Gravitation

1. What this chapter covers

This maps cleanly onto the CBSE 2026-27 syllabus, with one thing worth knowing before you revise.

Not in the 2026-27 chapter or syllabus

Older editions carried sections on all three, and most coaching handouts still do. The words "geostationary" and "polar satellite" do not occur anywhere in the 2026-27 chapter.

One more thing the chapter does not contain: the falling apple. It is a good story, but it is not in this book, and nothing in the syllabus rests on it.

2. How we got here: circles, then ellipses

This is the one chapter in Class 11 Physics where the history is worth a few minutes, because the sequence explains why the laws are numbered the way they are.

The starting picture put the planets on circles around a fixed central Sun. That model was discredited by the church, and its most famous supporter, Galileo, faced prosecution from the state for holding to it.

At about the same time, a Danish nobleman called Tycho Brahe (1546-1601) spent an entire working life recording planetary positions with the naked eye — no telescope. He never explained them. His assistant Johannes Kepler (1571-1640) analysed the compiled data afterwards and pulled three laws out of it.

Notice the order of events. Kepler's laws came from data, and described how planets move. They did not explain why. Newton supplied the why, decades later, and the fact that his single force law reproduces all three of Kepler's is the strongest evidence he had.

So when a question asks you to "derive Kepler's third law", it is asking you to run that history forwards: start from Newton's force, end at Kepler's observation.

3. Kepler's three laws

First law — the law of orbits

All planets move in elliptical orbits with the Sun at one of the two foci.

This was a genuine break from the Copernican model, which allowed only circles.

You should know the parts of the ellipse by name, because questions use the vocabulary without explaining it:

How to draw one, and why the definition works: pin the two ends of a string at the foci, pull it taut with a pencil tip and move the pencil around. The curve you get is an ellipse, because the string length never changes — so for every point on the curve, the sum of the distances to the two foci is constant.

A circle is a special case, not a different shape. Bring the two foci together and they merge into one point; the semi-major axis becomes the radius. This matters more than it looks: every circular-orbit result you derive later in the chapter is a special case of Kepler's laws, not a separate topic.

Second law — the law of areas

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Where this came from: the observation that planets appear to move slower when they are farther from the Sun and faster when nearer. Equal areas in equal times is exactly that statement made precise — far from the Sun the line is long, so a small angle sweeps a large area; close in, the line is short and the planet must move faster to sweep the same area.

The second law is angular momentum conservation in disguise. In a small time $\Delta t$ the area swept is

$\Delta \mathbf{A}=\frac{1}{2}(\mathbf{r}\times \mathbf{v})\Delta t⟹\frac{d\mathbf{A}}{dt}=\frac{\mathbf{L}}{2m}$

Gravity always points along the line joining planet to Sun, so it exerts no torque about the Sun, so $\mathbf{L}$ is constant, so $d\mathbf{A}/dt$ is constant.

Carry this consequence: the area law holds for any central force, not only an inverse square one. So the second law tells you gravity points along the line joining the two bodies — but it does not tell you the force goes as $1/r^2$.

Third law — the law of periods

$T^2\propto a^3$

where $a$ is the semi-major axis. This is the law that pins down the inverse square. Newton's insight was that a $1/r^2$ force, and essentially only that, reproduces $T^2\propto a^3$.

Two points the chapter makes explicitly, and both get asked:

• The constant of proportionality is the same for every planet orbiting the same Sun. That is what makes the law useful — you can compare two planets without knowing either mass.
• It applies to Earth satellites too. The chapter derives $T^2=k(R_E+h{)}^3$ for a satellite, which is the third law again with the Earth in place of the Sun.

4. The universal law of gravitation

Every particle attracts every other particle along the line joining them, with a force

$F=G\frac{m_1{m}_2}{r^2}$

Two properties you must be able to use, not just quote.

The principle of superposition

Gravitational forces add as vectors. If several masses act on one body, the resultant is the vector sum of the individual forces:

${\mathbf{F}}_R={\mathbf{F}}_1+{\mathbf{F}}_2+{\mathbf{F}}_3+\dots$

Each pair is worked out as if the others were not there, and only then are the results added.

Use symmetry before you use algebra. The chapter's Example 7.1 places equal masses at the corners of an equilateral triangle and asks for the force on a mass at the centre. The full vector sum comes out zero — but you can see it is zero by symmetry in one line, without computing anything. Then when one corner mass is doubled, only the extra mass contributes, and the problem collapses to a single force.

The two shell results — why a planet can be treated as a point

An extended body is not a point, so strictly you would have to add up the pull of every particle in it. Calculus does this, and for a uniform spherical shell the answer takes two remarkably clean forms:

These two results carry the rest of the chapter. The first is the reason you are allowed to write $F=G{M}_{E}m/r^2$ for the Earth at all, and the reason Cavendish could treat his lead spheres as points. The second is the reason $g$ falls as you descend, and it is what Exercise 7.11 turns on.

5. Measuring G — the Cavendish experiment

G had to be measured, not derived. No theory predicts its value. It was first determined by the English scientist Henry Cavendish in 1798, and the accepted value is

$G=6.67\times 10^{-11}{\text{N m}}^2{\text{kg}}^{-2}$

How the apparatus works, since this is a standard 3-mark description:

1. A bar carries two small lead spheres at its ends, suspended at its centre from a fine wire.
2. Two large lead spheres are brought close to the small ones, on opposite sides.
3. Each big sphere attracts its neighbour. The two forces are equal and opposite, so there is no net force on the bar — only a torque, of magnitude $F\times L$ where $L$ is the bar's length.
4. The wire twists until its restoring torque balances the gravitational torque. If $\theta$ is the angle of twist and $\tau$ the restoring couple per unit angle:

$\tau \theta =G\frac{Mm}{d^2}L$

5. $\tau$ is found separately, by applying a known torque and measuring the twist. Measure $\theta$, and $G$ follows.

Read step 3 again — it is the design insight. Putting the large spheres on opposite sides deliberately cancels the net force and leaves a pure torque, which a fine wire can register as a visible rotation. A force that small could not be weighed; a twist can be watched.

That an experiment this delicate was needed tells you how weak gravity is between ordinary objects. It only becomes obvious when one of the masses is planet-sized.

6. Why the Earth behaves like a point, and what g really is

Picture the Earth as a large number of concentric spherical shells, the smallest at the centre and the largest at the surface.

A point outside the Earth is outside every one of those shells. By the first shell result, each shell pulls as if its mass sat at the common centre — so the whole Earth pulls as if its entire mass were concentrated at its centre. Hence, for a mass $m$ at the surface:

$F=\frac{G{M}_{E}m}{R_E^2}=mg⟹g=\frac{G{M}_E}{R_E^2}$

This is where $g$ comes from. It is not a fundamental constant — it is $G$, the Earth's mass and the Earth's radius packaged into one number. Change any of the three and $g$ changes, which is why the Moon has a different $g$ and why the same object weighs slightly different amounts at different places.

The distinction that costs the most marks

7. Why g falls off in both directions

This is where students lose marks, because the two cases have different formulas and the result is counter-intuitive.

Above the surface, at height $h$, the distance from the centre is simply larger:

$g_h=\frac{G{M}_E}{(R_E+h{)}^2}=g{\left(1+\frac{h}{R_E}\right)}^{-2}\approx g\left(1-\frac{2h}{R_E}\right)(h\ll R_E)$

Below the surface, at depth $d$, something different happens. Stand at radius $r$ from the centre. Every shell of radius greater than $r$ has you inside it, and by the second shell result exerts no force at all. Only the sphere of radius $r$ beneath you pulls, as if its mass were at the centre.

For a uniform Earth that inner mass goes as $r^3$, while the force goes as $1/r^2$ — so the force goes as $r^3/r^2=r$, linear in the distance from the centre:

$g_d=g\left(1-\frac{d}{R_E}\right)$

So g is largest at the surface and falls off whichever way you move from it — but for entirely different reasons. Going up, the distance increases. Going down, the amount of mass pulling you decreases, because everything above you contributes nothing.

The trap: students assume that since gravity is "stronger nearer the centre", $g$ must keep rising as you descend. It does the opposite. Getting closer to the centre also means leaving mass behind you, and that effect wins.

8. Potential energy, and why the sign is negative

Take zero potential energy at infinity — the natural choice, since the force vanishes there. Bringing a mass in from infinity then releases energy, so the potential energy is negative:

$U=-\frac{GMm}{r}$

Gravitational potential is the same thing per unit mass, and CBSE lists it separately, so know both:

$V=-\frac{GM}{r}$

The negative sign is not a bookkeeping quirk. It is what "bound" means. A system with negative total energy cannot reach infinity, because reaching infinity requires getting to at least zero energy. Everything below zero is trapped.

And $mgh$ is only an approximation. It is the small-height limit of the change in this $U$, valid while $h\ll R_E$. Use it for a ball thrown off a roof; do not use it for a rocket.

9. Escape speed

Escape speed is the minimum launch speed for which the total energy reaches zero — the threshold of being unbound:

$\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0⟹v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\approx 11.2\text{km/s}$

Three things about escape speed that get asked:

• It does not depend on the escaping body's mass — $m$ cancels. A marble and a spacecraft need the same speed.
• It does not depend on the direction of projection, because energy is a scalar and $U$ depends only on distance.
• It does depend on where you launch from — height enters through $R$.

Why the Moon has no atmosphere

Put the Moon's own $g$ and radius into the same formula and escape speed comes out at about 2.3 km/s, roughly five times smaller than Earth's.

Gas molecules on the Moon's surface routinely reach speeds above that. So any atmosphere it once had simply leaked away into space, molecule by molecule. The Moon is airless because its escape speed is low — a fact about thermal speeds and gravity, not about how the Moon formed.

That is the kind of link worth having ready: one formula, one number, one visible consequence.

10. Satellites, orbital speed and period

A satellite in a circular orbit is in free fall — gravity supplies exactly the centripetal force needed, and nothing else acts:

$\frac{m{V}^2}{R_E+h}=\frac{G{M}_{E}m}{(R_E+h{)}^2}⟹V=\sqrt{\frac{G{M}_E}{R_E+h}}$

The period follows from going once round at that speed:

$T=\frac{2\pi (R_E+h)}{V}=2\pi \sqrt{\frac{(R_E+h{)}^3}{G{M}_E}}⟹T^2=\frac{4{\pi }^2}{G{M}_E}(R_E+h{)}^3$

That is Kepler's third law, derived rather than observed. Which is why this derivation is the most-set 5-mark question in the chapter.

Put $h=0$ into the speed formula and you get $V=\sqrt{g{R}_E}$, which works out to about 7.9 km/s — so

$v_e=\sqrt{2}V$

Escape speed is only about 41% more than orbital speed. Worth remembering as a sanity check: if a calculation gives you an escape speed double the orbital speed, you have made an error.

The one thing to take from the period formula: $T$ depends only on the orbital radius — not on the satellite's mass, not on how it got there. Two satellites at the same height keep the same period whatever they weigh.

11. The energy of an orbiting satellite

For a circular orbit of radius $r=R_E+h$, substitute the orbital speed into the kinetic energy:

$K=+\frac{G{M}_{E}m}{2r}U=-\frac{G{M}_{E}m}{r}E=K+U=-\frac{G{M}_{E}m}{2r}$

Three relations follow, and they are the ones worth memorising because they turn multi-step problems into one line:

$E=-KE=\frac{1}{2}U|U|=2K$

The total energy is negative, which is precisely the statement that the satellite is bound. If $E$ ever became positive the satellite would leave and not return.

A consequence that gets asked: to free an orbiting satellite you must supply $+G{M}_{E}m/2r$. That is less than launching the same object from rest on the ground, because the satellite already carries kinetic energy pointed the right way. Exercise 7.6 turns on exactly this point.

And a counter-intuitive one: raising a satellite to a higher orbit increases its total energy but decreases its speed, since $V=\sqrt{GM/r}$ falls as $r$ grows. Higher orbit, slower satellite, more energy. The extra energy went into potential energy, and more than paid for the kinetic energy lost.

12. Weightlessness — not the absence of gravity

The chapter no longer teaches this as a section, but Exercise 7.9 asks about the effects on an astronaut, so you need it.

At the height of the space station, gravity is close to its full surface strength. An astronaut floats not because gravity is weak, but because she and the station are both in free fall towards Earth, falling together at the same rate. With nothing pushing up on her, there is no sensation of weight.

The chapter's own Point to Ponder puts it plainly: this is free fall, not an absence of gravitational force.

What you actually feel as weight is the normal force — the push of the floor, which is what a bathroom scale reads. Remove the floor's push and the sensation goes, whatever gravity is doing.

The same reasoning covers the lift problems you will meet later: apparent weight drops to zero when floor and passenger accelerate downward together at $g$.

13. Two results the exercises lean on

Gravitational shielding is impossible. Electric fields can be screened by a conductor, because charges rearrange themselves to cancel the field inside. There is no negative mass, so nothing can rearrange to cancel gravity. Putting a body inside a hollow sphere does not shield it from outside matter — the sphere's own pull vanishes inside, but everything beyond it still reaches through. Exercise 7.1 asks this directly.

Tidal effects follow the gradient, not the force. The Sun pulls on the Earth far harder than the Moon does, yet the Moon dominates our tides. Tides depend on how much the pull differs across the Earth's diameter, and that difference falls off as $1/r^3$ — much faster than the $1/r^2$ force itself. On that measure the Moon's closeness beats the Sun's mass.

Summary

• Kepler's laws came from Tycho Brahe's naked-eye data and describe how planets move; Newton's law of gravitation explains why.
• Orbits are ellipses with the Sun at one focus; a circle is the special case where the two foci merge.
• The law of areas is angular momentum conservation, and holds for any central force — so it does not by itself imply an inverse square law. The third law does.
• $F=G{m}_1{m}_2/r^2$; forces from several masses add as vectors.
• A uniform shell pulls an outside point as if all its mass were at the centre, and pulls an inside point not at all. Everything else in the chapter rests on these two results.
• $G$ was measured by Cavendish in 1798 using a torsion balance, where opposed spheres give a pure torque and no net force.
• $g=G{M}_E/R_E^2$ is not fundamental — it is built from $G$, the Earth's mass and its radius.
• $g$ is maximum at the surface: it falls as an inverse square going up, and falls linearly going down, reaching zero at the centre.
• $U=-GMm/r$ and $V=-GM/r$, with zero taken at infinity. Negative total energy means bound.
• $v_e=\sqrt{2gR}\approx 11.2$ km/s — independent of the escaping body's mass and of direction. The Moon's is 2.3 km/s, which is why it has no atmosphere.
• $V=\sqrt{GM/r}$ and $T^2\propto r^3$: the period depends on orbital radius alone.
• A bound satellite has $E=-GMm/2r$, so $E=-K$ and $E=U/2$.
• Weightlessness is shared free fall, not absent gravity.