PhysicsCBSE Class 11 · Chapter 7

Gravitation Class 11: Notes, Satellites and Solved Numericals

Q: Are geostationary satellites and weightlessness in the 2026-27 syllabus?

No. Geostationary and polar satellites and weightlessness were removed in the NCERT rationalisation. The syllabus keeps orbital velocity and the energy of an orbiting satellite.

Q: Why is gravitational potential energy negative?

We choose U = 0 when the masses are infinitely far apart. Gravity is attractive, so bringing them closer releases energy and U falls below zero. A negative energy means the system is bound.

Q: Does escape speed depend on the angle of projection?

No (ignoring air resistance). Escape depends only on having enough kinetic energy, and energy is a scalar, so any direction works as long as the path does not hit the planet.

Q: Why is g zero at the centre of the Earth?

At the centre the mass of the Earth surrounds you equally in all directions, so the pulls cancel. Formally, g(d) = g(1 - d/R) gives zero at d = R.

Q: How does this chapter connect to Class 12?

Gravitation and electrostatics share the inverse-square law. Field, potential (-GM/r vs kq/r), potential energy and the shell result all reappear for charges in Class 12 Electric Charges and Fields and Electrostatic Potential.

Updated 14 October 2026 · Fact-checked

Gravitation is Chapter 7 of CBSE Class 11 Physics. It covers Kepler's three laws of planetary motion, Newton's universal law of gravitation, the acceleration due to gravity and its variation with altitude and depth, gravitational potential energy and potential, escape speed, orbital speed of a satellite and the energy of an orbiting satellite.

What Gravitation is about

The same force that pulls an apple to the ground keeps the Moon in its orbit. Newton's insight was that gravity is universal: every pair of masses attracts with a force $F = \frac{G m _{1} m _{2}}{r ^{2}}$ . This chapter starts with Kepler's laws, which describe planetary orbits (ellipses, equal areas in equal times, and $T^{2} \propto a^{3}$ ), and shows how Newton's law explains them.

Next comes g, the acceleration due to gravity, $g = \frac{G M _{E}}{R _{E}^{2}}$ , and how it falls as you go up ( $g (1 - \frac{2 h}{R})$ for small heights) or down ( $g (1 - \frac{d}{R})$ ) into the Earth. Because gravity weakens with distance, the potential energy is not simply mgh far from the Earth; instead $U = - \frac{GM m}{r}$ , and this leads to the escape speed $2 g R \approx 11.2$ km s $^{- 1}$ .

Finally you study satellites: the orbital speed $GM / r$ , the period, and the energy of an orbiting satellite, which is negative because it is bound. Geostationary satellites and weightlessness are not in the 2026-27 syllabus. The inverse-square law here has the same form as Coulomb's law in Class 12, so many results (potential, field, energy) carry over directly.

Weightage: This chapter is part of Unit VI, Gravitation. The syllabus groups Unit VI with Unit IV, Work, Energy and Power and Unit V, Motion of System of Particles and Rigid Body: together they carry 17 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit VI Gravitation

Key concepts

Kepler's laws of planetary motion

NCERT section 7.2

From Tycho Brahe's careful observations, Kepler found three laws:

1. Law of orbits: every planet moves in an ellipse with the Sun at one focus. The nearest point is the perihelion, the farthest the aphelion; the semi-major axis a is half their sum of distances.
2. Law of areas: the line joining the Sun to a planet sweeps out equal areas in equal times. So a planet moves faster when it is nearer the Sun.
3. Law of periods: the square of the period is proportional to the cube of the semi-major axis: $T^{2} \propto a^{3}$ .

Why the law of areas holds: gravity on the planet points along the line to the Sun, so its torque about the Sun is zero and the angular momentum $L = m v r_{⊥}$ is constant. The areal velocity is $\frac{Δ A}{Δ t} = \frac{L}{2 m}$ , a constant. At perihelion and aphelion (velocity perpendicular to radius), $v_{p} r_{p} = v_{a} r_{a}$ .

Worked example

A planet's average distance from the Sun is 4 times that of the Earth. Find its period in Earth years. [1 mark]

Try it first, then show the solutionSolution

$T^{2} \propto a^{3} \Rightarrow \frac{T}{1 y e a r} = (\frac{4}{1})^{3/2} = 8$
T = 8 years

Answer: 8 years

Newton's universal law of gravitation and G

NCERT section 7.3 and 7.4

Every particle attracts every other particle with a force along the line joining them, proportional to the product of their masses and inversely proportional to the square of their separation:
$F = \frac{G m _{1} m _{2}}{r ^{2}}$

In vector form the force on $m_{2}$ is $F = - \frac{G m _{1} m _{2}}{r ^{2}} \overset{r}{^}$ (attractive, towards $m_{1}$ ).

$G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$ , a universal constant, first measured by Cavendish with a torsion balance.
Gravitational forces obey the principle of superposition: the force from several masses is the vector sum.
A uniform spherical shell (or sphere) attracts an outside particle as if all its mass were at its centre; a shell exerts no net force on a particle inside it.
Gravity is the weakest fundamental force but always attractive and long-range, so it dominates on the scale of planets.

Worked example

Find the gravitational force between two uniform 100 kg spheres whose centres are 1 m apart. [1 mark]

Try it first, then show the solutionSolution

$F = \frac{6.67 \times 1 0 ^{- 11} \times 100 \times 100}{1 ^{2}}$
$F = 6.67 \times 1 0^{- 7} N$ , far too small to notice

Answer: $6.67 \times 1 0^{- 7}$ N

Acceleration due to gravity of the Earth

NCERT section 7.5

Treating the Earth as a uniform sphere of mass $M_{E}$ and radius $R_{E}$ , the force on a mass m at its surface is $\frac{G M _{E} m}{R _{E}^{2}} = m g$ , so
$g = \frac{G M _{E}}{R _{E}^{2}}$

g does not depend on the mass of the falling body: all bodies fall with the same acceleration (ignoring air).

Since g, $R_{E}$ and G can be measured, this 'weighs the Earth': $M_{E} = \frac{g R _{E}^{2}}{G} \approx 6 \times 1 0^{24}$ kg. The same formula gives g on any planet or moon.

Worked example

Taking g = 9.8 m s $^{- 2}$ and $R_{E} = 6.37 \times 1 0^{6}$ m, estimate the mass of the Earth. [2 marks]

Try it first, then show the solutionSolution

$M_{E} = \frac{g R _{E}^{2}}{G} = \frac{9.8 \times ( 6.37 \times 1 0 ^{6} ) ^{2}}{6.67 \times 1 0 ^{- 11}}$
$M_{E} \approx 5.96 \times 1 0^{24} k g$

Answer: About $6.0 \times 1 0^{24}$ kg

Variation of g with altitude and depth

NCERT section 7.6

Above the surface at height h:
$g (h) = \frac{G M _{E}}{( R _{E} + h ) ^{2}} = g (\frac{R _{E}}{R _{E} + h})^{2}$
For $h ≪ R_{E}$ : $g (h) \approx g (1 - \frac{2 h}{R _{E}})$ .

Below the surface at depth d: only the inner sphere of radius $(R_{E} - d)$ attracts (the outer shell gives zero net force). For uniform density its mass is $M_{E} \frac{( R _{E} - d ) ^{3}}{R _{E}^{3}}$ , so
$g (d) = g (1 - \frac{d}{R _{E}})$

So g is greatest at the surface, decreases as you go up (as $1/ r^{2}$ ) and decreases linearly as you go down, becoming zero at the centre.

For small distances, g falls twice as fast going up as going down: a height h causes the same drop as a depth 2h.

Worked example

Find the value of g (a) at a height equal to the Earth's radius, (b) at a depth equal to half the radius, in terms of the surface value g. [2 marks]

Try it first, then show the solutionSolution

(a) $g (R) = g (\frac{R}{2 R})^{2} = \frac{g}{4}$
(b) $g (R /2) = g (1 - \frac{1}{2}) = \frac{g}{2}$

Answer: (a) g/4 (b) g/2

Gravitational potential energy and potential

NCERT section 7.7

Gravity is a conservative force, so it has a potential energy. Taking U = 0 at infinite separation, the gravitational potential energy of masses M and m at distance r is
$U = - \frac{GM m}{r}$
It is negative: work must be done to pull the masses apart to infinity.

Gravitational potential at a point is the potential energy per unit mass there: $V = - \frac{GM}{r}$ (J kg $^{- 1}$ ).

Near the surface, the change in U from $R_{E}$ to $R_{E} + h$ is $G M_{E} m (\frac{1}{R _{E}} - \frac{1}{R _{E} + h}) \approx m g h$ when $h ≪ R_{E}$ : the familiar mgh is a special case.

For several particles, add the potential energies of every pair: $U = - G \sum \frac{m _{i} m _{j}}{r _{ij}}$ .

Worked example

How much work is needed to lift a 100 kg body from the Earth's surface to a height equal to the Earth's radius? ( $R_{E} = 6.4 \times 1 0^{6}$ m, g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$W = Δ U = - \frac{GM m}{2 R} - (- \frac{GM m}{R}) = \frac{GM m}{2 R} = \frac{m g R}{2}$ (using $GM = g R^{2}$ )
$W = \frac{100 \times 9.8 \times 6.4 \times 1 0 ^{6}}{2} \approx 3.1 \times 1 0^{9} J$

Answer: About $3.1 \times 1 0^{9}$ J (not mgR, which would be double)

Escape speed

NCERT section 7.8

The escape speed is the least speed with which a body must be projected from the surface so that it never returns (reaches infinity with zero speed).

Energy conservation: $\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0$ , so
$v_{e} = \frac{2 GM}{R} = 2 g R$

For the Earth, $v_{e} \approx 11.2$ km s $^{- 1}$ ; it does not depend on the mass of the body or the direction of projection (ignoring air).

On the Moon $v_{e}$ is only about 2.3 km s $^{- 1}$ . Gas molecules there easily reach such speeds, so the Moon has no atmosphere.

Worked example

Calculate the escape speed from the Earth. (g = 9.8 m s $^{- 2}$ , $R_{E} = 6.4 \times 1 0^{6}$ m) [1 mark]

Try it first, then show the solutionSolution

$v_{e} = 2 g R = 2 \times 9.8 \times 6.4 \times 1 0^{6}$
$v_{e} = 1.254 \times 1 0^{8} \approx 1.12 \times 1 0^{4} m s^{- 1}$

Answer: About 11.2 km s $^{- 1}$

Earth satellites: orbital speed and period

NCERT section 7.9

A satellite in a circular orbit of radius $r = R_{E} + h$ is held by gravity, which supplies the centripetal force:
$\frac{m v _{o}^{2}}{r} = \frac{G M _{E} m}{r ^{2}} \Rightarrow v_{o} = \frac{G M _{E}}{R _{E} + h}$

The orbital speed does not depend on the satellite's mass; it decreases with height.
Period: $T = \frac{2 π r}{v _{o}} = 2 π \frac{( R _{E} + h ) ^{3}}{G M _{E}}$ , so $T^{2} \propto r^{3}$ (Kepler's third law).
Close to the surface ( $h ≪ R_{E}$ ): $v_{o} \approx g R_{E} \approx 7.9$ km s $^{- 1}$ and $T \approx 2 π \frac{R _{E}}{g} \approx 85$ minutes.
Near the surface, $v_{e} = 2 v_{o}$ .

The Moon is the Earth's natural satellite; artificial satellites are used for communication, weather and navigation.

Worked example

Find the orbital speed and period of a satellite 600 km above the Earth. ( $G M_{E} = 4.0 \times 1 0^{14}$ N m $^{2}$ kg $^{- 1}$ , $R_{E} = 6.4 \times 1 0^{6}$ m) [2 marks]

Try it first, then show the solutionSolution

$r = 7.0 \times 1 0^{6}$ m; $v_{o} = \frac{4.0 \times 1 0 ^{14}}{7.0 \times 1 0 ^{6}} \approx 7.56 \times 1 0^{3} m s^{- 1}$
$T = \frac{2 π r}{v _{o}} = \frac{2 π \times 7.0 \times 1 0 ^{6}}{7.56 \times 1 0 ^{3}} \approx 5.8 \times 1 0^{3} s \approx 97$ min

Answer: About 7.6 km s $^{- 1}$ ; about 97 minutes

Energy of an orbiting satellite

NCERT section 7.10

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

Kinetic energy $K = \frac{1}{2} m v_{o}^{2} = \frac{GM m}{2 r}$
Potential energy $U = - \frac{GM m}{r}$
Total energy $E = K + U = - \frac{GM m}{2 r}$

So $K = - E$ and $U = 2 E$ . The total energy is negative: the satellite is bound to the Earth. Giving it energy $\frac{GM m}{2 r}$ (its binding energy) would free it.

To move a satellite to a higher orbit you must supply energy, yet it ends up moving slower (K falls while U rises by twice as much). If the total energy were zero or positive, the body would escape.

Worked example

A satellite of mass m orbits at radius r. How much energy is needed to shift it to an orbit of radius 2r? [2 marks]

Try it first, then show the solutionSolution

$E_{1} = - \frac{GM m}{2 r}$ ; $E_{2} = - \frac{GM m}{4 r}$
$Δ E = E_{2} - E_{1} = \frac{GM m}{4 r}$

Answer: $\frac{GM m}{4 r}$

Physics in real life: Gravitation

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

g at a mountain top

Suppose a climber stands on the summit of Mount Everest, about 8.8 km above sea level.

By what percentage is g less than at sea level, and what is its value? ( $R_{E} = 6400$ km, g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Fractional decrease $\approx \frac{2 h}{R _{E}} = \frac{2 \times 8.8}{6400} = 0.00275$ , i.e. 0.275%
$g^{'} = 9.8 \times (1 - 0.00275) \approx 9.77 m s^{- 2}$

Answer: About 0.28% less; about 9.77 m s $^{- 2}$

Memory hook: Going up, g drops by twice h over R.

Kepler's third law

Imagine astronomers discovered a second natural moon of the Earth orbiting at 4 times the Moon's distance.

Suppose the Moon's period is 27.3 days. What would the new moon's period be?

Try it first, then show the solutionSolution

$T \propto r^{3/2}$ , so $\frac{T _{2}}{T _{1}} = 4^{3/2} = 8$
$T_{2} = 8 \times 27.3 \approx 218$ days

Answer: About 218 days

Memory hook: Four times farther, eight times slower to go round.

Escape speed and atmospheres

Suppose a space agency is comparing a launch from the Earth with one from Mercury, where M = $3.3 \times 1 0^{23}$ kg and R = $2.44 \times 1 0^{6}$ m.

Find the escape speed from Mercury and compare it with the Earth's 11.2 km s $^{- 1}$ .

Try it first, then show the solutionSolution

$v_{e} = \frac{2 GM}{R} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 3.3 \times 1 0 ^{23}}{2.44 \times 1 0 ^{6}}$
$v_{e} = 1.8 \times 1 0^{7} \approx 4.2 \times 1 0^{3} m s^{- 1}$ , well under half the Earth's value

Answer: About 4.2 km s $^{- 1}$ ; much easier to launch from Mercury

Memory hook: Smaller planet, smaller escape speed, thinner atmosphere.

Topics in this chapter

Gravitation formula sheet

Kepler's third law: $T^{2} \propto a^{3}$; Circular orbit: $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ .
Areal velocity: $\frac{Δ A}{Δ t} = \frac{L}{2 m}, v_{p} r_{p} = v_{a} r_{a}$
Universal law of gravitation: $F = \frac{G m _{1} m _{2}}{r ^{2}}, G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$
g at the surface: $g = \frac{G M _{E}}{R _{E}^{2}}$
g at height h: $g (h) = g (\frac{R _{E}}{R _{E} + h})^{2} \approx g (1 - \frac{2 h}{R _{E}})$
g at depth d: $g (d) = g (1 - \frac{d}{R _{E}})$
Potential energy and potential: $U = - \frac{GM m}{r}, V = - \frac{GM}{r}$
Escape speed: $v_{e} = \frac{2 GM}{R} = 2 g R \approx 11.2 k m s^{- 1}$
Orbital speed and period: $v_{o} = \frac{GM}{R + h}, T = 2 π \frac{( R + h ) ^{3}}{GM}$
Energy of a satellite: $K = \frac{GM m}{2 r}, U = - \frac{GM m}{r}, E = - \frac{GM m}{2 r}$

How the school exam asks Gravitation

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Kepler's laws	State the laws; explain the law of areas by angular momentum; periods from $T^{2} \propto a^{3}$ ; perihelion and aphelion speeds.	Zero torque → L constant; use ratios $T_{1}^{2} / T_{2}^{2} = a_{1}^{3} / a_{2}^{3}$ .
Newton's law and g	Force between masses; g on a planet with given M and R; mass of the Earth; superposition of forces.	$g = GM / R^{2}$ ; scale with ratios: $g \propto M / R^{2}$ .
Variation of g	g at a height or depth; height and depth with the same g; weight of a body down a mine or up a mountain.	Exact formula for large h, $1 - 2 h / R$ for small h; $1 - d / R$ below.
Potential energy and escape speed	Work to raise a body far up; PE of a system of masses; derive $v_{e}$ ; why the Moon has no atmosphere.	Use $- GM m / r$ (not mgh) for large heights; $v_{e}$ from $K + U = 0$ .
Satellites	Derive $v_{o}$ and T; find them for a given height; energy of an orbit; energy to change orbit; relation $v_{e} = 2 v_{o}$ .	Gravity = centripetal force; $E = - GM m /2 r$ ; changes in orbit are differences of E.

Tricks to remember Gravitation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Up 2h, down d
Going up h: g falls by a fraction $2 h / R$ . Going down d: g falls by $d / R$ . Same drop at height h and depth 2h.
Use it when: Comparing g above and below.
Root 2 between orbit and escape
$v_{o} = g R$ , $v_{e} = 2 g R$ : escape is $2$ times orbit (7.9 → 11.2 km s $^{- 1}$ ).
Use it when: Satellite and escape MCQs.
Satellite energy: K = −E, U = 2E
Same pattern as the Bohr atom in Class 12.
Use it when: Energy-of-orbit questions.
ORB-AREA-PERIOD
Kepler 1: orbits are ellipses. Kepler 2: equal areas. Kepler 3: $T^{2} \propto a^{3}$ .
Use it when: Stating the laws in order.
Mass of the satellite cancels
Orbital speed, period and escape speed never contain the small body's mass.
Use it when: Assertion-reason items.

Gravitation: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: What escape speed depends on · 1 mark · MCQExam level

The escape speed of a body from the Earth's surface depends on:
(a) the mass of the body
(b) the direction of projection
(c) the mass and radius of the Earth
(d) the height of the body

Try it first, then show the solutionSolution

$v_{e} = 2 GM / R$ contains only the planet's mass and radius.

Answer: (c)

Example 2: No air on the Moon · 1 mark · Assertion-reasonExam level

Assertion (A): The Moon has no atmosphere.
Reason (R): The escape speed on the Moon is small, so gas molecules there can easily escape.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

The Moon's escape speed (about 2.3 km s $^{- 1}$ ) is within the speeds of gas molecules at its daytime temperatures, so any atmosphere leaks away.

Answer: (a)

Example 3: g at height R/2 · 2 marks · Very short answerHarder exam level

Find g at a height of R/2 above the Earth's surface (g = 9.8 m s $^{- 2}$ at the surface). Why can you not use $g (1 - 2 h / R)$ here? [2 marks]

Try it first, then show the solutionSolution

$g^{'} = g (\frac{R}{3 R /2})^{2} = \frac{4 g}{9} \approx 4.36 m s^{- 2}$
The approximation needs $h ≪ R$ ; here h = R/2 is not small (it would give g' = 0, which is wrong).

Answer: About 4.36 m s $^{- 2}$

Example 4: Weight up and down · 3 marks · Short answerHarder exam level

A 60 kg astronaut trains in a deep shaft 3200 km below the surface (imagined) and later in a station 3200 km above it. Find the gravitational force on her in each place. (g = 9.8 m s $^{- 2}$ , $R_{E} = 6400$ km) [3 marks]

Try it first, then show the solutionSolution

Depth: $g^{'} = 9.8 (1 - \frac{3200}{6400}) = 4.9 m s^{- 2}$ , force $= 60 \times 4.9 = 294 N$
Height: $g^{'} = 9.8 (\frac{6400}{9600})^{2} = 9.8 \times \frac{4}{9} \approx 4.36 m s^{- 2}$
Force $= 60 \times 4.36 \approx 261 N$

Answer: 294 N below; about 261 N above

Example 5: Satellite at 1000 km · 3 marks · Short answerHarder exam level

A satellite orbits 1000 km above the Earth. Find (a) its orbital speed, (b) its period, (c) how much faster it would need to go at that height to escape. ( $GM = 4.0 \times 1 0^{14}$ N m $^{2}$ kg $^{- 1}$ , $R_{E} = 6.4 \times 1 0^{6}$ m) [3 marks]

Try it first, then show the solutionSolution

(a) r = $7.4 \times 1 0^{6}$ m; $v_{o} = \frac{4.0 \times 1 0 ^{14}}{7.4 \times 1 0 ^{6}} \approx 7.35 k m s^{- 1}$
(b) $T = \frac{2 π \times 7.4 \times 1 0 ^{6}}{7.35 \times 1 0 ^{3}} \approx 6.3 \times 1 0^{3} s \approx 105$ min
(c) Escape speed from that orbit $= 2 \times 7.35 \approx 10.4 k m s^{- 1}$ , so about 3.0 km s $^{- 1}$ more

Answer: (a) about 7.35 km s $^{- 1}$ (b) about 105 min (c) about 3.0 km s $^{- 1}$ more

Example 6: Escape and orbital speed compared · 5 marks · Long answerHarder exam level

(a) Derive the escape speed from a planet of mass M and radius R. (b) Show that for a satellite orbiting just above the surface, $v_{e} = 2 v_{o}$ . (c) A planet has the same density as the Earth but twice its radius. Find its escape speed. (d) Why is it easier to launch rockets eastwards near the equator? (Earth's $v_{e}$ = 11.2 km s $^{- 1}$ ) [5 marks]

Try it first, then show the solutionSolution

(a) $\frac{1}{2} m v_{e}^{2} = \frac{GM m}{R} \Rightarrow v_{e} = \frac{2 GM}{R}$
(b) Just above the surface, $v_{o} = \frac{GM}{R}$ , so $\frac{v _{e}}{v _{o}} = 2$
(c) $M = \frac{4}{3} π R^{3} ρ$ , so $v_{e} = \frac{8 π Gρ}{3} R \propto R$ for the same density
Twice the radius gives twice the escape speed: about 22.4 km s $^{- 1}$
(d) The Earth's rotation already gives the rocket an eastward speed (about 0.46 km s $^{- 1}$ at the equator), so less fuel is needed.

Answer: $v_{e} = 2 GM / R$ ; ratio $2$ ; about 22.4 km s $^{- 1}$

Example 7: Comet speeds · 2 marks · Very short answerHOTS / competency

A comet's closest distance from the Sun is 0.5 AU and its farthest is 35 AU. Find the ratio of its speeds at these two points. Which law are you using? [2 marks]

Try it first, then show the solutionSolution

At both points the velocity is perpendicular to the radius, so angular momentum gives $v_{p} r_{p} = v_{a} r_{a}$ (Kepler's law of areas)
$\frac{v _{p}}{v _{a}} = \frac{r _{a}}{r _{p}} = \frac{35}{0.5} = 70$

Answer: 70 : 1 (law of areas / conservation of angular momentum)

Example 8: Raising a satellite · 3 marks · Short answerHOTS / competency

A 1000 kg satellite moves from a circular orbit of radius 2R to one of radius 3R (R = Earth's radius, measured from the centre). Find the energy needed. Does its speed increase or decrease? ( $R = 6.4 \times 1 0^{6}$ m, g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

$Δ E = - \frac{GM m}{6 R} - (- \frac{GM m}{4 R}) = \frac{GM m}{12 R} = \frac{m g R}{12}$
$Δ E = \frac{1000 \times 9.8 \times 6.4 \times 1 0 ^{6}}{12} \approx 5.2 \times 1 0^{9} J$
$v_{o} \propto \frac{1}{r}$ , so it moves more slowly in the higher orbit, even though energy was supplied

Answer: About $5.2 \times 1 0^{9}$ J; speed decreases

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

An Earth-observation satellite

Suppose a 500 kg weather satellite is placed in a circular orbit 1200 km above the Earth. Take $G M_{E} = 4.0 \times 1 0^{14}$ N m $^{2}$ kg $^{- 1}$ and $R_{E} = 6.4 \times 1 0^{6}$ m.

(i) · 1 mark
What is the radius of its orbit?
Answer
Answer: $7.6 \times 1 0^{6}$ m
R + h
(ii) · 1 mark
Find its orbital speed.
Answer
Answer: About 7.3 km s $^{- 1}$
$4.0 \times 1 0^{14} /7.6 \times 1 0^{6}$
(iii) · 1 mark
Find its total energy.
Answer
Answer: About $- 1.32 \times 1 0^{10}$ J
$- \frac{GM m}{2 r} = - \frac{4.0 \times 1 0 ^{14} \times 500}{2 \times 7.6 \times 1 0 ^{6}}$
(iv) · 1 mark
How much energy would free it completely from the Earth?
Answer
Answer: About $1.32 \times 1 0^{10}$ J
Binding energy = −E

Exploring Mars

Suppose a rover lands on Mars, whose mass is $6.4 \times 1 0^{23}$ kg and radius $3.4 \times 1 0^{6}$ m. Take $G = 6.67 \times 1 0^{- 11}$ N m $^{2}$ kg $^{- 2}$ .

(i) · 1 mark
Find g on Mars.
Answer
Answer: About 3.7 m s $^{- 2}$
$\frac{GM}{R ^{2}}$
(ii) · 1 mark
A rover weighs 1800 N on the Earth (g = 9.8). What does it weigh on Mars?
Answer
Answer: About 680 N
Mass ≈ 184 kg; $184 \times 3.7$
(iii) · 1 mark
Find the escape speed from Mars.
Answer
Answer: About 5.0 km s $^{- 1}$
$2 g R = 2 \times 3.69 \times 3.4 \times 1 0^{6}$
(iv) · 1 mark
Does the rover's mass change on Mars?
Answer
Answer: No
Mass is the amount of matter; only weight changes with g.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Derive an expression for the escape speed from the surface of the Earth. [2 marks] (2 marks)

Step	Marks
To just escape, total energy at the surface must be zero: $\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0$	1
$v_{e} = \frac{2 GM}{R} = 2 g R$ (using $GM = g R^{2}$ )	1

Derive the variation of g with depth, assuming the Earth to be a uniform sphere. [3 marks] (3 marks)

Step	Marks
At depth d, only the inner sphere of radius $(R - d)$ attracts; the outer shell gives zero net force	1
Its mass $M_{s} = M \frac{( R - d ) ^{3}}{R ^{3}}$ , so $g (d) = \frac{G M _{s}}{( R - d ) ^{2}} = \frac{GM ( R - d )}{R ^{3}}$	1
$g (d) = g (1 - \frac{d}{R})$ ; zero at the centre	1

(a) Derive the orbital speed and period of a satellite at height h above the Earth. (b) Show that its total energy is $- \frac{GM m}{2 ( R + h )}$ . (c) Find the orbital speed and period at h = 900 km. ( $GM = 4.0 \times 1 0^{14}$ N m $^{2}$ kg $^{- 1}$ , $R = 6.4 \times 1 0^{6}$ m) [5 marks] (5 marks)

Step	Marks
(a) $\frac{m v ^{2}}{R + h} = \frac{GM m}{( R + h ) ^{2}} \Rightarrow v_{o} = \frac{GM}{R + h}$	1
$T = \frac{2 π ( R + h )}{v _{o}} = 2 π \frac{( R + h ) ^{3}}{GM}$	1
(b) $K = \frac{1}{2} m v_{o}^{2} = \frac{GM m}{2 ( R + h )}$ , $U = - \frac{GM m}{R + h}$ , so $E = - \frac{GM m}{2 ( R + h )}$	1
(c) $r = 7.3 \times 1 0^{6}$ m; $v_{o} = \frac{4.0 \times 1 0 ^{14}}{7.3 \times 1 0 ^{6}} \approx 7.40 \times 1 0^{3} m s^{- 1}$	1
$T = \frac{2 π \times 7.3 \times 1 0 ^{6}}{7.40 \times 1 0 ^{3}} \approx 6.2 \times 1 0^{3} s \approx 103$ min	1

Where students lose marks in Gravitation

The slips examiners see most often, and what to write instead.

Mistake: Using mgh for large heights
Forgetting that g changes with height.
Wrong working
Work to lift to height R: mgR
Right working
$\frac{GM m}{2 R} = \frac{m g R}{2}$
Do this: Use $U = - GM m / r$ whenever h is comparable to R.
Usually costs: 1-2 marks.
Mistake: Taking r as the height above the surface
Forgetting to add the Earth's radius.
Do this: In every orbit formula, r = R + h measured from the Earth's centre.
Usually costs: Whole numerical.
Mistake: Using $1 - 2 h / R$ when h is not small
Applying the approximation everywhere.
Do this: For h = R/2 or R, use the exact $g (R / (R + h))^{2}$ .
Usually costs: 1 mark.
Mistake: Saying a satellite's total energy is positive
Adding K and |U|.
Do this: U is negative and twice K in size, so E = −K < 0.
Usually costs: 1 mark.
Mistake: Writing answers on geostationary satellites or weightlessness
Old syllabus notes.
Do this: Both topics were removed for 2026-27; focus on orbital speed, period and energy.
Usually costs: Time.

NCERT exercise guide: Gravitation

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 7.1-7.7 (concepts)

What it tests: Shielding from gravity, what changes with mass and height, Kepler's laws, and the variation of g.

How to approach it: Gravity cannot be shielded; quantities like g, $v_{e}$ and $v_{o}$ do not depend on the small body's mass.

Same kind as Exercise 7.4

Europa, a moon of Jupiter, has a period of 3.55 days and an orbital radius of $6.71 \times 1 0^{8}$ m. Show that Jupiter's mass is about a thousandth of the Sun's, given that the Earth orbits the Sun with period 365 days at $1.50 \times 1 0^{11}$ m.

Show the solutionSolution

$M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$ , so $\frac{M _{J}}{M _{S}} = \frac{r _{J}^{3}}{r _{E}^{3}} \cdot \frac{T _{E}^{2}}{T _{J}^{2}}$
$= (\frac{6.71 \times 1 0 ^{8}}{1.50 \times 1 0 ^{11}})^{3} \times (\frac{365}{3.55})^{2} = (4.47 \times 1 0^{- 3})^{3} \times (103)^{2}$
$\approx 8.95 \times 1 0^{- 8} \times 1.06 \times 1 0^{4} \approx 9.5 \times 1 0^{- 4}$

Answer: About $1 0^{- 3}$ of the Sun's mass

Exercises 7.8-7.16 (g, potential and escape)

What it tests: Escape speed with an initial speed, weight at a height, gravitational potential and field between masses, and g inside the Earth.

How to approach it: Energy conservation with $- GM m / r$ ; g formulas above and below the surface.

Same kind as Exercise 7.16

A body weighs 600 N on the Earth's surface. What would it weigh 1600 km below the surface and 1600 km above it? ( $R_{E} = 6400$ km)

Show the solutionSolution

Below: $W^{'} = 600 (1 - \frac{1600}{6400}) = 600 \times 0.75 = 450 N$
Above: $W^{'} = 600 (\frac{6400}{8000})^{2} = 600 \times 0.64 = 384 N$

Answer: 450 N below; 384 N above

Exercises 7.17-7.21 (energy and satellites)

What it tests: Height reached by a rocket, energy to escape from orbit, approaching stars, and the field between two spheres.

How to approach it: Total energy before = after; for orbits $E = - GM m /2 r$ .

Same kind as Exercise 7.17

A rocket is fired vertically from the Earth at 4 km s $^{- 1}$ . Ignoring air resistance, how high does it rise? ( $R_{E} = 6.4 \times 1 0^{6}$ m, g = 9.8 m s $^{- 2}$ )

Show the solutionSolution

$\frac{1}{2} v^{2} - \frac{GM}{R} = - \frac{GM}{R + h}$ , which gives $h = \frac{R v ^{2}}{v _{e}^{2} - v ^{2}}$ with $v_{e}^{2} = 2 g R = 1.254 \times 1 0^{8}$
$h = \frac{6.4 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{7}}{1.254 \times 1 0 ^{8} - 1.6 \times 1 0 ^{7}} = \frac{1.024 \times 1 0 ^{14}}{1.094 \times 1 0 ^{8}}$
$h \approx 9.4 \times 1 0^{5} m$

Answer: About 940 km

Practice questions on Gravitation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The value of g at the centre of the Earth is:
(a) 9.8 m s $^{- 2}$
(b) infinite
(c) zero
(d) 4.9 m s $^{- 2}$
Hint
$g (1 - d / R)$ with d = R.
Answer
(c) zero
Q2 · 1 mark · MCQExam level
Kepler's law of periods states that:
(a) $T \propto a$
(b) $T^{2} \propto a^{3}$
(c) $T^{3} \propto a^{2}$
(d) $T \propto a^{2}$
Hint
Square and cube.
Answer
(b) $T^{2} \propto a^{3}$
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The total mechanical energy of a satellite in orbit is negative.
Reason (R): The satellite is bound to the Earth by gravity.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
E = −GMm/2r.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
State Kepler's second law. Which conservation law does it follow from?
Hint
Central force, zero torque.
Answer
The line from the Sun to a planet sweeps equal areas in equal times. It follows from conservation of angular momentum, because gravity is a central force and exerts no torque about the Sun.
Q5 · 2 marks · Very short answerHarder exam level
A planet has twice the mass and twice the radius of the Earth. Find g on its surface in terms of the Earth's g.
Hint
$g \propto M / R^{2}$ .
Answer
$g^{'} = g \times \frac{2}{2 ^{2}} = \frac{g}{2}$ .
Q6 · 2 marks · Very short answerHarder exam level
At what height above the Earth does g fall by 1%? At what depth does the same happen? ( $R_{E} = 6400$ km)
Hint
Small changes.
Answer
Height: $\frac{2 h}{R} = 0.01 \Rightarrow h = 32$ km. Depth: $\frac{d}{R} = 0.01 \Rightarrow d = 64$ km.
Q7 · 2 marks · Very short answerHarder exam level
Three particles, each of mass 1 kg, sit at the corners of an equilateral triangle of side 1 m. Find the gravitational potential energy of the system.
Hint
Three pairs.
Answer
$U = - 3 \frac{G ( 1 ) ( 1 )}{1} = - 3 \times 6.67 \times 1 0^{- 11} \approx - 2.0 \times 1 0^{- 10}$ J.
Q8 · 3 marks · Short answerHarder exam level
Find the period of a satellite orbiting at a height equal to the Earth's radius. ( $R_{E} = 6.4 \times 1 0^{6}$ m, g = 9.8 m s $^{- 2}$ )
Hint
r = 2R and $GM = g R^{2}$ .
Answer
$T = 2 π \frac{( 2 R ) ^{3}}{g R ^{2}} = 2 π \frac{8 R}{g} = 2 π \frac{8 \times 6.4 \times 1 0 ^{6}}{9.8} \approx 1.44 \times 1 0^{4}$ s, about 4 hours.
Q9 · 2 marks · Very short answerHarder exam level
Explain why a satellite in a circular orbit needs no fuel to keep moving, and whether gravity does any work on it.
Hint
Force perpendicular to velocity.
Answer
There is no air resistance to overcome, and gravity, always perpendicular to the velocity, does no work. It only changes the direction of motion, supplying the centripetal force, so the speed and energy stay constant without fuel.
Q10 · 3 marks · Short answerHOTS / competency
If the Earth's radius shrank by 1% with its mass unchanged, by about what percentage would g at its surface change? What would happen to the escape speed?
Hint
$g \propto R^{- 2}$ , $v_{e} \propto R^{- 1/2}$ .
Answer
$\frac{Δ g}{g} = - 2 \frac{Δ R}{R} = + 2%$ : g increases by about 2%. $\frac{Δ v _{e}}{v _{e}} = - \frac{1}{2} \frac{Δ R}{R} = + 0.5%$ : the escape speed increases by about 0.5%.
Q11 · 3 marks · Short answerHOTS / competency
Two satellites A and B orbit the Earth at radii r and 4r. Find the ratios of their (a) orbital speeds, (b) periods, (c) total energies (same mass).
Hint
$v \propto r^{- 1/2}$ , $T \propto r^{3/2}$ , $E \propto - 1/ r$ .
Answer
(a) $v_{A} : v_{B} = 2 : 1$ (b) $T_{A} : T_{B} = 1 : 8$ (c) $E_{A} : E_{B} = 4 : 1$ (A is more tightly bound; both negative).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Kepler's law of areas.Show answer
The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.
2.What is the value of G?Show answer
$6.67 \times 1 0^{- 11}$ N m $^{2}$ kg $^{- 2}$ .
3.Write g in terms of the Earth's mass and radius.Show answer
$g = G M_{E} / R_{E}^{2}$ .
4.What is g at the centre of the Earth?Show answer
Zero.
5.Write the gravitational potential energy of two masses.Show answer
$U = - GM m / r$ .
6.What is the escape speed from the Earth?Show answer
About 11.2 km s $^{- 1}$ .
7.Write the orbital speed of a satellite at height h.Show answer
$v_{o} = GM / (R + h)$ .
8.What is the total energy of a satellite in orbit of radius r?Show answer
$- GM m /2 r$ .

Best order to study it

Learn Kepler's three laws and link the second to conservation of angular momentum.
Learn Newton's law of gravitation, G and the shell results.
Derive g = GM/R² and use it to weigh the Earth.
Derive the variation of g with height and depth.
Learn gravitational potential energy and potential, and why mgh is an approximation.
Derive the escape speed.
Derive the orbital speed and period of a satellite.
Finish with the energy of an orbiting satellite and orbit-change problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state Kepler's three laws?
Can you show the areal velocity is constant using angular momentum?
Can you find g on another planet from its mass and radius?
Can you derive g at height h and at depth d?
Can you find the height at which g falls by 1%?
Can you derive $U = - GM m / r$ is negative and explain why?
Can you derive the escape speed and explain why the Moon has no atmosphere?
Can you derive the orbital speed and period of a satellite?
Can you write K, U and E for a satellite?
Can you find the energy needed to raise a satellite to a higher orbit?

Gravitation: frequently asked questions

Are geostationary satellites and weightlessness in the 2026-27 syllabus?

No. Geostationary and polar satellites and weightlessness were removed in the NCERT rationalisation. The syllabus keeps orbital velocity and the energy of an orbiting satellite.

Why is gravitational potential energy negative?

We choose U = 0 when the masses are infinitely far apart. Gravity is attractive, so bringing them closer releases energy and U falls below zero. A negative energy means the system is bound.

Does escape speed depend on the angle of projection?

No (ignoring air resistance). Escape depends only on having enough kinetic energy, and energy is a scalar, so any direction works as long as the path does not hit the planet.

Why is g zero at the centre of the Earth?

At the centre the mass of the Earth surrounds you equally in all directions, so the pulls cancel. Formally, g(d) = g(1 - d/R) gives zero at d = R.

How does this chapter connect to Class 12?

Gravitation and electrostatics share the inverse-square law. Field, potential (-GM/r vs kq/r), potential energy and the shell result all reappear for charges in Class 12 Electric Charges and Fields and Electrostatic Potential.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 7 (Reprint 2026-27)