Class 11 Physics Chapter 7: Gravitation — Kepler's Laws & Satellites

01 / Concept Mastery

Comprehensive NCERT Theory

Complete derivations, rigorous physics definitions, historical context (Ptolemy to Kepler and Newton), Cavendish torsion balance, and stepwise worked examples 7.1 to 7.7.

10 SECTIONS

Section Directory

7.1 Introduction & History 7.2 Kepler's Laws 7.3 Universal Law of Gravitation 7.4 The Gravitational Constant ($G$) 7.5 Acceleration due to Gravity ($g$) 7.6 Variation of $g$ (Height & Depth) 7.7 Gravitational Potential Energy 7.8 Escape Speed ($v_e$) 7.9 Earth Satellites & Orbits 7.10 Energy of Orbiting Satellite

7.1

NCERT Section

Introduction & Historical Perspective

Historical Overview

7.1.1 From Geocentric to Heliocentric Model

From ancient times, humans observed the celestial motions of the Sun, Moon, and planets against the background of fixed stars:

Geocentric Model (Ptolemy, ~2000 years ago): Proposed that all celestial bodies move in circles around a motionless Earth located at the centre of the universe. Elaborate epicycles were introduced to account for planetary retrograde motion.
Heliocentric Model (Aryabhata 5th century CE, Nicolas Copernicus 1543): Proposed a definitive model in which all planets (including the Earth) revolve in circular orbits around a fixed central Sun. Galileo supported this view with telescopic observations of Jupiter's moons and phases of Venus.
Tycho Brahe (1546–1601): Recorded comprehensive, highly precise planetary positions over his lifetime using naked-eye sighting instruments.
Johannes Kepler (1571–1630): Analyzed Brahe's extensive Mars data and formulated three foundational empirical laws that eliminated circular epicycles.

7.2

NCERT Section

Kepler's Laws of Planetary Motion

Empirical Laws to Physical Principles

7.2.1 The Three Laws of Kepler

1. Kepler's First Law (Law of Orbits) All planets move in elliptical orbits with the Sun situated at one of the two foci of the ellipse. Major axis length $= 2a$, Semi-major axis $= a$. Minor axis length $= 2b$, Semi-minor axis $= b$. Perihelion distance (closest to Sun) $r_p = a(1 - e)$, Aphelion distance (farthest from Sun) $r_a = a(1 + e)$.

Figure 7.1: Anatomy of Keplerian Ellipse. Sun is at focus S. Semimajor axis is $a = OP = OA$. Closest point is Perihelion $P$, farthest is Aphelion $A$.

2. Kepler's Second Law (Law of Areas) The line that joins any planet to the Sun sweeps out equal areas in equal intervals of time. That is, areal velocity is constant: $$\frac{d\mathbf{A}}{dt} = \frac{1}{2}(\mathbf{r} \times \mathbf{v}) = \frac{\mathbf{L}}{2m} = \text{constant}$$ Since gravitational force $\mathbf{F} = -\frac{GMm}{r^2}\hat{\mathbf{r}}$ is purely central, Torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = 0 \implies \mathbf{L} = \text{constant}$. Consequently, at perihelion and aphelion: $r_p v_p = r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p} > 1$. (Planets move fastest at perihelion).

Figure 7.2: Kepler's 2nd Law. Both shaded sectors have equal area $\Delta A_1 = \Delta A_2$ when swept in identical time intervals $\Delta t_1 = \Delta t_2$.

3. Kepler's Third Law (Law of Periods) The square of the period of revolution ($T$) of a planet is directly proportional to the cube of the semi-major axis ($a$) of its orbit: $$T^2 = k\, a^3 \qquad \text{where } k = \frac{4\pi^2}{G M_S}$$ $$\frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}$$

Interactive 3D Keplerian Orbit & Areal Velocity Simulator

Three.js 3D Lab

Drag to orbit in 3D

Current Distance ($r$)

147.1 × 10⁶ km

Orbital Speed ($v$)

30.3 km/s

Areal Velocity ($dA/dt$)

Constant (L/2m)

Eccentricity ($e$): 0.55

Speed:

7.3

NCERT Section

Universal Law of Gravitation

Inverse Square Law & Superposition

7.3.1 Mathematical Statement & Shell Theorems

Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:

Universal Gravitation Vector Equations $$\mathbf{F}_{21} = - \frac{G m_1 m_2}{r_{21}^2} \hat{\mathbf{r}}_{21} = - \frac{G m_1 m_2}{|\mathbf{r}_2 - \mathbf{r}_1|^3} (\mathbf{r}_2 - \mathbf{r}_1)$$ $$\mathbf{F}_{12} = -\mathbf{F}_{21} \quad (\text{Action-Reaction Pair satisfying Newton's 3rd Law})$$ Superposition Principle for $n$ Particles: $$\mathbf{F}_1 = \sum_{i=2}^n \mathbf{F}_{1i} = \sum_{i=2}^n \frac{G m_1 m_i}{r_{1i}^2} \hat{\mathbf{r}}_{i1}$$

NCERT Shell Theorems (Calculus Integration of Spherical Mass):

Shell Theorem 1 (Outside Shell): The gravitational attraction between a uniform hollow spherical shell and an external point mass is exactly as if the entire mass of the shell were concentrated at its geometric centre ($F = \frac{G M m}{r^2}$ for $r \ge R$).
Shell Theorem 2 (Inside Shell): The gravitational force exerted by a uniform spherical shell on a point mass placed anywhere inside it is strictly zero ($F = 0$ for $r < R$).

Example 7.1 • Gravitational Force on Mass at Centroid of Triangle

NCERT Solved

Three equal masses $m$ are placed at the vertices of an equilateral triangle $\Delta ABC$ of side $l$. What is the gravitational force acting on a mass $2m$ placed at the centroid $G$ of the triangle? What is the force if the mass at vertex $A$ is doubled ($2m$)?

Answers: (a) Net Force $\mathbf{F}_R = 0$ | (b) Doubled mass at $A \implies \mathbf{F}_R' = \frac{2Gm^2}{l^2}\hat{\mathbf{j}}$ (Directed towards $A$)

Step 1: Distance from centroid $G$ to each vertex: In equilateral triangle of side $l$, median length $= \frac{\sqrt{3}}{2}l$. Centroid divides median in $2:1$ ratio $\implies AG = BG = CG = \frac{2}{3}\left(\frac{\sqrt{3}}{2}l\right) = \frac{l}{\sqrt{3}} \equiv r_0$.

Step 2: Vector forces on mass $2m$ at centroid: $$\mathbf{F}_{GA} = \frac{G(2m)(m)}{(l/\sqrt{3})^2}\hat{\mathbf{j}} = \frac{6Gm^2}{l^2}\hat{\mathbf{j}}$$ $$\mathbf{F}_{GB} = \frac{6Gm^2}{l^2}(-\cos 30^\circ\hat{\mathbf{i}} - \sin 30^\circ\hat{\mathbf{j}}) = \frac{6Gm^2}{l^2}\left(-\frac{\sqrt{3}}{2}\hat{\mathbf{i}} - \frac{1}{2}\hat{\mathbf{j}}\right)$$ $$\mathbf{F}_{GC} = \frac{6Gm^2}{l^2}(+\cos 30^\circ\hat{\mathbf{i}} - \sin 30^\circ\hat{\mathbf{j}}) = \frac{6Gm^2}{l^2}\left(+\frac{\sqrt{3}}{2}\hat{\mathbf{i}} - \frac{1}{2}\hat{\mathbf{j}}\right)$$

Step 3: Vector Sum: $$\mathbf{F}_R = \mathbf{F}_{GA} + \mathbf{F}_{GB} + \mathbf{F}_{GC} = \frac{6Gm^2}{l^2}\left[ \left(-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right)\hat{\mathbf{i}} + \left(1 - \frac{1}{2} - \frac{1}{2}\right)\hat{\mathbf{j}} \right] = \mathbf{0}$$ (By 3-fold rotational symmetry, resultant is identically zero).

Step 4: Mass at vertex $A$ doubled to $2m$: $$\mathbf{F}_{GA}' = \frac{G(2m)(2m)}{(l/\sqrt{3})^2}\hat{\mathbf{j}} = \frac{12Gm^2}{l^2}\hat{\mathbf{j}}$$ $$\mathbf{F}_R' = \mathbf{F}_{GA}' + (\mathbf{F}_{GB} + \mathbf{F}_{GC}) = \frac{12Gm^2}{l^2}\hat{\mathbf{j}} - \frac{6Gm^2}{l^2}\hat{\mathbf{j}} = \mathbf{\frac{6Gm^2}{l^2}\hat{\mathbf{j}}}$$

7.4

NCERT Section

The Gravitational Constant ($G$) & Cavendish Experiment

Experimental Physics

7.4.1 Cavendish Torsion Balance (1798)

The universal constant $G$ was first measured experimentally by English physicist Henry Cavendish using a high-sensitivity torsion balance:

Cavendish Equilibrium Formulation Gravitational attractive force between large lead sphere ($M$) and small lead sphere ($m$) separated by distance $d$: $$F = \frac{G M m}{d^2}$$ Deflecting gravitational torque on bar of length $L$: $$\tau_\text{grav} = F \cdot L = \frac{G M m L}{d^2}$$ At equilibrium, gravitational torque balances restoring torque of suspension wire with restoring couple per unit twist $\kappa$: $$\tau_\text{restoring} = \kappa \theta \implies \frac{G M m L}{d^2} = \kappa \theta \implies G = \frac{\kappa \theta d^2}{M m L}$$ $$\text{Standard Accepted Value: } G = 6.6743 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \quad ([\text{M}^{-1}\text{L}^3\text{T}^{-2}])$$

"Cavendish Weighed the Earth": Once $G$ was experimentally measured, using $g = \frac{G M_E}{R_E^2}$ allowed the mass of the Earth $M_E = \frac{g R_E^2}{G} \approx 5.97 \times 10^{24}\text{ kg}$ to be calculated for the very first time in human history!

7.5 - 7.6

NCERT Sections

Acceleration due to Gravity ($g$) and its Variations

Gravitational Field Dynamics

7.5.1 Surface Gravity & Variations with Altitude & Depth

Acceleration due to Gravity Equations 1. On the Surface of Earth ($r = R_E$): $$g = \frac{G M_E}{R_E^2} = \frac{G \left(\frac{4}{3}\pi R_E^3 \rho\right)}{R_E^2} = \frac{4}{3}\pi G \rho R_E \approx 9.8\text{ m s}^{-2}$$ 2. Variation of $g$ with Altitude $h$ (Above Earth's Surface): $$g(h) = \frac{G M_E}{(R_E + h)^2} = g \left(1 + \frac{h}{R_E}\right)^{-2}$$ $$\text{For } h \ll R_E \text{ (Binomial Approximation): } g(h) \approx g \left(1 - \frac{2h}{R_E}\right)$$ 3. Variation of $g$ with Depth $d$ (Below Earth's Surface): Mass of active core inside radius $(R_E - d)$ is $M_d = M_E \frac{(R_E - d)^3}{R_E^3}$. Outer shell exerts zero net force. $$g(d) = \frac{G M_d}{(R_E - d)^2} = \frac{G M_E (R_E - d)}{R_E^3} = g \left(1 - \frac{d}{R_E}\right)$$ $$\text{At the Centre of the Earth } (d = R_E): \quad g_\text{centre} = 0$$

Figure 7.8: Variation of $g(r)$ from the centre of the Earth ($r=0$) to outer space ($r \to \infty$). Peak gravity occurs at the surface $r = R_E$.

Interactive 3D Earth Cutaway • Variation of $g$ (Depth & Altitude)

3D Field Probe

Rotate Earth to view internal layers

Local Gravity ($g$)

9.80 m/s²

Regime & Mathematical Formula

Surface of Earth ($r = R_E$) ➔ $g = 9.80\text{ m/s}^2$

Radial Position ($r/R_E$): 0 = Centre | 1.0 = Surface | 2.5 = 9,600 km Alt

7.7

NCERT Section

Gravitational Potential Energy & Potential

Work & Potential Energy

7.7.1 Derivation from Work Done against Central Gravity

The gravitational potential energy $V(r)$ of a mass $m$ at distance $r$ from the centre of Earth of mass $M_E$ is defined as the work done by an external agent in bringing mass $m$ from infinity ($\infty$) to distance $r$ without acceleration:

Potential Energy & Potential Formulations $$W = \int_\infty^r \mathbf{F}_\text{ext}\cdot d\mathbf{r} = \int_\infty^r \frac{G M_E m}{r'^2} dr' = \left[ -\frac{G M_E m}{r'} \right]_\infty^r = -\frac{G M_E m}{r}$$ $$V(r) = -\frac{G M_E m}{r} \quad (\text{Taking reference } V(\infty) = 0)$$ Gravitational Potential ($U$ or $V_g$ per unit mass): $$U(r) = \frac{V(r)}{m} = -\frac{G M_E}{r} \quad (\text{SI Unit: } \text{J kg}^{-1})$$ Multi-Particle System Total Potential Energy: $$V_\text{total} = -\sum_{i < j} \frac{G m_i m_j}{r_{ij}}$$

Example 7.3 • Potential Energy of 4 Particles on Square Vertices

NCERT Solved

Find the potential energy of a system of four particles, each of mass $m$, placed at the vertices of a square of side $l$. Also obtain the gravitational potential at the centre of the square.

Answers: System P.E. $V = -5.41\frac{Gm^2}{l}$ | Potential at centre $U(O) = -4\sqrt{2}\frac{Gm}{l} \approx -5.66\frac{Gm}{l}$

Step 1: Count all unique interaction pairs for 4 masses: Total pairs $= \frac{4 \times 3}{2} = 6$ pairs. - $4$ side pairs at distance $l$. - $2$ diagonal pairs at distance $\sqrt{2}l$.

Step 2: Sum Potential Energies: $$V = - \left[ 4 \left(\frac{G m^2}{l}\right) + 2 \left(\frac{G m^2}{\sqrt{2}l}\right) \right] = -\frac{G m^2}{l} (4 + \sqrt{2}) = -\frac{Gm^2}{l}(4 + 1.414) = \mathbf{-5.41\frac{Gm^2}{l}}$$

Step 3: Potential at the centre of the square: Distance from centre $O$ to each vertex $r = \frac{\sqrt{2}l}{2} = \frac{l}{\sqrt{2}}$. $$U(O) = \sum_{i=1}^4 \left(-\frac{G m}{r}\right) = - 4 \frac{G m}{l/\sqrt{2}} = \mathbf{- 4\sqrt{2}\frac{Gm}{l}} \approx \mathbf{-5.66\frac{Gm}{l}}$$

7.8

NCERT Section

Escape Speed ($v_e$)

Energy Conservation & Unbound Trajectories

7.8.1 Escape Speed Derivation from Mechanical Energy Conservation

The escape speed is the minimum speed with which an object must be projected from the surface of a celestial body so that it escapes its gravitational pull and never falls back.

Escape Speed Master Equations At the surface: Total Mechanical Energy $E_i = \frac{1}{2}m v_i^2 - \frac{G M_E m}{R_E}$. At infinity ($r \to \infty$): Potential energy $V(\infty) = 0 \implies E_f = \frac{1}{2}m v_f^2 \ge 0$. By Conservation of Energy ($E_i = E_f \ge 0$): $$\frac{1}{2}m v_e^2 - \frac{G M_E m}{R_E} \ge 0 \implies v_e = \sqrt{\frac{2 G M_E}{R_E}} = \sqrt{2 g R_E}$$ On Earth: $v_e = \sqrt{2 \times 9.8\text{ m s}^{-2} \times 6.4 \times 10^6\text{ m}} \approx \mathbf{11.2\text{ km s}^{-1}}$ On Moon: $v_e \approx \mathbf{2.3\text{ km s}^{-1}}$ (Moon has no atmosphere because thermal root-mean-square speed of gas molecules exceeds moon's escape speed).

Interactive 3D Launch Cannon & Escape Velocity Simulator

Energy Barrier Lab

Drag to change perspective

Launch Velocity ($v_0$)

11.2 km/s

Trajectory State & Energy Status

Parabolic Escape Trajectory ($v_0 = 11.2\text{ km/s}$) ➔ $E_\text{total} = 0$, reaches infinity

Launch Speed: Orbital: 7.9 km/s | Escape: 11.2 km/s

Example 7.4 • Neutral Point and Minimum Speed Between Two Fixed Spheres

NCERT Solved

Two uniform solid spheres of equal radii $R$, but masses $M$ and $4M$ have a centre-to-centre separation $6R$. A projectile of mass $m$ is projected from the surface of sphere $M$ directly towards the centre of the second sphere. Obtain the minimum launch speed $v$ so that it reaches the surface of the second sphere.

Final Answer: $v_\text{min} = \sqrt{\frac{3 G M}{5 R}}$

Step 1: Locate Neutral Point $N$ where net gravitational force is zero: Let $r$ be distance from centre of mass $M$: $$\frac{G M m}{r^2} = \frac{G(4M)m}{(6R - r)^2} \implies (6R - r)^2 = 4r^2 \implies 6R - r = 2r \implies \mathbf{r = 2R}$$

Step 2: Total Mechanical Energy at surface of sphere $M$ ($r = R$): Distance to sphere $M$ is $R$, and distance to sphere $4M$ is $6R - R = 5R$: $$E_i = \frac{1}{2}m v^2 - \frac{GMm}{R} - \frac{G(4M)m}{5R} = \frac{1}{2}m v^2 - \frac{9 GMm}{5R}$$

Step 3: Total Mechanical Energy at Neutral Point $N$ ($r = 2R$, speed $\to 0$): Distance to $M$ is $2R$, and distance to $4M$ is $4R$: $$E_N = 0 - \frac{GMm}{2R} - \frac{G(4M)m}{4R} = -\frac{GMm}{2R} - \frac{GMm}{R} = -\frac{3 GMm}{2R}$$

Step 4: Apply Conservation of Mechanical Energy ($E_i = E_N$): $$\frac{1}{2}m v^2 - \frac{9 GMm}{5R} = -\frac{3 GMm}{2R} \implies \frac{1}{2}v^2 = \frac{GM}{R}\left(\frac{9}{5} - \frac{3}{2}\right) = \frac{GM}{R}\left(\frac{3}{10}\right)$$ $$v^2 = \frac{3 GM}{5 R} \implies \mathbf{v = \sqrt{\frac{3 GM}{5 R}}}$$

7.9 - 7.10

NCERT Sections

Earth Satellites, Orbital Speed & Satellite Energy

Orbital Mechanics

7.9.1 Circular Orbit Kinematics & Binding Energy

For a satellite of mass $m$ orbiting in a circular path at height $h$ above Earth's surface ($r = R_E + h$):

Satellite Orbital Master Formulations 1. Centripetal Force = Gravitational Force: $$\frac{m v^2}{R_E + h} = \frac{G M_E m}{(R_E + h)^2} \implies v = \sqrt{\frac{G M_E}{R_E + h}}$$ $$\text{Close to Earth } (h \approx 0): \quad v_0 = \sqrt{\frac{G M_E}{R_E}} = \sqrt{g R_E} \approx \mathbf{7.92\text{ km s}^{-1}}$$ 2. Time Period of Orbit ($T$): $$T = \frac{2\pi (R_E + h)}{v} = \frac{2\pi (R_E + h)^{3/2}}{\sqrt{G M_E}} \implies T^2 = \left(\frac{4\pi^2}{G M_E}\right)(R_E + h)^3$$ $$\text{For low-Earth orbit } (h \approx 0): \quad T_0 = 2\pi\sqrt{\frac{R_E}{g}} \approx \mathbf{85\text{ minutes}}$$ 3. Energy Distribution of Orbiting Satellite: $$K = \frac{1}{2}m v^2 = \frac{G M_E m}{2(R_E + h)}$$ $$V = -\frac{G M_E m}{R_E + h}$$ $$E_\text{total} = K + V = -\frac{G M_E m}{2(R_E + h)} = -K = \frac{V}{2}$$ $$\text{Binding Energy } E_B = -E_\text{total} = +\frac{G M_E m}{2(R_E + h)}$$

Interactive 3D Earth Satellite System (LEO, Polar & Geostationary)

Multi-Orbit Lab

Drag to view orbital planes

Geostationary (GEO)

$h = 35,800\text{ km}$, $T = 24\text{ h}$ (Fixed)

Polar Satellite

$h \approx 500\text{–}800\text{ km}$, $T \approx 100\text{ min}$

Low-Earth Orbit (LEO)

$v \approx 7.92\text{ km/s}$, $T \approx 85\text{ min}$

Example 7.5 • Martian Moon Phobos & Orbital Calculations

NCERT Solved

(i) Mars moon Phobos has period $T = 7\text{ h } 39\text{ min}$ ($459\text{ min}$) and orbital radius $R = 9.4 \times 10^3\text{ km}$. Calculate mass of Mars $M_m$.
(ii) If Mars orbital radius is $1.52$ times Earth's orbital radius around the Sun, calculate Martian year in days.

Answers: (i) $M_m = 6.48 \times 10^{23}\text{ kg}$ | (ii) $T_\text{Mars} \approx 684\text{ days}$

(i) Mass of Mars using Kepler's 3rd Law: $$T = 459 \times 60 = 27540\text{ s}, \quad R = 9.4 \times 10^6\text{ m}$$ $$M_m = \frac{4\pi^2 R^3}{G T^2} = \frac{4(3.1416)^2 (9.4 \times 10^6)^3}{(6.67 \times 10^{-11})(27540)^2} = \mathbf{6.48 \times 10^{23}\text{ kg}}$$

(ii) Martian Year: $$\frac{T_\text{Mars}}{T_\text{Earth}} = \left(\frac{R_\text{Mars-Sun}}{R_\text{Earth-Sun}}\right)^{3/2} = (1.52)^{3/2} \approx 1.874$$ $$T_\text{Mars} = 1.874 \times 365\text{ days} = \mathbf{684\text{ days}}$$

Example 7.6 • Weighing the Earth in Two Different Ways

NCERT Solved

Given $g = 9.81\text{ m s}^{-2}$, $R_E = 6.37 \times 10^6\text{ m}$, lunar distance $R = 3.84 \times 10^8\text{ m}$, and Moon period $T = 27.3\text{ days}$. Calculate $M_E$ using two independent methods.

Answers: Method 1: $M_E = 5.97 \times 10^{24}\text{ kg}$ | Method 2: $M_E = 6.02 \times 10^{24}\text{ kg}$ (Agreement within $< 1\%$)

Method 1: Surface gravity relation: $$M_E = \frac{g R_E^2}{G} = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = \mathbf{5.97 \times 10^{24}\text{ kg}}$$

Method 2: Lunar orbit period: $$T = 27.3 \times 24 \times 3600 = 2.3587 \times 10^6\text{ s}$$ $$M_E = \frac{4\pi^2 R^3}{G T^2} = \frac{4\pi^2 (3.84 \times 10^8)^3}{(6.67 \times 10^{-11})(2.3587 \times 10^6)^2} = \mathbf{6.02 \times 10^{24}\text{ kg}}$$

Example 7.7 • Kepler's Constant $k$ in SI and Astronomical Units

NCERT Solved

Express constant $k = 10^{-13}\text{ s}^2\text{ m}^{-3}$ in $\text{d}^2\text{ km}^{-3}$. If Moon is at $3.84 \times 10^5\text{ km}$, find its period in days.

Answers: $k = 1.33 \times 10^{-14}\text{ d}^2\text{ km}^{-3}$ | $T = 27.3\text{ days}$

Step 1: Unit conversion: $$1\text{ s} = \frac{1}{86400}\text{ d}, \quad 1\text{ m} = 10^{-3}\text{ km}$$ $$k = 10^{-13} \times \frac{(1/86400)^2}{(10^{-3})^3} = 10^{-13} \times \frac{10^9}{(86400)^2} = \mathbf{1.33 \times 10^{-14}\text{ d}^2\text{ km}^{-3}}$$

Step 2: Calculate lunar period $T$: $$T^2 = k R^3 = (1.33 \times 10^{-14})(3.84 \times 10^5)^3 = (1.33 \times 10^{-14})(56.623 \times 10^{15}) = 753.08\text{ d}^2$$ $$T = \sqrt{753.08} = \mathbf{27.3\text{ days}}$$

02 / Self-Assessment Lab

Concept Check Questions

Interactive questions designed to build rock-solid conceptual mastery. Click any option to inspect comprehensive explanations for all choices.

15 MCQS

0 / 15 Answered Score: 0

Q1 Pending

Kepler's Second Law (Law of Areas, $d\mathbf{A}/dt = \text{constant}$) is a direct consequence of the conservation of which physical quantity?

Incorrect. Linear momentum continuously changes direction as the planet curves around the elliptical orbit.

Correct! Since gravity is a strictly central force ($\mathbf{F} \parallel \mathbf{r}$), the net torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = 0$. Hence angular momentum $\mathbf{L} = \text{constant}$, leading directly to $\frac{d\mathbf{A}}{dt} = \frac{\mathbf{L}}{2m} = \text{constant}$.

Incorrect. Kinetic energy changes continuously, peaking at perihelion and reaching a minimum at aphelion.

Incorrect. Gravitational potential varies with distance $r$.

Takeaway: The Law of Areas holds true for any central force field where torque is zero.

Q2 Pending

If the distance between two point masses is doubled, how does the gravitational attractive force between them change?

Incorrect. Force decreases with distance, not increases.

Incorrect. Gravity follows an inverse square law ($1/r^2$), not an inverse linear law.

Correct! Since $F \propto 1/r^2$, doubling $r \to 2r$ makes $F' = \frac{G m_1 m_2}{(2r)^2} = \frac{1}{4}F$.

Incorrect. Gravitational interaction depends directly on spatial separation.

Takeaway: Universal gravitation strictly obeys the Inverse Square Law ($F \propto 1/r^2$).

Q3 Pending

At what depth $d$ below the surface of the Earth will the acceleration due to gravity become zero?

Incorrect. At half radius, $g(d) = g(1 - 0.5) = g/2$.

Correct! Using $g(d) = g\left(1 - \frac{d}{R_E}\right)$, setting $d = R_E$ gives $g(R_E) = g(1 - 1) = 0$. By spherical symmetry, all surrounding shells exert zero net force.

Incorrect. $d$ cannot exceed the Earth's radius $R_E$.

Incorrect. At the centre, symmetric pulls in all directions cancel completely.

Takeaway: Acceleration due to gravity is exactly zero at the centre of the Earth.

Q4 Pending

What is the ratio of escape speed from Earth ($v_e$) to orbital speed in a close low-Earth orbit ($v_0$)?

Incorrect. A satellite in orbit is bound; escaping requires more kinetic energy.

Correct! $v_e = \sqrt{2 g R_E}$ and $v_0 = \sqrt{g R_E} \implies \frac{v_e}{v_0} = \sqrt{2}$. An increase in speed of only $\approx 41.4\%$ allows a low-orbit satellite to escape completely.

Incorrect. $v_e^2 / v_0^2 = 2$, so velocity ratio is $\sqrt{2}$.

Incorrect. Escape speed is strictly greater than orbital speed.

Takeaway: $v_e = \sqrt{2} v_0 \approx 1.414 v_0$.

Q5 Pending

Why does the Moon have no atmosphere?

Incorrect. Solar wind plays a role over eons, but the primary fundamental physics cause is gravitational.

Incorrect. $G$ is a universal physical constant identical everywhere.

Correct! Due to smaller mass and radius, Moon's escape speed is only $v_e \approx 2.3\text{ km s}^{-1}$. The thermal rms speed $v_\text{rms} = \sqrt{3kT/m}$ of atmospheric gases exceeds this, allowing molecules to escape into space.

Incorrect. Gravity is the primary confining mechanism for planetary atmosphere.

Takeaway: Low escape velocity ($2.3\text{ km s}^{-1}$) cannot permanently trap lighter gas molecules against thermal agitation.

03 / Textbook Solutions

Complete NCERT Exercises 7.1 – 7.21

Step-by-step mathematical derivations, conceptual reasoning, and numerical answers for all 21 textbook problems.

21 PROBLEMS

All Questions (7.1-7.21)

Conceptual & Reasoning

Variation of g

Escape & Satellite Energy

Collisions & Double Stars

Ex 7.1 Conceptual Foundations

7.1 Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

Prerequisite: ↗ 7.3 Gravitational Shielding & Tidal Gradients

(a) Gravitational Shielding: No. Gravitational force is an intrinsic property of mass and is completely independent of the intervening medium. Unlike electromagnetism (where induced charges redistribute on a conductor to cancel external electric fields), mass possesses only positive gravitational charge (always attractive). Hence, no gravitational shielding is possible.

(b) Gravity Detection in Large Space Station: Yes. In a small space ship, the entire vessel is in uniform free-fall with identical acceleration $g$. In a very large space station, different parts of the station are at different distances from the centre of the Earth and along slightly non-parallel radial lines. This produces detectable tidal gradient forces ($\Delta g \approx \frac{2GM}{r^3}\Delta r$), allowing the astronaut to detect gravity.

(c) Tidal Effect of Moon vs Sun: Tidal force is governed by the spatial gradient of gravitational force ($\frac{dF}{dr} \propto \frac{M}{r^3}$), not the force $F \propto \frac{M}{r^2}$ itself: $$\frac{\text{Tidal Force}_\text{Moon}}{\text{Tidal Force}_\text{Sun}} = \frac{M_\text{Moon} / r_\text{Moon}^3}{M_\text{Sun} / r_\text{Sun}^3} = \left(\frac{M_\text{Moon}}{M_\text{Sun}}\right)\left(\frac{r_\text{Sun}}{r_\text{Moon}}\right)^3 \approx 2.18$$ Because the Moon is $\sim 390$ times closer to Earth than the Sun, the inverse-cube term dominates, making lunar tides more than twice as strong as solar tides.

Ex 7.16 Weight Inside Earth

7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed $250\text{ N}$ on the surface?

Prerequisite: ↗ 7.6 Variation of g with Depth

Answer: Weight halfway down $= 125\text{ N}$

Step 1: Formula for acceleration due to gravity at depth $d$: $$g(d) = g\left(1 - \frac{d}{R_E}\right)$$

Step 2: Halfway down depth $d = R_E / 2$: $$g(d) = g\left(1 - \frac{R_E/2}{R_E}\right) = g\left(1 - \frac{1}{2}\right) = \frac{g}{2}$$

Step 3: Calculate weight: $$W' = m g(d) = m\left(\frac{g}{2}\right) = \frac{W}{2} = \frac{250\text{ N}}{2} = \mathbf{125\text{ N}}$$

Ex 7.17 Rocket Vertical Apex Height

7.17 A rocket is fired vertically with a speed of $5\text{ km s}^{-1}$ from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth $M_E = 6.0 \times 10^{24}\text{ kg}$; mean radius of the earth $R_E = 6.4 \times 10^6\text{ m}$; $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$.

Prerequisite: ↗ 7.7 Conservation of Mechanical Energy

Answers: Distance from Earth's centre $r = 8.0 \times 10^6\text{ m}$ | Height above surface $h = 1.6 \times 10^6\text{ m} = 1600\text{ km}$

Step 1: Conservation of Energy between surface and apex $r = R_E + h$ ($v = 0$): $$\frac{1}{2}m v^2 - \frac{G M_E m}{R_E} = 0 - \frac{G M_E m}{r}$$ $$\frac{G M_E}{r} = \frac{G M_E}{R_E} - \frac{1}{2}v^2 \implies \frac{1}{r} = \frac{1}{R_E} - \frac{v^2}{2 G M_E}$$

Step 2: Numerical evaluation: $$2 G M_E = 2(6.67 \times 10^{-11})(6.0 \times 10^{24}) = 8.004 \times 10^{14}\text{ m}^3\text{ s}^{-2}$$ $$\frac{v^2}{2 G M_E} = \frac{(5000)^2}{8.004 \times 10^{14}} = \frac{25 \times 10^6}{8.004 \times 10^{14}} = 3.123 \times 10^{-8}\text{ m}^{-1}$$ $$\frac{1}{R_E} = \frac{1}{6.4 \times 10^6} = 15.625 \times 10^{-8}\text{ m}^{-1}$$ $$\frac{1}{r} = (15.625 - 3.123) \times 10^{-8} = 12.502 \times 10^{-8}\text{ m}^{-1}$$ $$r = \frac{1}{12.502 \times 10^{-8}} = \mathbf{8.0 \times 10^6\text{ m}}$$ $$h = r - R_E = 8.0 \times 10^6 - 6.4 \times 10^6 = \mathbf{1.6 \times 10^6\text{ m}} = \mathbf{1600\text{ km}}$$

Ex 7.18 Hyperbolic Trajectory Final Speed

7.18 The escape speed of a projectile on the earth’s surface is $11.2\text{ km s}^{-1}$. A body is projected out with thrice this speed ($v = 3 v_e$). What is the speed of the body far away from the earth (at infinity)? Ignore the presence of the sun and other planets.

Prerequisite: ↗ 7.8 Energy Conservation for $v > v_e$

Answer: Speed at infinity $v_\infty = 2\sqrt{2} v_e \approx 31.68\text{ km s}^{-1}$

Step 1: Energy balance from surface to infinity: $$E_\text{surface} = \frac{1}{2}m v^2 - \frac{G M_E m}{R_E} = \frac{1}{2}m v^2 - \frac{1}{2}m v_e^2 \quad \left(\text{since } \frac{G M_E}{R_E} = \frac{1}{2}v_e^2\right)$$ $$E_\infty = \frac{1}{2}m v_\infty^2 + 0$$

Step 2: Relate speeds: $$\frac{1}{2}m v_\infty^2 = \frac{1}{2}m (v^2 - v_e^2) \implies v_\infty = \sqrt{v^2 - v_e^2}$$ Given $v = 3 v_e$: $$v_\infty = \sqrt{(3 v_e)^2 - v_e^2} = \sqrt{9 v_e^2 - v_e^2} = \sqrt{8 v_e^2} = \mathbf{2\sqrt{2} v_e}$$ $$v_\infty = 2\sqrt{2} \times 11.2\text{ km s}^{-1} \approx 2.8284 \times 11.2 = \mathbf{31.68\text{ km s}^{-1}}$$

Ex 7.19 Satellite Escape Energy

7.19 A satellite orbits the earth at a height of $400\text{ km}$ above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of satellite $m = 200\text{ kg}$; mass of earth $M_E = 6.0 \times 10^{24}\text{ kg}$; radius of earth $R_E = 6.4 \times 10^6\text{ m}$; $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$.

Prerequisite: ↗ 7.10 Binding Energy of Satellite

Answer: Energy Required $E = 5.89 \times 10^9\text{ J} = 5.89\text{ GJ}$

Step 1: Orbital radius: $$r = R_E + h = 6.4 \times 10^6 + 0.4 \times 10^6 = 6.8 \times 10^6\text{ m}$$

Step 2: Total Energy of orbiting satellite: $$E_\text{orbit} = -\frac{G M_E m}{2 r}$$

Step 3: Energy expended to reach infinity ($E_\infty = 0$): $$\Delta E = E_\infty - E_\text{orbit} = 0 - \left(-\frac{G M_E m}{2 r}\right) = \frac{G M_E m}{2 r}$$ $$\Delta E = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(200)}{2(6.8 \times 10^6)} = \frac{8.004 \times 10^{16}}{13.6 \times 10^6} = \mathbf{5.89 \times 10^9\text{ J}}$$

Ex 7.20 Head-on Colliding Binary Stars

7.20 Two stars each of one solar mass ($M = 2 \times 10^{30}\text{ kg}$) are approaching each other for a head-on collision. When they are at distance $10^9\text{ km}$, their speeds are negligible. What is the speed with which they collide? Radius of each star $R = 10^4\text{ km} = 10^7\text{ m}$. Assume stars remain undistorted until they collide ($G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$).

Prerequisite: ↗ 7.7 Two-Body Conservation of Energy

Answer: Collision speed $v \approx 2.58 \times 10^6\text{ m s}^{-1} = 2580\text{ km s}^{-1}$

Step 1: Initial State ($r_1 = 10^{12}\text{ m}$, speeds $\approx 0$): $$E_i = 0 - \frac{G M^2}{r_1} \approx 0 \quad (\text{since } r_1 \gg 2R)$$

Step 2: Collision State ($r_2 = 2R = 2 \times 10^7\text{ m}$): By symmetry, both identical stars have equal speed $v$: $$E_f = \frac{1}{2}M v^2 + \frac{1}{2}M v^2 - \frac{G M^2}{2R} = M v^2 - \frac{G M^2}{2R}$$

Step 3: Energy Conservation ($E_i = E_f$): $$M v^2 - \frac{G M^2}{2R} = 0 \implies v^2 = \frac{G M}{2R}$$ $$v = \sqrt{\frac{G M}{2R}} = \sqrt{\frac{(6.67 \times 10^{-11})(2 \times 10^{30})}{2(10^7)}} = \sqrt{\frac{1.334 \times 10^{20}}{2 \times 10^7}} = \sqrt{6.67 \times 10^{12}}$$ $$v \approx \mathbf{2.58 \times 10^6\text{ m s}^{-1}} = \mathbf{2580\text{ km s}^{-1}}$$

Ex 7.21 Equilibrium & Potential at Midpoint

7.21 Two heavy spheres each of mass $100\text{ kg}$ and radius $0.10\text{ m}$ are placed $1.0\text{ m}$ apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Prerequisite: ↗ 7.7 Superposition & Earnshaw's Theorem

Answers: Force $F = 0$ | Potential $V = -2.67 \times 10^{-8}\text{ J kg}^{-1}$ | Equilibrium is UNSTABLE

Step 1: Force at midpoint $O$ ($r = 0.5\text{ m}$ from each): The two equal masses $M = 100\text{ kg}$ pull with equal and opposite gravitational forces: $$\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2 = \frac{G M m}{(0.5)^2}\hat{\mathbf{i}} - \frac{G M m}{(0.5)^2}\hat{\mathbf{i}} = \mathbf{0}$$

Step 2: Gravitational Potential at midpoint: Potential is a scalar, so potentials add algebraically: $$V = V_1 + V_2 = - \frac{G M}{0.5} - \frac{G M}{0.5} = - \frac{2 G M}{0.5} = - 4 G M$$ $$V = - 4(6.67 \times 10^{-11})(100) = \mathbf{- 2.67 \times 10^{-8}\text{ J kg}^{-1}}$$

Step 3: Nature of Equilibrium: Since net force is zero, the point mass is in equilibrium. If displaced slightly along the line joining the spheres towards sphere 1, the attractive force towards sphere 1 increases while the force towards sphere 2 decreases. The net force pulls the particle further away from the midpoint (no restoring force). Therefore, the equilibrium is UNSTABLE along the axial line.

04 / Rapid Reference

Chapter Summary & Formulas

Newton's law of gravitation, Kepler's laws, variations in $g$ with height/depth/rotation, escape velocity, satellite orbital energy ratios, and geostationary requirements.

100% SYLLABUS

7.1 - 7.3 • Planetary Gravitation

Gravitation Law & Kepler's Three Laws

Universal Law of Gravitation: The attractive force between two point masses: $$F = G \frac{M m}{r^2}$$ $G \approx 6.67 \times 10^{-11}\text{ N m}^2\text{kg}^{-2}$ (Universal constant, dimension $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$).

Kepler's Laws of Planetary Motion

Kepler's Law	Formulation / Law	Crucial Principles & Constant
1st: Law of Orbits	All planets move in elliptical orbits with the Sun at one of the foci.	Distance of closest approach (perihelion): $r_p = a(1-e)$. Farthest (aphelion): $r_a = a(1+e)$.
2nd: Law of Areas	A line joining planet and Sun sweeps out equal areas in equal intervals: $\frac{dA}{dt} = \text{constant}$.	Based on the Conservation of Angular Momentum: $\frac{dA}{dt} = \frac{L}{2m}$. Sector speed is constant.
3rd: Law of Periods	$$T^2 \propto a^3$$	$T^2 = \left(\frac{4\pi^2}{GM_{\text{sun}}}\right) a^3$. $a$ is the semi-major axis.

7.4 & 7.5 • Acceleration due to Gravity

Variations in Acceleration due to Gravity ($g$)

At Earth's surface: $g_0 = \frac{G M_E}{R_E^2} \approx 9.8\text{ m/s}^2$. Acceleration decreases as you go above or below the surface.

Variation Factor	Mathematical Formula	Conditions & Tricks
Altitude (exact)	$g(h) = g_0 \left(\frac{R_E}{R_E + h}\right)^2$	Always valid for any height $h$.
Altitude (approx)	$g(h) \approx g_0 \left(1 - \frac{2h}{R_E}\right)$	Condition: Valid only if $h \ll R_E$ (typically $h < 320\text{ km}$).
Depth ($d$)	$g(d) = g_0 \left(1 - \frac{d}{R_E}\right)$	Always valid. At center ($d=R_E$), gravity $g=0$.
Latitude ($\lambda$)	$g' = g_0 - \omega^2 R_E \cos^2\lambda$	Due to Earth's rotation. Equator ($\lambda=0^\circ$): $g$ is minimum. Poles ($\lambda=90^\circ$): $g$ is maximum.

Double-Drop Rule: For a small height $h$ and depth $d$, the reduction in $g$ is identical when $d = 2h$.

7.6 - 7.8 • Orbital Mechanics

Escape Velocity, Orbital Speed & Satellite Energies

Escape Velocity ($v_e$) vs. Orbital Speed ($v_o$)

Escape Velocity ($v_e$): Minimum speed required to escape Earth's gravitational pull: $$v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E} \approx 11.2\text{ km/s}$$ Condition: Completely independent of the mass of the escaping body and the angle of projection.
Orbital Velocity ($v_o$): Speed of a satellite in a circular orbit at altitude $h$: $$v_o = \sqrt{\frac{GM_E}{R_E + h}}$$ For a satellite close to the surface ($h \ll R_E$), $v_o = \sqrt{gR_E} \approx 7.9\text{ km/s}$.

Speed Link: $v_e = \sqrt{2}v_o \approx 1.414 v_o$.
Trick: Increasing the orbital speed of a low-Earth satellite by $41.4\%$ ($\sqrt{2}-1$) unbinds it, causing it to escape.

Satellite Energy Relationships

Energy Form	Formula (Radius $r$)	Energy Ratios Shortcut
Kinetic Energy ($K$)	$K = +\frac{GMm}{2r}$	Cosmic Ratio: $$E = -K = \frac{U}{2}$$ $$E_{\text{binding}} = +K = -E$$
Potential Energy ($U$)	$U = -\frac{GMm}{r}$
Total Energy ($E$)	$E = -\frac{GMm}{2r}$

7.8 • Geostationary Conditions

Geostationary Satellite Requirements

A satellite that appears stationary relative to an observer on Earth must satisfy these strict conditions:

Time Period: Exactly $24\text{ hours}$ (matches Earth's rotation).
Orbit Plane: Must lie in the equatorial plane of the Earth.
Direction of Rotation: West to East (same as Earth).
Altitude ($h$): Approximately $35,800\text{ km}$ (or orbital radius $r \approx 6.4 R_E \approx 42,200\text{ km}$) above Earth's surface.

NCERT Official

Points to Ponder & Exam Tips

The gravitational force is a central force and is conservative. Work done in any closed path is zero.
At the center of the Earth, gravitational potential is minimum (most negative) but gravitational force is zero ($g=0$).
Inside a hollow spherical shell, the gravitational field intensity is zero everywhere, but the gravitational potential is constant and non-zero.
Astronauts in space stations experience weightlessness because the station and the astronauts are in a state of continuous free fall towards the Earth.

05 / Timed Exam Prep

Chapter Mastery Tests

Simulated multi-tier exams covering fundamental laws, numerical variations, and orbital dynamics.

3 LEVELS

1. What are the dimensional units of Universal Gravitational Constant $G$?

2. The value of $g$ on the surface of the Moon is approximately:

1. At what height $h$ above the Earth's surface does $g$ reduce to $g/4$?

2. What is the orbital period of a geostationary satellite?

1. If the radius of Earth shrinks by $1\%$ keeping mass constant, the acceleration due to gravity on its surface:

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