Class 11 Physics Chapter 7: Gravitation | NCERT Notes & Solutions

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Study Kepler’s laws, gravitation, variation in g, escape speed and satellite motion.

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: $$d\mathbf{A}dt = 12(\mathbf{r} × \mathbf{v}) = \mathbf{L}2m = constant$$ Since gravitational force $\mathbf{F} = -GMmr^2\hat{\mathbf{r}}$ is purely central, Torque $\boldsymbol{τ} = \mathbf{r} × \mathbf{F} = 0 \implies \mathbf{L} = constant$. Consequently, at perihelion and aphelion: $r_p v_p = r_a v_a \implies v_pv_a = r_ar_p > 1$. (Planets move fastest at perihelion).

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

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: $$T^2 = k\, a^3 \qquad where k = 4π^2G M_S$$ $$T_1^2T_2^2 = a_1^3a_2^3$$

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 $$\mathbf{F}_{21} = - G m_1 m_2r_{21}^2 \hat{\mathbf{r}}_{21} = - G m_1 m_2|\mathbf{r}_2 - \mathbf{r}_1|^3 (\mathbf{r}_2 - \mathbf{r}_1)$$ $$\mathbf{F}_{12} = -\mathbf{F}_{21} \quad (Action-Reaction Pair satisfying Newton's 3rd Law)$$ Superposition Principle for $n$ Particles: $$\mathbf{F}_1 = ∑_{i=2}^n \mathbf{F}_{1i} = ∑_{i=2}^n G m_1 m_ir_{1i}^2 \hat{\mathbf{r}}_{i1}$$

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 ($F = G M mr^2$ for $r ≥ R$).
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 $Δ ABC$ 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 $A \implies \mathbf{F}_R' = 2Gm^2l^2\hat{\mathbf{j}}$ (Directed towards $A$)

Step 1: Distance from centroid $G$ to each vertex: In equilateral triangle of side $l$, median length $= √32l$. Centroid divides median in $2:1$ ratio $\implies AG = BG = CG = 23(√32l) = l√3 \equiv r_0$.

Step 2: Vector forces on mass $2m$ at centroid: $$\mathbf{F}_{GA} = G(2m)(m)(l/√3)^2\hat{\mathbf{j}} = 6Gm^2l^2\hat{\mathbf{j}}$$ $$\mathbf{F}_{GB} = 6Gm^2l^2(-\cos 30^°\hat{\mathbf{i}} - \sin 30^°\hat{\mathbf{j}}) = 6Gm^2l^2(-√32\hat{\mathbf{i}} - 12\hat{\mathbf{j}})$$ $$\mathbf{F}_{GC} = 6Gm^2l^2(+\cos 30^°\hat{\mathbf{i}} - \sin 30^°\hat{\mathbf{j}}) = 6Gm^2l^2(+√32\hat{\mathbf{i}} - 12\hat{\mathbf{j}})$$

Step 3: Vector Sum: $$\mathbf{F}_R = \mathbf{F}_{GA} + \mathbf{F}_{GB} + \mathbf{F}_{GC} = 6Gm^2l^2[ (-√32 + √32)\hat{\mathbf{i}} + (1 - 12 - 12)\hat{\mathbf{j}} ] = \mathbf{0}$$ (By 3-fold rotational symmetry, resultant is identically zero).

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

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 $d$: $$F = G M md^2$$ Deflecting gravitational torque on bar of length $L$: $$τ_grav = F · L = G M m Ld^2$$ At equilibrium, gravitational torque balances restoring torque of suspension wire with restoring couple per unit twist $\kappa$: $$τ_restoring = \kappa θ \implies G M m Ld^2 = \kappa θ \implies G = \kappa θ d^2M m L$$ $$Standard Accepted Value: G = 6.6743 × 10^{-11} N m^2 kg^{-2} \quad ([M^{-1}L^3T^{-2}])$$

"Cavendish Weighed the Earth": Once $G$ was experimentally measured, using $g = G M_ER_E^2$ allowed the mass of the Earth $M_E = g R_E^2G ≈ 5.97 × 10^{24} kg$ 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 ($r = R_E$): $$g = G M_ER_E^2 = G (43π R_E^3 ρ)R_E^2 = 43π G ρ R_E ≈ 9.8 m s^{-2}$$ 2. Variation of $g$ with Altitude $h$ (Above Earth's Surface): $$g(h) = G M_E(R_E + h)^2 = g (1 + hR_E)^{-2}$$ $$For h \ll R_E (Binomial Approximation): g(h) ≈ g (1 - 2hR_E)$$ 3. Variation of $g$ with Depth $d$ (Below Earth's Surface): Mass of active core inside radius $(R_E - d)$ is $M_d = M_E (R_E - d)^3R_E^3$. Outer shell exerts zero net force. $$g(d) = G M_d(R_E - d)^2 = G M_E (R_E - d)R_E^3 = g (1 - dR_E)$$ $$At the Centre of the Earth (d = R_E): \quad g_centre = 0$$

Figure 7.8: Variation of $g(r)$ from the centre of the Earth ($r=0$) to outer space ($r \to ∞$). 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 ($r = R_E$) ➔ $g = 9.80 m/s^2$

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 ($∞$) to distance $r$ without acceleration:

Potential Energy & Potential Formulations $$W = ∫_∞^r \mathbf{F}_ext· d\mathbf{r} = ∫_∞^r G M_E mr'^2 dr' = [ -G M_E mr' ]_∞^r = -G M_E mr$$ $$V(r) = -G M_E mr \quad (Taking reference V(∞) = 0)$$ Gravitational Potential ($U$ or $V_g$ per unit mass): $$U(r) = V(r)m = -G M_Er \quad (SI Unit: J kg^{-1})$$ Multi-Particle System Total Potential Energy: $$V_total = -∑i < j G m_i m_jr_{ij}$$

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. $V = -5.41Gm^2l$ | Potential at centre $U(O) = -4√2Gml ≈ -5.66Gml$

Step 1: Count all unique interaction pairs for 4 masses: Total pairs $= 4 × 32 = 6$ pairs. - $4$ side pairs at distance $l$. - $2$ diagonal pairs at distance $√2l$.

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

Step 3: Potential at the centre of the square: Distance from centre $O$ to each vertex $r = √2l2 = l√2$. $$U(O) = ∑_{i=1}^4 (-G mr) = - 4 G ml/√2 = \mathbf{- 4√2Gml} ≈ \mathbf{-5.66Gml}$$

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 $E_i = 12m v_i^2 - G M_E mR_E$. At infinity ($r \to ∞$): Potential energy $V(∞) = 0 \implies E_f = 12m v_f^2 ≥ 0$. By Conservation of Energy ($E_i = E_f ≥ 0$): $$12m v_e^2 - G M_E mR_E ≥ 0 \implies v_e = √2 G M_ER_E = √2 g R_E$$ On Earth: $v_e = \sqrt{2 × 9.8 m s^{-2} × 6.4 × 10^6 m} ≈ \mathbf{11.2 km s^{-1}}$ On Moon: $v_e ≈ \mathbf{2.3 km s^{-1}}$ (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 ($v_0 = 11.2 km/s$) ➔ $E_total = 0$, 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: $v_min = √3 G M5 R$

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

Step 2: Total Mechanical Energy at surface of sphere $M$ ($r = R$): Distance to sphere $M$ is $R$, and distance to sphere $4M$ is $6R - R = 5R$: $$E_i = 12m v^2 - GMmR - G(4M)m5R = 12m v^2 - 9 GMm5R$$

Step 3: Total Mechanical Energy at Neutral Point $N$ ($r = 2R$, speed $\to 0$): Distance to $M$ is $2R$, and distance to $4M$ is $4R$: $$E_N = 0 - GMm2R - G(4M)m4R = -GMm2R - GMmR = -3 GMm2R$$

Step 4: Apply Conservation of Mechanical Energy ($E_i = E_N$): $$12m v^2 - 9 GMm5R = -3 GMm2R \implies 12v^2 = GMR(95 - 32) = GMR(310)$$ $$v^2 = 3 GM5 R \implies \mathbf{v = √3 GM5 R}$$

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: $$m v^2R_E + h = G M_E m(R_E + h)^2 \implies v = √G M_ER_E + h$$ $$Close to Earth (h ≈ 0): \quad v_0 = √G M_ER_E = √g R_E ≈ \mathbf{7.92 km s^{-1}}$$ 2. Time Period of Orbit ($T$): $$T = 2π (R_E + h)v = 2π (R_E + h)^{3/2}√G M_E \implies T^2 = (4π^2G M_E)(R_E + h)^3$$ $$For low-Earth orbit (h ≈ 0): \quad T_0 = 2π√R_Eg ≈ \mathbf{85 minutes}$$ 3. Energy Distribution of Orbiting Satellite: $$K = 12m v^2 = G M_E m2(R_E + h)$$ $$V = -G M_E mR_E + h$$ $$E_total = K + V = -G M_E m2(R_E + h) = -K = V2$$ $$Binding Energy E_B = -E_total = +G M_E m2(R_E + h)$$

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

Multi-Orbit Lab

Drag to view orbital planes

Geostationary (GEO)

$h = 35,800 km$, $T = 24 h$ (Fixed)

Polar Satellite

$h ≈ 500–800 km$, $T ≈ 100 min$

Low-Earth Orbit (LEO)

$v ≈ 7.92 km/s$, $T ≈ 85 min$

Example 7.5 • Martian Moon Phobos & Orbital Calculations

NCERT Solved

(i) Mars moon Phobos has period $T = 7 h 39 min$ ($459 min$) and orbital radius $R = 9.4 × 10^3 km$. 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) $M_m = 6.48 × 10^{23} kg$ | (ii) $T_Mars ≈ 684 days$

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

(ii) Martian Year: $$T_MarsT_Earth = (R_Mars-SunR_Earth-Sun)^{3/2} = (1.52)^{3/2} ≈ 1.874$$ $$T_Mars = 1.874 × 365 days = \mathbf{684 days}$$

Example 7.6 • Weighing the Earth in Two Different Ways

NCERT Solved

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

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

Method 1: Surface gravity relation: $$M_E = g R_E^2G = 9.81 × (6.37 × 10^6)^26.67 × 10^{-11} = \mathbf{5.97 × 10^{24} kg}$$

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

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

NCERT Solved

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

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

Step 1: Unit conversion: $$1 s = 186400 d, \quad 1 m = 10^{-3} km$$ $$k = 10^{-13} × (1/86400)^2(10^{-3})^3 = 10^{-13} × 10^9(86400)^2 = \mathbf{1.33 × 10^{-14} d^2 km^{-3}}$$

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

Quiz

Concept quiz

Choose an answer to see the reasoning and key idea.

15 MCQS

0 / 15 Answered Score: 0

Q1 Pending

Kepler's Second Law (Law of Areas, $d\mathbf{A}/dt = constant$) 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 $\boldsymbol{τ} = \mathbf{r} × \mathbf{F} = 0$. Hence angular momentum $\mathbf{L} = constant$, leading directly to $d\mathbf{A}dt = \mathbf{L}2m = constant$.

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 $F ∝ 1/r^2$, doubling $r \to 2r$ makes $F' = G m_1 m_2(2r)^2 = 14F$.

Incorrect. Gravitational interaction depends directly on spatial separation.

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

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 $g(d) = g(1 - dR_E)$, 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! $v_e = √2 g R_E$ and $v_0 = √g R_E \implies v_ev_0 = √2$. An increase in speed of only $≈ 41.4%$ allows a low-orbit satellite to escape completely.

Incorrect. $v_e^2 / v_0^2 = 2$, so velocity ratio is $√2$.

Incorrect. Escape speed is strictly greater than orbital speed.

Takeaway: $v_e = √2 v_0 ≈ 1.414 v_0$.

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 $v_e ≈ 2.3 km s^{-1}$. The thermal rms speed $v_rms = √3kT/m$ 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 ($2.3 km s^{-1}$) cannot permanently trap lighter gas molecules against thermal agitation.

Exercises

NCERT Exercises

Try each question before opening its worked solution.

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 ($Δ g ≈ 2GMr^3Δ r$), allowing the astronaut to detect gravity.

(c) Tidal Effect of Moon vs Sun: Tidal force is governed by the spatial gradient of gravitational force ($dFdr ∝ Mr^3$), not the force $F ∝ Mr^2$ itself: $$Tidal Force_MoonTidal Force_Sun = M_Moon / r_Moon^3M_Sun / r_Sun^3 = (M_MoonM_Sun)(r_Sunr_Moon)^3 ≈ 2.18$$ 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 $250 N$ on the surface?

Prerequisite: ↗ 7.6 Variation of g with Depth

Answer: Weight halfway down $= 125 N$

Step 1: Formula for acceleration due to gravity at depth $d$: $$g(d) = g(1 - dR_E)$$

Step 2: Halfway down depth $d = R_E / 2$: $$g(d) = g(1 - R_E/2R_E) = g(1 - 12) = g2$$

Step 3: Calculate weight: $$W' = m g(d) = m(g2) = W2 = 250 N2 = \mathbf{125 N}$$

Ex 7.17 Rocket Vertical Apex Height

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

Prerequisite: ↗ 7.7 Conservation of Mechanical Energy

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

Step 1: Conservation of Energy between surface and apex $r = R_E + h$ ($v = 0$): $$12m v^2 - G M_E mR_E = 0 - G M_E mr$$ $$G M_Er = G M_ER_E - 12v^2 \implies 1r = 1R_E - v^22 G M_E$$

Step 2: Numerical evaluation: $$2 G M_E = 2(6.67 × 10^{-11})(6.0 × 10^{24}) = 8.004 × 10^{14} m^3 s^{-2}$$ $$v^22 G M_E = (5000)^28.004 × 10^{14} = 25 × 10^68.004 × 10^{14} = 3.123 × 10^{-8} m^{-1}$$ $$1R_E = 16.4 × 10^6 = 15.625 × 10^{-8} m^{-1}$$ $$1r = (15.625 - 3.123) × 10^{-8} = 12.502 × 10^{-8} m^{-1}$$ $$r = 112.502 × 10^{-8} = \mathbf{8.0 × 10^6 m}$$ $$h = r - R_E = 8.0 × 10^6 - 6.4 × 10^6 = \mathbf{1.6 × 10^6 m} = \mathbf{1600 km}$$

Ex 7.18 Hyperbolic Trajectory Final Speed

7.18 The escape speed of a projectile on the earth’s surface is $11.2 km s^{-1}$. 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 $v_∞ = 2√2 v_e ≈ 31.68 km s^{-1}$

Step 1: Energy balance from surface to infinity: $$E_surface = 12m v^2 - G M_E mR_E = 12m v^2 - 12m v_e^2 \quad (since G M_ER_E = 12v_e^2)$$ $$E_∞ = 12m v_∞^2 + 0$$

Step 2: Relate speeds: $$12m v_∞^2 = 12m (v^2 - v_e^2) \implies v_∞ = √v^2 - v_e^2$$ Given v = 3v_e: $$v_∞ = √(3 v_e)^2 - v_e^2 = √9 v_e^2 - v_e^2 = √8 v_e^2 = \mathbf{2√2 v_e}$$ $$v_∞ = 2√2 × 11.2 km s^{-1} ≈ 2.8284 × 11.2 = \mathbf{31.68 km s^{-1}}$$

Ex 7.19 Satellite Escape Energy

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

Prerequisite: ↗ 7.10 Binding Energy of Satellite

Answer: Energy Required $E = 5.89 × 10^9 J = 5.89 GJ$

Step 1: Orbital radius: $$r = R_E + h = 6.4 × 10^6 + 0.4 × 10^6 = 6.8 × 10^6 m$$

Step 2: Total Energy of orbiting satellite: $$E_orbit = -G M_E m2 r$$

Step 3: Energy expended to reach infinity ($E_∞ = 0$): $$Δ E = E_∞ - E_orbit = 0 - (-G M_E m2 r) = G M_E m2 r$$ $$Δ E = (6.67 × 10^{-11})(6.0 × 10^{24})(200)2(6.8 × 10^6) = 8.004 × 10^{16}13.6 × 10^6 = \mathbf{5.89 × 10^9 J}$$

Ex 7.20 Head-on Colliding Binary Stars

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

Prerequisite: ↗ 7.7 Two-Body Conservation of Energy

Answer: Collision speed $v ≈ 2.58 × 10^6 m s^{-1} = 2580 km s^{-1}$

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

Step 2: Collision State ($r_2 = 2R = 2 × 10^7 m$): By symmetry, both identical stars have equal speed $v$: $$E_f = 12M v^2 + 12M v^2 - G M^22R = M v^2 - G M^22R$$

Step 3: Energy Conservation ($E_i = E_f$): $$M v^2 - G M^22R = 0 \implies v^2 = G M2R$$ $$v = √G M2R = √(6.67 × 10^{-11})(2 × 10^{30})2(10^7) = √1.334 × 10^{20}2 × 10^7 = \sqrt{6.67 × 10^{12}}$$ $$v ≈ \mathbf{2.58 × 10^6 m s^{-1}} = \mathbf{2580 km s^{-1}}$$

Ex 7.21 Equilibrium & Potential at Midpoint

7.21 Two heavy spheres each of mass $100 kg$ and radius $0.10 m$ are placed $1.0 m$ 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 $V = -2.67 × 10^{-8} J kg^{-1}$ | Equilibrium is UNSTABLE

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

Step 2: Gravitational Potential at midpoint: Potential is a scalar, so potentials add algebraically: $$V = V_1 + V_2 = - G M0.5 - G M0.5 = - 2 G M0.5 = - 4 G M$$ $$V = - 4(6.67 × 10^{-11})(100) = \mathbf{- 2.67 × 10^{-8} J kg^{-1}}$$

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.

Revision

Review gravitational fields, potential energy, escape speed and satellite orbits.

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: $$F = G M mr^2$$ $G ≈ 6.67 × 10^{-11} N m^2kg^{-2}$ (Universal constant, dimension $[M^{-1}L^3T^{-2}]$).

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: $dAdt = constant$.	Based on the Conservation of Angular Momentum: $dAdt = L2m$. Sector speed is constant.
3rd: Law of Periods	$$T^2 ∝ a^3$$	$T^2 = (4π^2GM_{sun}) a^3$. $a$ is the semi-major axis.

7.4 & 7.5 • Acceleration due to Gravity

Variations in Acceleration due to Gravity ($g$)

At Earth's surface: $g_0 = G M_ER_E^2 ≈ 9.8 m/s^2$. Acceleration decreases as you go above or below the surface.

Variation Factor	Mathematical Formula	Conditions & Tricks
Altitude (exact)	$g(h) = g_0 (R_ER_E + h)^2$	Always valid for any height $h$.
Altitude (approx)	$g(h) ≈ g_0 (1 - 2hR_E)$	Condition: Valid only if $h \ll R_E$ (typically $h < 320 km$).
Depth ($d$)	$g(d) = g_0 (1 - dR_E)$	Always valid. At center ($d=R_E$), gravity $g=0$.
Latitude ($λ$)	$g' = g_0 - ω^2 R_E \cos^2λ$	Due to Earth's rotation. Equator ($λ=0^°$): $g$ is minimum. Poles ($λ=90^°$): $g$ 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: $$v_e = √2GM_ER_E = √2gR_E ≈ 11.2 km/s$$ 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 $h$: $$v_o = √GM_ER_E + h$$ For a satellite close to the surface ($h \ll R_E$), $v_o = √gR_E ≈ 7.9 km/s$.

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

Satellite Energy Relationships

Energy Form	Formula (Radius $r$)	Energy Ratios Shortcut
Kinetic Energy ($K$)	$K = +GMm2r$	Cosmic Ratio: $$E = -K = U2$$ $$E_{binding} = +K = -E$$
Potential Energy ($U$)	$U = -GMmr$
Total Energy ($E$)	$E = -GMm2r$

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 $24 hours$ (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 $35,800 km$ (or orbital radius $r ≈ 6.4 R_E ≈ 42,200 km$) 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.

Tests

Practice tests

Choose a level, answer the questions and submit to check your score.

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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