radius of planet, \( G \) = gravitational constant.

• Using density \( \rho \) and volume: \( v_e = \sqrt{\frac{8\pi G \rho R^2}{3}} \)

• For Earth, \( v_e = 11.2\ \text{km/s} \)

• If projection speed \( v < v_e \), body returns or orbits.

• If \( v > v_e \), body escapes with speed \( v' = \sqrt{v^2 - v_e^2} \)

Solved Examples

Ex 1: Calculate escape speed from a planet of mass \( 5 \times 10^{24} \) kg and radius \( 6 \times 10^6 \) m.

• Use \( v_e = \sqrt{\frac{2GM}{R}} \)

• \( G = 6.67 \times 10^{-11} \) N·m²/kg²

• \( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5 \times 10^{24}}{6 \times 10^6}} \approx 11.8 \times 10^3 \) m/s

• \( v_e = 11.8 \) km/s

Practice Set

Q1: Escape speed depends on

• A) Mass only

• B) Radius only

• C) Both mass and radius

• D) Neither mass nor radius

Q2: If escape speed from a planet is 10 km/s, what is the speed of a body projected at 15 km/s after escaping?

• A) 5 km/s

• B) 11.18 km/s

• C) 18.3 km/s

• D) 25 km/s

Q3: A planet has twice the radius and same density as Earth. Escape speed from this planet is (Earth's escape speed = 11.2 km/s):

• A) 22.4 km/s

• B) 15.8 km/s

• C) 11.2 km/s

• D) 7.9 km/s

Answer Key

Q1: C) Both mass and radius — Escape speed formula depends on \( M \) and \( R \).

Q2: B) 11.18 km/s — \( v' = \sqrt{15^2 - 10^2} = \sqrt{225 - 100} = \sqrt{125} = 11.18 \) km/s.

Q3: B) 15.8 km/s — \( v_e \propto R \) for constant density, so \( v_e = 11.2 \times \sqrt{2} = 15.8 \) km/s.

Memory Aids

• Escape speed \( v_e = \sqrt{2GM/R} \) — remember "2GM over R".

• For constant density, \( v_e \propto R \).

• Speed after escape: \( v' = \sqrt{v^2 - v_e^2} \).

Earth Satellites

Definition and Examples

• A satellite is a lighter body revolving continuously in an orbit around a much larger body.

• Example: Earth is a satellite of the Sun; Moon is a satellite of Earth.

Solved Examples

• Ex 1: Identify the satellite in the Earth-Sun system.
  Answer: Earth is the satellite revolving around the Sun.

Practice Set

• Q1: Which of the following is a natural satellite?
  A) Earth
  B) Moon
  C) Sun
  D) Mars

• Q2: The Moon revolves around which body?
  A) Sun
  B) Earth
  C) Mars
  D) Jupiter

• Q3: Which of the following is NOT a satellite?
  A) Earth
  B) Moon
  C) Artificial satellite
  D) Mars

Answer Key

• Q1: B - Moon is Earth's natural satellite.

• Q2: B - Moon revolves around Earth.

• Q3: D - Mars is a planet, not a satellite.

Memory Aids

• Satellite = smaller body orbiting a larger one.

• Moon → Earth → Sun (satellite chain).

Motion of a Satellite

Key Points

• Satellites move in circular orbits around Earth.

• Velocity direction is tangent to the orbit.

• Acceleration (centripetal) points towards the center of the orbit.

Solved Examples

• Ex 1: Direction of acceleration of a satellite?
  Answer: Towards the center of Earth (centripetal acceleration).

Practice Set

• Q1: The velocity of a satellite is always:
  A) Radial
  B) Tangential
  C) Zero
  D) Opposite to acceleration

• Q2: The acceleration of a satellite is directed:
  A) Tangent to orbit
  B) Away from Earth
  C) Towards Earth center
  D) Perpendicular to velocity

• Q3: Which force provides centripetal acceleration to a satellite?
  A) Gravitational force
  B) Frictional force
  C) Magnetic force
  D) Electrostatic force

Answer Key

• Q1: B - Velocity is tangent to the orbit.

• Q2: C - Acceleration is centripetal, towards Earth center.

• Q3: A - Gravity provides centripetal force.

Memory Aids

• Velocity tangent, acceleration radial inward.

• Gravity = centripetal force for satellites.

Orbital Speed of a Satellite

Formulas and Facts

• Orbital speed at height h: \( v_0 = \sqrt{\frac{GM_E}{R_E + h}} = R_E \sqrt{\frac{g}{R_E + h}} \)

• Near Earth's surface (\( h \ll R_E \)): \( v_0 = \sqrt{g R_E} \)

• Orbital speed is independent of satellite's mass.

Solved Examples

• Ex 1: Calculate orbital speed near Earth's surface.
  Given: \( g = 9.8\,m/s^2 \), \( R_E = 6.4 \times 10^6 m \)
  \( v_0 = \sqrt{9.8 \times 6.4 \times 10^6} = 7.9 \times 10^3 m/s \)

Practice Set

• Q1: Orbital speed depends on:
  A) Mass of satellite
  B) Height of orbit
  C) Both A and B
  D) None

• Q2: Orbital speed near Earth's surface is approximately:
  A) 3 km/s
  B) 8 km/s
  C) 15 km/s
  D) 30 km/s

• Q3: Orbital speed at height \( h = R_E \) is:
  A) \( \sqrt{\frac{g R_E}{2}} \)
  B) \( \sqrt{\frac{g R_E}{3}} \)
  C) \( \sqrt{\frac{g R_E}{4}} \)
  D) \( \sqrt{\frac{g R_E}{1}} \)

Answer Key

• Q1: B - Depends on height, not mass.

• Q2: B - Approximately 8 km/s near surface.

• Q3: A - At \( h = R_E \), speed is \( \sqrt{\frac{g R_E}{2}} \).

Memory Aids

• \( v_0 = \sqrt{\frac{GM_E}{R_E + h}} \)

• Near surface: \( v_0 = \sqrt{g R_E} \)

• Speed independent of satellite mass.

Time Period of a Satellite

Formulas and Notes

• Time period: \( T = 2\pi \sqrt{\frac{(R_E + h)^3}{GM_E}} = \frac{2\pi}{\sqrt{g}} \sqrt{\frac{(R_E + h)^3}{R_E^2}} \)

• Near Earth's surface (\( h \ll R_E \)): \( T = 2\pi \sqrt{\frac{R_E}{g}} = 84.6 \text{ minutes} \)

Solved Examples

• Ex 1: Calculate time period near Earth's surface.
  Given: \( R_E = 6.4 \times 10^6 m \), \( g = 9.8 m/s^2 \)
  \( T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}} = 5076 s = 84.6 min \)

Practice Set

• Q1: Time period depends on:
  A) Mass of satellite
  B) Height of orbit
  C) Both A and B
  D) None

• Q2: Time period near Earth's surface is about:
  A) 45 min
  B) 60 min
  C) 85 min
  D) 120 min

• Q3: If height doubles, time period:
  A) Increases by \( \sqrt{8} \)
  B) Doubles
  C) Triples
  D) Remains same

Answer Key

• Q1: B - Depends on orbit height, not mass.

• Q2: C - Approximately 85 minutes.

• Q3: A - Time period \( \propto (R_E + h)^{3/2} \), so doubling height increases \( T \) by \( \sqrt{8} \).

Memory Aids

• \( T = 2\pi \sqrt{\frac{(R_E + h)^3}{GM_E}} \)

• Near surface: \( T \approx 84.6 \text{ min} \)

Height of Satellite

Formula

• Height above Earth surface:
  \( h = \left( \frac{T^2 R_E^2 g}{4 \pi^2} \right)^{1/3} - R_E \)

Solved Examples

• Ex 1: Find height for \( T = 2 \text{ hours} = 7200 s \).
  Use formula to calculate \( h \).

Practice Set

• Q1: Height depends on:
  A) Time period
  B) Mass of satellite
  C) Both A and B
  D) None

• Q2: For \( T = 84.6 \text{ min} \), height is:
  A) 0 km
  B) 100 km
  C) 500 km
  D) 1000 km

• Q3: If time period increases, height:
  A) Increases
  B) Decreases
  C) Remains same
  D) Zero

Answer Key

• Q1: A - Height depends on time period.

• Q2: A - At 84.6 min, satellite is near surface, so height ~ 0 km.

• Q3: A - Height increases with time period.

Memory Aids

• Height from \( T \): \( h = \left( \frac{T^2 R_E^2 g}{4 \pi^2} \right)^{1/3} - R_E \)

• Longer period → higher orbit.

Energy of an Orbiting Satellite

Potential Energy

• Potential energy (U) of satellite at height h: U = -\(\frac{GM_Em}{R_E + h}\)

• G = universal gravitational constant (6.67 × 10-11 N·m²/kg²)

• ME = mass of Earth, m = mass of satellite

• RE = radius of Earth, h = height above Earth surface (m)

Kinetic Energy

• Kinetic energy (K) of satellite: K = \(\frac{1}{2}mv_0^2 = \frac{1}{2} \frac{GM_Em}{R_E + h} = \frac{|U|}{2}\)

• v0 = orbital speed of satellite

Total Mechanical Energy

• Total energy (E) = K + U = −\(\frac{GM_Em}{2(R_E + h)}\) = \(\frac{U}{2}\)

• Energy is negative, indicating a bound system

Near Earth Surface Orbit

• For h << RE, total energy: E = −\(\frac{GM_Em}{2R_E}\)

Energy Relations

• Kinetic energy equals negative total energy: K = −E

• Potential energy equals twice total energy: U = 2E

Binding Energy

• Energy required to remove satellite to infinity: E_B = −E = \(\frac{GM_Em}{2(R_E + h)}\)

Elliptic Orbit

• Kinetic and potential energies vary along orbit

• Total energy remains constant and negative

Solved Examples

• Ex 1: Calculate total energy of satellite of mass 1000 kg orbiting at 300 km above Earth surface (RE = 6.4 × 106 m, ME = 6 × 1024 kg, G = 6.67 × 10-11 N·m²/kg²).

  • Use formula: E = −GMm/2(RE + h)

  • Calculate denominator: 6.4 × 106 + 3 × 105 = 6.7 × 106 m

  • E = −(6.67×10-11)(6×1024)(1000) / (2 × 6.7×106)

  • E ≈ −2.99 × 1010 J

• Ex 2: Find kinetic energy of satellite in above example.

  • K = −E = 2.99 × 1010 J

Practice Set

• Q1: Potential energy of satellite at height h is given by:

  • A) \(\frac{GM_Em}{R_E + h}\)

  • B) −\(\frac{GM_Em}{R_E + h}\)

  • C) \(\frac{1}{2} \frac{GM_Em}{R_E + h}\)

  • D) −\(\frac{1}{2} \frac{GM_Em}{R_E + h}\)

• Q2: Total mechanical energy of satellite is:

  • A) Equal to kinetic energy

  • B) Equal to potential energy

  • C) Half of potential energy

  • D) Twice kinetic energy

• Q3: A satellite of mass 500 kg orbits Earth at 200 km height. Calculate binding energy. (Use RE = 6.4 × 106 m, ME = 6 × 1024 kg, G = 6.67 × 10-11 N·m²/kg²)

  • A) 1.5 × 1010 J

  • B) 3.0 × 1010 J

  • C) 7.5 × 109 J

  • D) 6.0 × 1010 J

Answer Key

• Q1: B

  • Potential energy is negative due to gravitational attraction: U = −GMm/(R + h)

• Q2: C

  • Total energy E = U/2, i.e., half of potential energy

• Q3: A

  • Binding energy E_B = GMm/2(R + h) = (6.67×10-11)(6×1024)(500) / (2 × 6.6×106) ≈ 1.5 × 1010 J

Memory Aids

• Potential energy is negative: U = −GMm/(R + h)

• Kinetic energy is half magnitude of |U|: K = |U|/2

• Total energy is half of potential: E = U/2

• Binding energy equals negative total energy