Gravitation: Class 11 Physics · Chapter 7

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

90 min
1 simulation
26 practice questions
28 MCQs

What you will learn in Gravitation

State Kepler's three laws, derive the law of areas from conservation of angular momentum, and use T² ∝ a³ in calculations
State Newton's law of gravitation in scalar and vector form, add forces by superposition, and use the two shell results
Explain how Cavendish measured G and how G, g and the Earth's radius give the Earth's mass and density
Derive and apply the variation of g with height and with depth
Derive the gravitational potential energy −GMm/r, define gravitational potential and explain why mgh is an approximation
Derive the escape speed and calculate it for the Earth, the Moon and other bodies
Derive the orbital speed, period and energy of a satellite and solve problems on moving between orbits

Original companion text aligned to NCERT Physics Class 11, Chapter 7, pp. 127–143. Not affiliated with or endorsed by NCERT.

7.1 Introduction

A ripe mango drops from its branch. A river runs downhill to the sea and never uphill. A high-jumper who clears two metres is back on the mat in under a second. Behind all three is one pull: the Earth attracts every object near it. In this chapter we study that pull, gravitation (गुरुत्वाकर्षण), and find that it also steers the Moon and the planets.

Galileo Galilei was the first to study falling bodies with careful measurements. Free fall is too quick to time by eye, so he slowed it down by rolling balls along gentle slopes. He concluded that, when air resistance is negligible, every body speeds up towards the ground at the same steady rate, whatever its mass. Today we call that rate the acceleration due to gravity, $g \approx 9.8 m / s^{2} .$ (The famous story of two balls dropped from the Leaning Tower of Pisa may never have happened.)

A second puzzle looked unrelated. Night after night the stars hold their patterns; the constellations your grandparents saw are the ones you see. A handful of bright points do not behave: over weeks they creep across the constellations, and every so often they reverse for a while. The Greeks called them planetai, wanderers. Explaining their paths took nearly two thousand years of models, sketched in Fig. 7.1 and the timeline below.

2/3To copy the backward loops, each planet rides a small circle (an epicycle) carried round a big one.

Fig. 7.1Two ways of placing the centre of the solar system.NCERT p. 127

Three pictures of the sky. In the first step the Earth sits at the centre and the Sun and a planet move on circles around it. In the second step the planet rides a small circle, an epicycle, whose centre moves along the big circle, so the planet's path traces loops. In the third step the Sun sits at the centre and the Earth and the planet both move on circles around it.

Timeline

From wandering lights to Kepler's laws

c. 150 CE
Ptolemy's Earth-centred model
Claudius Ptolemy of Alexandria put the Earth at the centre and used epicycles to match the planets' loops.
499 CE
Aryabhata's spinning Earth
In the Aryabhatiya, Aryabhata explained the daily turning of the sky as the Earth rotating on its axis; some scholars also read a Sun-centred scheme in his work.
1543
Copernicus puts the Sun at the centre
Nicolaus Copernicus's book, published in the year he died, had the planets circling the Sun.
1576–1597
Tycho Brahe's measurements
The Danish nobleman measured planetary positions without a telescope, more precisely than anyone before him.
1609–1619
Kepler's three laws
Johannes Kepler, Tycho's assistant, found three rules hidden in Tycho's data.
1633
Galileo's trial
Galileo, whose telescope had supported the Sun-centred view, was tried by the Roman Inquisition and kept under house arrest.
1687
Newton's Principia
Isaac Newton showed that one law of gravitation explains both falling bodies and Kepler's laws.

Kepler's rules describe how the planets move; Newton's law explains why. We follow the same road: Kepler's laws, Newton's law, the measurement of $G$ , how $g$ changes from place to place, gravitational energy, escape from a planet, and satellites.

Quick check

1 of 2

Galileo concluded that, with air resistance negligible, a heavy stone and a light pebble dropped together from the same height

land togetherland with the stone firstland with the pebble firstland in an order that depends on the height

Show answer

Answer (A) land together

Both have the same acceleration

g

, so they land together. The heavier stone does feel a larger force, but it also has more mass to accelerate: the ratio is the same.

7.2 Kepler's Laws

Kepler's laws are empirical: rules squeezed out of measurements, with no theory behind them yet. Each one turned out to be a clue to the force that holds the planets.

LawKepler's first law (law of orbits)

Every planet moves along an ellipse (दीर्घवृत्त) with the Sun at one focus of that ellipse, not at its centre.

An ellipse is a stretched circle. Choose two fixed points, the foci; the ellipse is the set of all points whose distances from the two foci add up to the same total. The line through the foci cuts the ellipse at the two ends of its long axis. Half of that long axis is the semi-major axis $a$ , the natural measure of an orbit's size, and half the short axis is the semi-minor axis $b$ .

For a planet, the end of the long axis nearest the Sun is its perihelion (उपसौर) and the far end is its aphelion (अपसौर). If $c$ is the distance from the centre to a focus, these distances are $a - c$ and $a + c$ , so $a$ is simply their average. The ratio $e = c / a$ , the eccentricity, says how stretched the ellipse is: $e = 0$ for a circle, close to 1 for a long thin one. A circle is the special ellipse whose foci coincide, with $a$ equal to its radius. The other focus of a planet's orbit is empty: nothing sits there.

4/4With the Sun at one focus, perihelion is the nearest point and aphelion the farthest; $a$ and $b$ are the semi-axes.

Fig. 7.2A loop of thread and two pins draw an ellipse.NCERT Fig. 7.1 · p. 128

Two pins stand on a board with a loop of thread around them. In the first step a pencil pulls the loop tight into a triangle. In the second step the pencil has moved round and the full ellipse is drawn, with the triangle at a new position. In the third step the two straight pieces from the pencil to the pins are labelled r1 and r2; their sum stays the same. In the last step the thread is gone, the Sun sits at the left focus, the left end of the long axis is labelled perihelion, the right end aphelion, and arrows from the centre show the semi-major axis a and the semi-minor axis b.

Try this

Draw an orbit at home

You’ll need

a sheet of thick cardboard
two drawing pins
a $20 c m$ loop of thread
a pencil and a ruler

Predict first

If you move the pins closer together and draw again with the same loop, the new curve will be

more stretchedrounder, closer to a circleexactly the same

Push two drawing pins into the cardboard $8 c m$ apart and drop the loop over them.
Pull the loop tight with the pencil tip and draw all the way round, keeping the thread taut.
Measure the long and short axes of your ellipse and record them.
Repeat with the pins $4 c m$ apart, and then with both pins in the same hole.

Table 7.1Your ellipses

Pin separation	Long axis (cm)	Short axis (cm)
8 cm
4 cm
0 cm (one hole)

Your entries are saved automatically.

LawKepler's second law (law of areas)

In equal times, the line from the Sun to a planet sweeps out equal areas.

So a planet cannot keep a steady pace. Near perihelion the Sun–planet line is short, so to sweep its month's worth of area the planet must travel a long way along the orbit. Near aphelion the line is long, and a short stretch of orbit is enough (Fig. 7.3). A planet is fastest at perihelion and slowest at aphelion. The Earth, for example, moves at about $30.3 k m / s$ in early January, when it is nearest the Sun, and at about $29.3 k m / s$ in early July.

3/3The two wedges have equal areas, so the planet moves fast near the Sun and slowly far away.

Fig. 7.3Equal areas in equal times (orbit drawn more stretched than any planet's).NCERT Fig. 7.2 · p. 128

An elliptical orbit with the Sun at the left focus. In the first step a short, wide wedge near the Sun is shaded; the planet covers a long arc of orbit in one month. In the second step a long, thin wedge on the far side is shaded; in one month the planet covers only a short arc. In the last step both wedges show together with the words equal areas, equal times, and the labels fast near the Sun and slow far away.

LawKepler's third law (law of periods)

A planet's orbital period $T$ and the semi-major axis $a$ of its orbit are related by $T^{2} \propto a^{3}$ : the period squared grows as the size cubed.

Measure $T$ in years and $a$ in astronomical units, where $1 A U = 1.496 \times 10^{11} m$ is the mean Earth–Sun distance. The Earth then has $T = 1$ and $a = 1$ , so the law predicts $T^{2} / a^{3} = 1$ for every planet. Table 7.2 shows that it holds from Mercury out to Neptune, and even for Halley's comet on its long, thin orbit.

Table 7.2Kepler's third law checked in AU and yearsNCERT Table 7.1 · p. 129

Body	$a$ (AU)	$T$ (years)	$T^{2} / a^{3}$
Mercury	0.387	0.241	1.002
Venus	0.723	0.615	1.001
Earth	1.000	1.000	1.000
Mars	1.524	1.881	1.000
Jupiter	5.203	11.86	0.999
Saturn	9.537	29.46	1.001
Uranus	19.19	84.01	0.999
Neptune	30.07	164.8	0.999
Halley's comet	17.83	75.32	1.001

One constant fits all the planets because, as you will prove in Section 7.9, it depends only on the mass of the Sun. Bodies orbiting something else obey $T^{2} \propto a^{3}$ with a different constant: Jupiter's moons share one value, the Earth's satellites another.

SimulationThe inner planets, to scale

The Sun with Mercury, Venus, the Earth and Mars on their orbits drawn to true scale. Pressing Play runs time forward at twenty days per second; the inner planets go round much faster than the outer ones, and the scale toggle switches between true and squeezed distances.

Fig. 7.4Press Play and watch the inner planets lap the outer ones.

Try this

Count Mercury's laps while the Earth goes round once. Does $T = a^{3 / 2}$ with $a = 0.387 A U$ predict about four?
How many Earth years does one lap of Mars take? Check it against ${1.52}^{3 / 2}$ .

Why should equal areas be swept in equal times? The answer is a conservation law you met in Chapter 6.

DerivationWhy the law of areas holds

Put the Sun at the origin. Let the planet, of mass $m$ , be at position $𝐫$ with velocity $𝐯$ . In a short time $Δ t$ it moves $𝐯 Δ t,$ and the thin triangle swept out has area

Δ 𝐀 = \frac{1}{2} 𝐫 \times 𝐯 Δ t

(7.1)

since the size of a cross product is the area of a parallelogram, and the triangle is half of it. With $𝐩 = m 𝐯$ and angular momentum $𝐋 = 𝐫 \times 𝐩,$

\frac{Δ 𝐀}{Δ t} = \frac{𝐫 \times 𝐩}{2 m} = \frac{𝐋}{2 m}

(7.2)

The Sun's pull points along $- 𝐫$ , straight at the Sun. A force along the position vector exerts no torque about the origin, because $𝐫 \times 𝐅 = 0,$ so $𝐋$ stays constant and the rate of sweeping area is constant too. That is the second law. A force like this, always along the line to a fixed point, is a central force. Constant $𝐋$ also keeps $𝐫$ and $𝐯$ in one fixed plane, so every orbit is flat.

Example 7.1NCERT Example 7.1 · p. 129

Mercury at its nearest and farthest

Mercury comes within $4.60 \times 10^{10} m$ of the Sun at perihelion, where it moves at $59.0 k m / s$ , and recedes to $6.98 \times 10^{10} m$ at aphelion. (a) Find its speed at aphelion. (b) Find the semi-major axis of its orbit. (c) A line through the Sun, perpendicular to the long axis, splits the orbit into a perihelion side and an aphelion side. Does Mercury spend more or less than half its year on the perihelion side?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A comet's aphelion distance is 9 times its perihelion distance. Its speed at perihelion is

3 times its speed at aphelion81 times its speed at aphelion9 times its speed at aphelionone-ninth of its speed at aphelion

Show answer

Answer (C) 9 times its speed at aphelion

Angular momentum

m r v

is the same at both ends, so

v_{p} / v_{a} = r_{a} / r_{p} = 9.

Taking a square root (3) or squaring (81) has no basis here; one-ninth gets the ratio upside down.

7.3 Universal Law of Gravitation

Legend says a falling apple set Newton thinking. Whatever the trigger, his key idea was bold: the force that pulls the apple down might also reach out to the Moon and hold it in its orbit, only weaker because the Moon is so far away.

He could check this with numbers known in his day. Treat the Moon's orbit as a circle of radius $R_{m} \approx 3.84 \times 10^{8} m,$ covered in $T \approx 27.3$ days. Its centripetal acceleration is then

a_{m} = \frac{v^{2}}{R_{m}} = \frac{4 π^{2} R_{m}}{T^{2}}

(7.3)

which works out to $a_{m} \approx 2.72 \times 10^{- 3} m / s^{2} .$ Is that what an inverse-square pull predicts? The apple is one Earth radius from the centre and the Moon about sixty, so a pull falling off as $1 / r^{2}$ should be $60^{2}$ times weaker at the Moon:

\frac{g}{a_{m}} = \frac{R_{m}^{2}}{R_{E}^{2}} \approx 3600

(7.4)

And indeed $9.8 / 0.00272 \approx 3600.$ Fig. 7.5 puts it another way: in one second an apple near the ground drops $4.9 m$ , while the Moon drops only about $1.4 m m$ below the straight line it would follow if gravity vanished.

3/3The Moon is 60 times farther from the Earth's centre and falls $60^{2} = 3600$ times less.

Fig. 7.5The same pull, 3600 times weaker at the Moon (gap exaggerated).NCERT p. 130

Two panels. In the first step an apple falls from a tree, an arrow showing it drops 4.9 metres in one second. In the second step the right panel shows the Earth and part of the Moon's orbit; a dashed straight line leaves the Moon, and the Moon's real path curves below it, with a short arrow marking how far it falls. In the last step a banner reads 60 times farther, 3600 times weaker.

LawNewton's law of universal gravitation

Any two particles attract each other. The force acts along the line joining them; its size is proportional to each of the two masses and falls off as the inverse square of their separation.

Law of gravitation

F = G \frac{m_{1} m_{2}}{r^{2}}

(7.5)

where

$G$: the universal gravitational constant
$r$: the distance between the two particles

Newton published the law in 1687 in his Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy). The bold word is universal: one law, with one constant $G$ , for apples, moons, planets and, as far as we can tell, galaxies.

In vector form, let $𝐫$ be the position of $m_{2}$ relative to $m_{1}$ , with unit vector $\hat{𝐫} = 𝐫 / r$ pointing from $m_{1}$ to $m_{2}$ . The force on $m_{2}$ due to $m_{1}$ points back towards $m_{1}$ :

𝐅_{21} = - G \frac{m_{1} m_{2}}{r^{2}} \hat{𝐫}

By Newton's third law the force on $m_{1}$ due to $m_{2}$ is $𝐅_{12} = - 𝐅_{21} .$ The Earth pulls a $1 k g$ mango with about $9.8 N$ , and the mango pulls the whole Earth with $9.8 N$ too; on the Earth's huge mass that force produces an acceleration far too small to notice.

PrinciplePrinciple of superposition

The net gravitational force on a particle is the vector sum of the forces that each of the other particles exerts on it, each pair interacting as though the others were absent.

𝐅_{1} = \sum_{i = 2}^{n} \frac{G m_{1} m_{i}}{r_{1 i}^{2}} {\hat{𝐫}}_{1 i}

where

${\hat{𝐫}}_{1 i}$: the unit vector from $m_{1}$ towards $m_{i}$

Example 7.2NCERT Example 7.2 · p. 130

Adding gravitational forces

(a) Three small lead spheres lie on a table: A ( $10 k g$ ) at the corner of a right angle, B ( $10 k g$ ) $0.40 m$ from A along the $x$ -axis, and C ( $20 k g$ ) $0.30 m$ from A along the $y$ -axis. Find the net gravitational force on A. (b) Six equal masses $M$ sit at the corners of a regular hexagon, each a distance $d$ from the centre, where a particle of mass $m$ rests. One corner mass is removed. Find the net force on $m$ now.

Try it yourself first. Then reveal the solution one step at a time.

Newton's law is about particles. A real body is a crowd of particles, each pulling in its own direction, so the forces must be added as vectors, a job for integral calculus. For a uniform spherical shell the result is remarkably simple.

TheoremShell theorem

(1) For a particle outside a thin uniform spherical shell, the shell pulls exactly like a point mass, equal to the shell's mass, placed at its centre. (2) For a particle anywhere inside such a shell, the shell's net pull is zero.

Outside. Pair every small patch of the shell with its mirror image across the line from the particle to the centre. The sideways parts of their pulls cancel, leaving parts along that line (Fig. 7.6). Adding all such pairs exactly, with calculus, gives $G M m / r^{2}$ towards the centre, where $M$ is the shell's mass and $r$ the particle's distance from the centre.

Inside. Draw a narrow double cone with its tip at the particle. It cuts two patches from the shell. The far patch is larger, its area growing as the square of its distance, but its pull weakens as the inverse square of the same distance. The two effects cancel exactly, so the near and far patches pull equally hard in opposite directions. Every direction pairs up like this, so the net force is zero everywhere inside.

A solid ball whose density depends only on the distance from its centre is a nest of such shells, so outside it, too, acts like a point mass at its centre.

2/4Their sideways pulls cancel; what is left points at the centre.

Fig. 7.6The two shell results, by symmetry.NCERT p. 131

A thin spherical shell. In the first step a particle m outside the shell is pulled by two matching patches, one above and one below the line to the centre. In the second step the up and down parts of those pulls are shown cancelling, and a single arrow points from m towards the centre. In the third step the particle is inside the shell; a narrow double cone through it cuts a small near patch and a large far patch, which pull it in opposite directions. In the last step the label says the pulls cancel and the force is zero.

7.4 The Gravitational Constant

Kepler's laws and the Moon test fix how gravity depends on mass and distance, but they cannot supply the number $G$ . Astronomy only ever measures the product $G M$ for a central body. To find $G$ alone you must measure the pull between two known masses in a laboratory, a force so feeble that you need great care to detect it at all.

Henry Cavendish managed it in 1798 with a torsion balance (Fig. 7.7). Its heart is a light horizontal rod, hung at its middle by a thin wire, with a small lead ball fixed to each end. Heavy lead balls are then set near the small ones: one by the left end on one side of the rod, the other by the right end on the far side. Each heavy ball tugs its neighbour, the two tugs point in opposite directions at opposite ends, and together they make a couple. The rod feels no net force, only a torque that twists the wire through a tiny angle.

2/3The large spheres pull opposite ends of the rod opposite ways: the wire twists and the spot moves.

Fig. 7.7A torsion balance seen from above (twist exaggerated).NCERT Fig. 7.6 · p. 132

A top view of a torsion balance. A rod with a small sphere at each end hangs from a wire at its centre, which carries a small mirror; a beam of light from the mirror falls on a scale at the top. In the first step there are no large spheres and the light spot is at the middle of the scale. In the second step large spheres sit beside the small ones on opposite sides, the rod twists a little and the light spot moves to one side. In the last step the large spheres are moved to the other sides, the rod twists the other way and the spot moves to the opposite side.

DerivationReading

G

from the twist

Let $M$ and $m$ be the large and small masses, $d$ the distance between the centres of a large sphere and its neighbouring small sphere, and $L$ the length of the rod. Spheres attract as if their masses sat at their centres, so each pair attracts with

F = G \frac{M m}{d^{2}}

(7.6)

The torque of a couple is one force times the distance between the two lines of action, here $F L$ . As the wire twists it pushes back with a torque $κ θ$ that grows with the angle; $κ$ , the torque per radian of twist, is found separately by timing the balance's slow free oscillations. The rod settles where the two torques match:

G \frac{M m}{d^{2}} L = κ θ

(7.7)

so measuring $θ$ gives $G$ . Modern versions fix a small mirror to the wire and read the twist from a spot of light on a distant scale: when the mirror turns through $θ$ , the reflected beam turns through $2 θ$ . Moving the large spheres to the opposite sides reverses the torque and doubles the swing, which also cancels any error in where the rod would rest on its own. The accepted value today is

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

(7.8)

(more precisely $6.674 \times 10^{- 11}$ in these units). It is still the least precisely known of the fundamental constants.

Example

How small is the twist?

In a classroom torsion balance, $M = 1.50 k g$ , $m = 20.0 g$ , $d = 5.00 c m$ , the rod length $L = 15.0 c m$ and the torsion constant $κ = 2.00 \times 10^{- 8} N m / r a d .$ When the large spheres are put in place, the light spot on a scale $2.00 m$ from the mirror moves $2.40 c m$ . Find $G$ , and the force between one pair of spheres.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Why are the two large spheres placed on opposite sides of the rod?

to double the pull on each small sphereto make a couple: a torque on the rod with no net forceto cancel the Earth's pull on the rodto stop the wire from stretching

Show answer

Answer (B) to make a couple: a torque on the rod with no net force

Opposite forces at the two ends give zero net force but a torque

F L

, which twists the wire. Putting both spheres on the same side would push the rod sideways instead of twisting it.

7.5 Acceleration due to Gravity of the Earth

Shell result (1) applies to every layer of an onion-like Earth at once. Stand anywhere above the ground and each layer, from the core out to the crust, pulls as though its mass sat at the centre. Add up the layers and the whole Earth behaves like one particle of mass $M_{E}$ at its centre.

Underground the bookkeeping changes. Go down a mine to a point at distance $r$ from the centre (Fig. 7.8). The layers above you now surround you, and by shell result (2) they add nothing to the pull. Only the ball of radius $r$ beneath your feet counts, acting like a particle of mass $M_{r}$ at the centre.

3/3Only the ball of radius $r$ pulls, as if its mass $M_{r}$ were at the centre.

Fig. 7.8Only the mass nearer the centre than you pulls you.NCERT Fig. 7.7 · p. 132

A cut-away circle for the Earth with a mine shaft running in from the surface to a body m at distance r from the centre. In the first step the whole Earth is shaded alike. In the second step the shell outside radius r is greyed out with the note that it gives no net pull. In the last step the inner ball of radius r is highlighted and an arrow points from the body towards the centre, with the note that the inner ball acts as if all its mass were at the centre.

DerivationThe force inside a uniform Earth

The ball beneath the body pulls like a point mass:

F = \frac{G m M_{r}}{r^{2}}

(7.9)

If the Earth's density $ρ$ is the same throughout, mass is proportional to volume:

M_{r} = \frac{4}{3} π r^{3} ρ = M_{E} \frac{r^{3}}{R_{E}^{3}}

F = \frac{G m M_{E}}{R_{E}^{3}} r

(7.10)

Inside a uniform Earth the force grows in proportion to the distance from the centre and vanishes at the centre itself. At the surface, $r = R_{E}$ :

F = \frac{G M_{E} m}{R_{E}^{2}}

(7.11)

By Newton's second law this force equals $m g$ , so the acceleration due to gravity at the surface is

g = \frac{F}{m} = \frac{G M_{E}}{R_{E}^{2}}

(7.12)

There is no trace of $m$ : every body falls with the same $g$ , just as Galileo found.

The force per unit mass at a point, $𝐅 / m$ , is called the gravitational field (or gravitational intensity) there. Near the Earth it points towards the centre, and at the surface its size is $g$ . Its unit is $N / k g$ , which is the same as $m / s^{2}$ .

With $G$ known, Eq. (7.12) turns two easy measurements, $g = 9.8 m / s^{2}$ and $R_{E} = 6.37 \times 10^{6} m,$ into the mass of the planet: $M_{E} = g R_{E}^{2} / G \approx 6.0 \times 10^{24} k g .$ Hence the saying that Cavendish 'weighed the Earth'. Dividing by the Earth's volume gives its mean density:

\begin{aligned} ρ & = \frac{3 g}{4 π G R_{E}} \\ \approx 5.5 \times 10^{3} k g / m^{3} \end{aligned}

Surface rocks have only about half this density, roughly $2.7 \times 10^{3} k g / m^{3},$ so the deep interior must be much denser: an early, strong clue to the Earth's iron core.

Quick check

1 of 2

Mars has mass

6.42 \times 10^{23} k g

and radius

3.39 \times 10^{6} m .

Find

g

at its surface, in

m / s^{2}

m/s²

Show answer

Answer 3.73 m/s²

g = G M / R^{2} = (6.67 \times 10^{- 11}) (6.42 \times 10^{23}) / (3.39 \times 10^{6})^{2} = 3.73 m / s^{2},

a little over a third of the Earth's.

7.6 Acceleration due to Gravity below and above the Surface of Earth

Equation (7.12) gives $g$ at the surface. How does it change as you climb a mountain, fly in a plane or go down a mine?

DerivationAt a height

h

above the surface

At height $h$ the body is outside the Earth, a distance $R_{E} + h$ from the centre:

F (h) = \frac{G M_{E} m}{(R_{E} + h)^{2}}

(7.13)

\begin{aligned} g (h) & = \frac{G M_{E}}{(R_{E} + h)^{2}} \\ = \frac{g}{(1 + h / R_{E})^{2}} \end{aligned}

(7.14)

For heights small compared with the Earth's radius ( $h ≪ R_{E}$ ), the binomial approximation $(1 + x)^{- 2} \approx 1 - 2 x$ for $∣ x ∣ ≪ 1$ gives

g (h) \approx g (1 - \frac{2 h}{R_{E}})

(7.15)

DerivationAt a depth

d

below the surface

At depth $d$ the body is $R_{E} - d$ from the centre. The shell of thickness $d$ above it exerts no net force, and the inner ball of radius $R_{E} - d$ pulls as if its mass $M_{s}$ were at the centre. With uniform density,

\frac{M_{s}}{M_{E}} = \frac{(R_{E} - d)^{3}}{R_{E}^{3}}

(7.16)

F (d) = \frac{G M_{s} m}{(R_{E} - d)^{2}}

(7.17)

F (d) = \frac{G M_{E} m}{R_{E}^{3}} (R_{E} - d)

(7.18)

\begin{aligned} g (d) & = \frac{F (d)}{m} \\ = g (1 - \frac{d}{R_{E}}) \end{aligned}

(7.19)

No approximation was needed this time: for a uniform Earth the result is exact.

Taken together (Fig. 7.9), the two results make the surface a peak: move away from it in either direction, up into the sky or down into the ground, and $g$ gets smaller. Close to the surface the drop is twice as fast going up, so a small height $h$ costs as much $g$ as a depth $2 h$ .

3/3Just above the surface the curve falls twice as steeply as the inside line rises.

Fig. 7.9How

g

varies with distance

r

from the Earth's centre (uniform Earth).NCERT Fig. 7.8 · p. 134

A graph of g, as a fraction of its surface value, against distance from the centre in units of the Earth's radius, from 0 to 4. In the first step a straight line rises from zero at the centre to 1 at the surface. In the second step a curve falls from 1 at the surface to one-sixteenth at four radii, labelled falls as one over r squared. In the last step a dashed tangent at the surface shows the outside curve falling twice as steeply as the inside line rises.

Example

Where does g drop by 1 %?

Take $R_{E} = 6370 k m .$ (a) At what height above the surface is $g$ 1 % less than at the surface? (b) At what depth? (c) By what percentage is $g$ smaller at the summit of Mount Everest, $8.85 k m$ above sea level?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

At what height above the surface is

g

one quarter of its surface value? Take

R_{E} = 6400 k m

and give the height in km.

Show answer

Answer 6400 km

g (h) / g = 1 / (1 + h / R_{E})^{2} = 1 / 4,

1 + h / R_{E} = 2

and

h = R_{E} = 6400 k m .

The approximate formula (7.15) cannot be used:

h

is not small.

7.7 Gravitational Potential Energy

Lift a book onto a shelf and the work you do is not lost: let the book fall and gravity returns it as kinetic energy. Energy stored like this, because of position, is potential energy. It can be defined only for conservative forces, whose work between two points is the same along every route. Gravity is conservative, so it has a potential energy.

For a lift of a few metres the pull hardly changes, so treat it as a constant $m g$ . Raising a body of mass $m$ from height $h_{1}$ to $h_{2}$ then takes work

W_{12} = m g (h_{2} - h_{1})

(7.20)

Any function whose difference reproduces this work can serve as the potential energy. The simplest is the one below, and it does give the work as a difference:

U (h) = m g h + U_{0}

(7.21)

W_{12} = U (h_{2}) - U (h_{1})

(7.22)

The additive constant $U_{0}$ is free: it is just the value you choose for ground level, and most problems set it to zero.

DerivationPotential energy far from the Earth

Over large distances the full inverse-square force must be used. For $r \geq R_{E}$ it is

F = \frac{G M_{E} m}{r^{2}}

(7.23)

Raise the body slowly from $r_{1}$ to a larger distance $r_{2}$ . The agent's force balances gravity at every point on the way, so the work it does is

W_{12} = \int_{r_{1}}^{r_{2}} \frac{G M_{E} m}{r^{2}} d r

W_{12} = G M_{E} m (\frac{1}{r_{1}} - \frac{1}{r_{2}})

(7.24)

A function whose difference $U (r_{2}) - U (r_{1})$ gives exactly this is

U (r) = - \frac{G M_{E} m}{r} + U_{\infty}

(7.25)

with a free constant $U_{\infty}$ , the value at infinite distance. Only differences are ever measured, so we may fix it; the universal choice $U_{\infty} = 0$ gives

U (r) = - \frac{G M_{E} m}{r}

for $r \geq R_{E}$ . Read it as a statement about work: bringing the body in slowly from infinity, the agent holds it back against a pull that would speed it up, so the agent's work, $U (r) - 0$ , is negative. To pull the body away again you must supply $G M_{E} m / r$ . Moving the zero would shift every value of $U$ by the same constant and change no force and no energy difference.

Where does $m g h$ come from, then? Raise a body from the surface to a small height $h$ and use Eq. (7.24) with $r_{1} = R_{E}$ and $r_{2} = R_{E} + h$ :

\begin{aligned} Δ U & = \frac{G M_{E} m h}{R_{E} (R_{E} + h)} \\ \approx \frac{G M_{E} m}{R_{E}^{2}} h = m g h \end{aligned}

for $h ≪ R_{E}$ . So $m g h$ is an approximation to a difference of the exact potential energy. It is excellent for a shelf or a mountain and badly wrong at satellite heights (Fig. 7.10).

3/3Close to the surface the curve is nearly straight, with slope $m g$ : that is where $m g h$ works.

Fig. 7.10

U = - G M_{E} m / r

: zero at infinity, most negative at the surface.NCERT p. 135

A graph of potential energy against distance r from the Earth's centre, with the zero line at the top and the Earth's interior shaded on the left. In the first step the curve starts deep and negative at the surface and rises with r. In the second step a note shows the curve flattening towards zero far away. In the last step a short straight tangent at the surface shows that, close to the surface, the curve is almost a straight line.

Definition

Gravitational potential

The gravitational potential at a point is the gravitational potential energy per unit mass of a body placed there: the work done per kilogram in bringing a body slowly from infinity to that point.

ExampleNear a mass $M$ , $V = - G M / r$ ; its SI unit is $J / k g$ .

Gravitational potential energy belongs to a pair of bodies. For two particles $m_{1}$ and $m_{2}$ a distance $r$ apart, with zero at infinite separation,

U = - \frac{G m_{1} m_{2}}{r}

For several particles, add one such term for every pair, by superposition: three particles have three pairs, four have six.

Example 7.3NCERT Example 7.3 · p. 135

Three masses on a triangle

Three particles, each of mass $2.0 k g$ , sit at the corners of an equilateral triangle of side $0.50 m$ . (a) Find the gravitational potential energy of the system. (b) Find the gravitational potential at the centroid of the triangle. (c) How much energy is released when a fourth particle of mass $1.0 k g$ is brought from far away to the centroid?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

With

U = 0

at infinity, the gravitational potential energy of a satellite in orbit is

positivezeronegativepositive or negative, depending on its speed

Show answer

Answer (C) negative

U = - G M m / r

is negative at every finite distance. The speed affects the kinetic energy, not

U

7.8 Escape Speed

A cricket ball skied high comes down within seconds; a probe sent to Mars never returns. Somewhere between the two lies a threshold launch speed, and conservation of energy finds it in a few lines.

DerivationThe least speed that never returns

Launch a body of mass $m$ with speed $v_{i}$ from a point a distance $R_{E} + h$ from the Earth's centre, ignoring air resistance and every other body. Ask what is left if it ever gets infinitely far away, where its potential energy is zero and its speed is some $v_{f}$ . Mechanical energy is conserved, so

\begin{aligned} \frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{R_{E} + h} \\ = \frac{1}{2} m v_{f}^{2} \end{aligned}

(7.28)

The right-hand side can never be negative, so the body reaches infinity only if

\frac{1}{2} m v_{i}^{2} - \frac{G M_{E} m}{R_{E} + h} \geq 0

(7.29)

The smallest launch speed makes the left side exactly zero, so that the body creeps out to infinity with its speed tending to zero:

\frac{1}{2} m (v_{i})_{\min}^{} = \frac{G M_{E} m}{R_{E} + h}

(7.30)

For a launch from the surface ( $h = 0$ ) this minimum is the escape speed (पलायन चाल):

v_{e} = \sqrt{\frac{2 G M_{E}}{R_{E}}}

(7.31)

and since $G M_{E} = g R_{E}^{2}$ by Eq. (7.12),

v_{e} = \sqrt{2 g R_{E}}

(7.32)

With $g = 9.8 m / s^{2}$ and $R_{E} = 6.37 \times 10^{6} m,$ $v_{e} = 11.2 k m / s,$ about $40 000$ km/h. The mass $m$ cancelled, so a pebble and a spacecraft need the same speed. The launch direction did not enter either: energy is a scalar, so any direction works as long as the path does not run into the ground (and we ignore the air). That is why careful books say escape speed rather than escape velocity.

2/3With $E = 0$ it keeps climbing; its speed reaches zero only at infinity. This is the escape speed.

Fig. 7.11Total energy decides whether a launched body returns.NCERT p. 136

The curve of potential energy against distance from the Earth's centre, starting deep at the surface and rising towards zero. In the first step a horizontal line for a negative total energy meets the curve at a turning point, and a bar at the surface shows the kinetic energy at launch: the body turns back. In the second step the total-energy line lies on the zero axis: the body just escapes, with its speed reaching zero only at infinity. In the last step the line lies above zero: the body escapes and keeps some kinetic energy far away.

The same formula works for any planet or moon. The Moon, with $g = 1.6 m / s^{2}$ and radius $1740 k m$ , has an escape speed of only about $2.4 k m / s$ .

Example 7.4NCERT Example 7.4 · p. 136

A launch between two spheres

Two uniform spheres are held fixed with their centres $10 R$ apart: A has mass $M$ and radius $R$ ; B has mass $16 M$ and radius $2 R$ . A small probe is fired from the surface of A straight towards the centre of B. (a) Where between them is the net force on the probe zero? (b) Find the least launch speed with which the probe reaches B. (c) With that launch speed, how fast does it hit B?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A planet has the same mean density as the Earth but twice its radius. Its escape speed is how many times the Earth's?

Show answer

Answer 2

v_{e} = \sqrt{2 G M / R}

with

M = \frac{4}{3} π R^{3} ρ

gives

v_{e} \propto R \sqrt{ρ},

so doubling

R

doubles

v_{e}

. Answering

\sqrt{2}

forgets that the mass grows as

R^{3}

7.9 Earth Satellites

Newton pictured a cannon on a very high mountain firing horizontally (Fig. 7.12). A slow ball curves down and lands nearby. A faster one lands farther away, because the ground curves away beneath it. At one particular speed the ball falls just as fast as the ground curves away, so it never lands: it is in orbit, falling round the Earth for ever. Faster still, the orbit stretches into an ellipse, and at the escape speed the ball never returns.

Anything that falls round a planet like this is a satellite (उपग्रह). The Moon has done it for billions of years, taking about $27.3$ days per lap; it also turns once on its axis per lap, so one face always looks at us. The first artificial satellite, Sputnik 1, went up in 1957, and today thousands relay phone calls and television, guide navigation, and watch weather and crops. Kepler's laws carry over unchanged, with the Earth in place of the Sun: satellite orbits are circles or ellipses with the Earth's centre at a focus.

4/5A little faster than $v_{o}$ and the orbit becomes an ellipse that swings far out.

Fig. 7.12Newton's cannon (mountain and speeds hugely exaggerated).NCERT p. 137

The Earth with a tall mountain on top and a cannon firing horizontally to the right. Each step adds a path at a higher launch speed and earlier paths fade. In the first step the ball curves down and lands close to the mountain. In the second it lands about a quarter of the way round. In the third it goes all the way round on a circle just above the mountain top. In the fourth it follows an ellipse that swings far below the Earth before returning. In the last step the path opens out and leaves the picture: the ball escapes.

DerivationSpeed and period in a circular orbit

A satellite of mass $m$ moves with speed $v$ on a circle of radius $R_{E} + h$ , where $h$ is its height above the surface. Moving in a circle needs a centripetal force

F_{c} = \frac{m v^{2}}{R_{E} + h}

(7.33)

directed towards the centre, and gravity supplies exactly this:

F_{g} = \frac{G M_{E} m}{(R_{E} + h)^{2}}

(7.34)

Setting $F_{c} = F_{g}$ and cancelling $m$ gives the orbital speed:

v = \sqrt{\frac{G M_{E}}{R_{E} + h}}

(7.35)

Higher orbits are slower. For an orbit skimming the surface, with air ignored and $h$ set to zero, this becomes

\begin{aligned} v_{o} & = \sqrt{\frac{G M_{E}}{R_{E}}} \\ = \sqrt{g R_{E}} \end{aligned}

(7.36)

which is about $7.9 k m / s$ . Compare Eq. (7.31): $v_{e} = \sqrt{2} v_{o}$ , so escaping takes only 41 % more speed than skimming round the Earth. One lap covers $2 π (R_{E} + h)$ , so the period is

\begin{aligned} T & = \frac{2 π (R_{E} + h)}{v} \\ = \frac{2 π (R_{E} + h)^{3 / 2}}{\sqrt{G M_{E}}} \end{aligned}

(7.37)

Squaring both sides,

\begin{aligned} T^{2} & = k (R_{E} + h)^{3}, \\ k & = \frac{4 π^{2}}{G M_{E}} \end{aligned}

(7.38)

This is Kepler's third law for the Earth's satellites. The constant depends only on the central mass; with the Sun's mass $M_{S}$ the same steps give the planets' law $T^{2} = K_{S} a^{3}$ , where $K_{S} = 4 π^{2} / G M_{S}$ . For a satellite skimming the surface,

\begin{aligned} T_{0} & = 2 π \sqrt{\frac{R_{E}}{g}} \\ \approx 84 m i n \end{aligned}

(7.39)

Real low orbits are a few hundred kilometres up, so they take a little longer: the International Space Station, about $420 k m$ high, goes round in about 93 minutes. Why do circle formulas work so well for planets too? Because their ellipses are barely stretched. The Earth's orbit has $e \approx 0.017$ ; drawn to scale it would look perfectly round, with $b$ and $a$ differing by under 0.02 %. Mercury ( $e \approx 0.21$ ) is the clear exception, with Mars ( $e \approx 0.09$ ) a distant second.

InteractiveOrbit launcher

A satellite is launched horizontally from a chosen height above a drawn Earth, at a speed you set with a slider. The predicted path appears at once: it may crash into the Earth, form a circle, form an ellipse or open out and escape. Pressing Play moves the satellite along the path. Readouts show the circular-orbit and escape speeds at that height, the total energy per kilogram, the farthest distance reached and, for closed orbits, the period. Buttons snap the speed to the circular or escape value.

Fig. 7.13Choose a height and a horizontal launch speed, then press Play. The panel names the orbit and gives its energy and period.

Example 7.5NCERT Example 7.5 · p. 137

Weighing Saturn, timing Ceres

(a) Saturn's largest moon, Titan, orbits $1.22 \times 10^{9} m$ from Saturn's centre with a period of $15.9$ days. Estimate Saturn's mass. (b) The dwarf planet Ceres orbits the Sun at 2.77 times the Earth's orbital radius. How long is its year, in Earth days? Treat both orbits as circles.

Try it yourself first. Then reveal the solution one step at a time.

Example 7.6NCERT Example 7.6 · p. 138

Two routes to the Earth's mass

Find the Earth's mass (a) from $g = 9.80 m / s^{2}$ at the surface and $R_{E} = 6.37 \times 10^{6} m,$ and (b) from the International Space Station's orbit, of radius $6.79 \times 10^{6} m$ and period $92.8$ min. Take $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2} .$

Try it yourself first. Then reveal the solution one step at a time.

Example 7.7NCERT Example 7.7 · p. 138

Kepler's constant in handy units

For orbits around the Sun, $T^{2} = K_{S} a^{3}$ with $K_{S} = 4 π^{2} / G M_{S}$ . (a) Find $K_{S}$ in $s^{2} m^{- 3},$ taking $M_{S} = 1.99 \times 10^{30} k g .$ (b) Express it in years² per AU³, using $1 A U = 1.496 \times 10^{11} m$ and $1 y r = 3.156 \times 10^{7} s .$ (c) An asteroid's orbit has a semi-major axis of $4.0 A U$ . Find its period.

Try it yourself first. Then reveal the solution one step at a time.

Kepler's third law in the form $T^{2} = (4 π^{2} / G M_{E}) a^{3}$ covers elliptical orbits too, with $a$ the semi-major axis and the Earth at a focus. A satellite whose distance from the Earth's centre swings between $7000 k m$ and $13 000 k m$ therefore has the same period as one on a circle of radius $10 000 k m$ : about 2.8 hours.

Quick check

1 of 3

Satellite A orbits a planet at radius

r

and satellite B at radius

4 r

. The ratio of their periods

T_{B} / T_{A}

41682

Show answer

Answer (C) 8

T \propto r^{3 / 2}

, so

T_{B} / T_{A} = 4^{3 / 2} = 8.

Answering 4 assumes

T \propto r

; 16 uses

r^{2}

; 2 uses

\sqrt{r}

7.10 Energy of an Orbiting Satellite

DerivationKinetic, potential and total energy

For a circular orbit of radius $r = R_{E} + h$ , Eq. (7.35) gives $v^{2} = G M_{E} / r$ , so the kinetic energy is

\begin{aligned} K & = \frac{1}{2} m v^{2} \\ = \frac{G M_{E} m}{2 r} \end{aligned}

(7.40)

With zero at infinity the potential energy is

U = - \frac{G M_{E} m}{r}

(7.41)

and the total mechanical energy is

\begin{aligned} E & = K + U \\ = - \frac{G M_{E} m}{2 r} \end{aligned}

(7.42)

Three tidy relations follow: $K = - E$ , $U = 2 E$ and $∣ U ∣ = 2 K$ . The total energy is negative: the satellite is bound to the Earth.

A negative total energy has a clear meaning. To free the satellite completely, taking it to infinity with zero speed where $E = 0$ , you must supply $G M_{E} m / 2 r$ . This is its binding energy. A body with $E \geq 0$ would not be a satellite at all: it would escape, as you saw in Section 7.8.

On an ellipse the split between $K$ and $U$ shifts around the orbit: the satellite gains speed as it drops towards perigee, its point nearest the Earth, and loses it as it climbs back out. The total does not shift. It stays at $E = - G M_{E} m / 2 a$ , with the semi-major axis $a$ in place of $r$ , negative as for a circle.

There is a surprise here (Fig. 7.14). Raising a satellite to a higher orbit needs energy, since $E$ becomes less negative. Yet in the higher orbit it moves more slowly: $K$ falls by exactly as much as $E$ rises, while $U$ rises by twice that amount. The energy supplied, together with the kinetic energy given up, all goes into potential energy.

3/3Moving up needs energy equal to the rise in $E$ : one quarter of $G M_{E} m / r$ .

Fig. 7.14Energies in units of

G M_{E} m / r

.NCERT p. 138

Bar charts about a zero line. In the first step, for an orbit of radius r, a kinetic-energy bar rises half a unit above zero, a potential-energy bar falls one unit below and a total-energy bar falls half a unit below. In the second step bars for an orbit of radius 2r appear, each half as long: kinetic energy a quarter of a unit, potential energy minus a half, total energy minus a quarter. In the last step an arrow shows the rise in total energy, a quarter of a unit, which is the energy to be supplied.

Example 7.8NCERT Example 7.8 · p. 139

Raising a satellite

A $1500 k g$ satellite is moved from a circular orbit of radius $7.00 \times 10^{6} m,$ about $630 k m$ up, to one of radius $4.22 \times 10^{7} m .$ Find the changes in its total, kinetic and potential energy. Use $G M_{E} = 3.98 \times 10^{14} m^{3} / s^{2} .$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

For a satellite in a circular orbit, which relation is correct?

E = K

U = - K

E = U

E = - K

Show answer

Answer (D) $E = - K$

K = G M m / 2 r

and

E = - G M m / 2 r

, so

E = - K

U = - K

is a common slip: in fact

U = - 2 K

Recap

Gravitation: key points and summary

Kepler's laws: orbits are ellipses with the Sun at a focus; the Sun–planet line sweeps equal areas in equal times; $T^{2} \propto a^{3}$ (Section 7.2).
The law of areas is angular momentum conservation, $d A / d t = L / 2 m$ , and holds for any central force (Derivation).
Newton's law: $F = G m_{1} m_{2} / r^{2}$ along the line joining two particles; forces from several bodies add as vectors (Section 7.3).
A uniform shell pulls an outside particle as if its mass were at its centre and exerts no force inside; gravity cannot be shielded (Theorem).
Cavendish's torsion balance gave $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ (Section 7.4).
At the surface $g = G M_{E} / R_{E}^{2}$ ; inside a uniform Earth the pull grows in proportion to $r$ (Section 7.5).
$g (h) \approx g (1 - 2 h / R_{E})$ for $h ≪ R_{E}$ and $g (d) = g (1 - d / R_{E});$ $g$ is greatest at the surface (Section 7.6).
$U = - G m_{1} m_{2} / r$ with $U = 0$ at infinity; $V = - G M / r$ ; $m g h$ is the near-surface approximation; for many bodies add over pairs (Section 7.7).
Escape speed $v_{e} = \sqrt{2 G M_{E} / R_{E}} = \sqrt{2 g R_{E}} \approx 11.2 k m / s,$ the same for every mass and direction (Section 7.8).
Circular orbits: $v = \sqrt{G M_{E} / r}$ , $T^{2} = (4 π^{2} / G M_{E}) r^{3};$ near the surface $v_{o} \approx 7.9 k m / s$ and $T_{0} \approx 84 m i n$ (Section 7.9).
Orbiting astronauts feel weightless because they are in free fall, not because gravity is absent (Floating is falling).
Satellite energy: $K = G M_{E} m / 2 r$ , $U = - G M_{E} m / r$ , $E = - G M_{E} m / 2 r$ (or $- G M_{E} m / 2 a$ for an ellipse); bound orbits have $E < 0$ (Section 7.10).

Table 7.3Gravitational quantities at a glanceNCERT p. 140

Quantity	Symbol	Expression	SI unit	Dimensions
Gravitational constant	$G$	$6.67 \times 10^{- 11}$	$N m^{2} k g^{- 2}$	$[M^{- 1} L^{3} T^{- 2}]$
Gravitational potential energy	$U$	$- G m_{1} m_{2} / r$	$J$	$[M L^{2} T^{- 2}]$
Gravitational potential	$V$	$- G M / r$	$J k g^{- 1}$	$[L^{2} T^{- 2}]$
Gravitational field (intensity)	$𝐠$	$- (G M / r^{2}) \hat{𝐫}$	$N k g^{- 1}$	$[L T^{- 2}]$
Escape speed	$v_{e}$	$\sqrt{2 G M / R}$	$m s^{- 1}$	$[L T^{- 1}]$
Orbital speed	$v$	$\sqrt{G M / r}$	$m s^{- 1}$	$[L T^{- 1}]$

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7.1 Introduction2 questions
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7.6 Acceleration due to Gravity below and above the Surface of Earth3 questions
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Glossary · 14

Key terms (English–Hindi)

acceleration due to gravityगुरुत्वीय त्वरणgurutviya tvaran: The acceleration $g$ of a freely falling body; at the Earth's surface $g = G M_{E} / R_{E}^{2} \approx 9.8 m / s^{2} .$ First seen in the chapter →

binding energyबंधन ऊर्जाbandhan oorja: The energy needed to free a bound body completely; for a satellite in a circular orbit it is $G M m / 2 r$ .First seen in the chapter →

central forceकेंद्रीय बलkendriya bal: A force that always points along the line joining a body to a fixed centre.First seen in the chapter →

ellipseदीर्घवृत्तdeerghvritt: A closed curve whose points all have the same sum of distances from two fixed points, the foci.First seen in the chapter →
escape speedपलायन चालpalayan chaal: The least launch speed with which a body can leave a planet's surface and never return, $v_{e} = \sqrt{2 G M / R} .$ First seen in the chapter →

gravitationगुरुत्वाकर्षणgurutvakarshan: The attraction that every mass exerts on every other mass.First seen in the chapter →
gravitational constantगुरुत्वीय नियतांकgurutviya niyatank: The constant $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$ in Newton's law of gravitation; it is the same everywhere.
gravitational potentialगुरुत्वीय विभवgurutviya vibhav: Gravitational potential energy per unit mass at a point, $V = - G M / r$ near a mass $M$ , in $J / k g$ .First seen in the chapter →
gravitational potential energyगुरुत्वीय स्थितिज ऊर्जाgurutviya sthitij oorja: The work done in bringing masses together slowly from infinite separation; for two masses $U = - G m_{1} m_{2} / r$ .First seen in the chapter →

orbital speedकक्षीय चालkakshiya chaal: The speed of a satellite in a circular orbit of radius $r$ , $v = \sqrt{G M / r}$ .First seen in the chapter →

perihelionउपसौरupsaur: The point of a planet's orbit nearest the Sun; the farthest point is the aphelion.First seen in the chapter →

satelliteउपग्रहupagrah: A body that orbits a larger body, such as the Moon or an artificial satellite around the Earth.First seen in the chapter →
semi-major axisअर्ध-दीर्घ अक्षardh-deergh aksh: Half the longest diameter of an ellipse, written $a$ ; for an orbit it is the average of the nearest and farthest distances.First seen in the chapter →

weightlessnessभारहीनताbharheenta: The state of a freely falling body, which feels no support force, although gravity still acts on it.First seen in the chapter →

Gravitation: Class 11 Physics · Chapter 7

What you will learn in Gravitation

7.1 Introduction

7.2 Kepler's Laws

7.3 Universal Law of Gravitation

7.4 The Gravitational Constant

7.5 Acceleration due to Gravity of the Earth

7.6 Acceleration due to Gravity below and above the Surface of Earth

7.7 Gravitational Potential Energy

7.8 Escape Speed

7.9 Earth Satellites

7.10 Energy of an Orbiting Satellite

Gravitation: key points and summary

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