JEE Mains Physics · Gravitation

Newton's Law, Gravitational Field and Potential

Gravitational forces add as vectors, a sphere or shell acts as a point mass from outside, and potential energy is a sum of −Gm₁m₂/r over every pair.

All Gravitation notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, nineteen of them multiple choice, and three from 2026. Seven add the pulls of several masses, seven ask for the field or potential of a sphere, a shell or a sphere with a cavity, and six ask for potential energy or the work needed to move masses. Six of the twenty come with a figure, so draw the arrangement before you add anything.

Concept 1 of 3: Newton's law and adding the pulls of several masses

Each mass pulls along the line joining it to the test mass, with a strength

Gm_1m_2/r^{2}

. Several pulls add as vectors. Symmetry does most of the work: equal masses on opposite sides cancel, so look for pairs before you resolve anything.

Definition

$F = \dfrac{Gm_1m_2}{r^{2}}$ , attractive, along the line of centres; $G = 6.67 \times 10^{-11}\ \text{N m}^{2}/\text{kg}^{2}$ .
Forces from several masses add as vectors. Pair masses that sit opposite each other first: only the DIFFERENCE of their pulls survives, along their line.
Equal masses at the corners of a square, or spread round a full ring, give zero force at the centre.
On a mass at one corner of a square of side $a$ with equal masses m at the other three: $\left(\sqrt{2} + \tfrac{1}{2}\right)\dfrac{Gm^{2}}{a^{2}}$ along the diagonal.
Splitting a total mass between two bodies: the product $m_1m_2$ , and so the force, is largest for an equal split.
Semicircular wire of mass M and length L, at its centre: $R = L/\pi$ and $F = \dfrac{2GMm}{\pi R^{2}} = \dfrac{2\pi GMm}{L^{2}}$ .
Ring of mass M and radius R, on its axis at distance x: $E = \dfrac{GMx}{(R^{2} + x^{2})^{3/2}}$ .

Newton's law of gravitation

F = \frac{Gm_1m_2}{r^{2}} \qquad \vec F_{\text{net}} = \sum_i \vec F_i

Worked example

Three particles of mass 2 kg each sit at the corners of an equilateral triangle of side 1 m. Find the net gravitational force on one of them.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q7Moderate

Example 1 · Gravitation · Newton's Law, Gravitational Field and Potential

Four identical particles of mass

m

are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is

\left( \frac{2\sqrt{2}+ 1}{32} \right)\frac{Gm^{2}}{{\text{ }L}^{2}}

, the length of the sides of the square is

Opposite masses leave only their difference

Masses M and 3M at opposite corners pull a test mass at the centre in opposite directions. The net pull is G(3M − M)m/r² towards 3M. Do the pairs first, then add the two diagonal results at 90°.

A semicircle does not cancel

A full ring gives zero force at its centre. A half ring does not: the components along the diameter cancel, but the ones towards the arc add to 2GMm/(πR²).

Moving mass between two bodies weakens the pull

The total mass stays the same, but the product m₁m₂ falls as the split becomes unequal. An equal split always gives the largest force.

Drill 6 more on newton's law and adding the pulls of several masses

Concept 2 of 3: Field and potential of a sphere, a shell and a cavity

Outside a uniform sphere or shell, all its mass acts as if it were at the centre. Inside a shell the pulls cancel, so the field is zero, but the potential is not: it keeps the surface value. Inside a solid sphere only the mass nearer the centre pulls, so the field grows in proportion to r.

Definition

Outside (r ≥ R), sphere or shell: $E = \dfrac{GM}{r^{2}}$ , $V = -\dfrac{GM}{r}$ . Dividing, $r = \dfrac{|V|}{E}$ .
Inside a shell: $E = 0$ and $V = -\dfrac{GM}{R}$ , the same everywhere inside. The potential is constant, not zero.
Inside a uniform solid sphere: $E = \dfrac{GMr}{R^{3}}$ and $V = -\dfrac{GM(3R^{2} - r^{2})}{2R^{3}}$ . At the centre $V = -\dfrac{3GM}{2R}$ , which is 1.5 times the surface value.
Field and potential are linked by $V(r_2) - V(r_1) = -\int_{r_1}^{r_2} E\,dr$ .
A cavity: treat it as the whole sphere plus a NEGATIVE sphere of mass $M(r/R)^{3}$ at the cavity's centre. Subtract its pull or potential.
Potentials are scalars and simply add: a mass at the centre of a shell gives $V = -\dfrac{Gm}{r} - \dfrac{GM}{R}$ inside the shell.

Uniform sphere of mass M and radius R

r \ge R:\ E = \frac{GM}{r^{2}},\ V = -\frac{GM}{r} \qquad r < R:\ E = \frac{GMr}{R^{3}},\ V = -\frac{GM(3R^{2} - r^{2})}{2R^{3}}

Worked example

A uniform solid sphere of mass M and radius R pulls a particle at distance 4R from its centre with force

F_1

. A spherical cavity of radius

R/2

is cut out; its centre is at

R/2

from O, on the particle's side. The force becomes

F_2

. Find

F_1 : F_2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q12Moderate

Example 2 · Gravitation · Newton's Law, Gravitational Field and Potential

The gravitational potential at a point above the surface of earth is

- 5.12 \times10^{7}\text{ }J/kg

and the acceleration due to gravity at that point is

6.4\text{ }m/s^{2}

. Assume that the mean radius of earth to be

6400\text{ }km

. The height of this point above the earth's surface is:

Zero field inside a shell, not zero potential

Inside a uniform shell the field is zero, so the potential does not change. It stays at −GM/R, the surface value. 'Potential is zero inside' is the wrong option.

A cavity ratio is always more than 1

F₁ (whole sphere) is larger than F₂ (sphere with a hole). If your F₁ : F₂ comes out below 1, you have inverted it, and the inverted ratio is usually one of the options.

Measure the cavity's distance to its own centre

The negative sphere sits at the cavity's centre, not at O. Check from the figure whether the cavity is on the particle's side or the far side; the distance changes.

Drill 6 more on field and potential of a sphere, a shell and a cavity

Concept 3 of 3: Potential energy of a system and work to rearrange it

Gravitational potential energy belongs to a pair of masses, and it is negative: it is zero when they are infinitely far apart. A system's energy is the sum over every pair. The work an outside agent does to rearrange the masses slowly is the final energy minus the initial energy.

Definition

One pair: $U = -\dfrac{Gm_1m_2}{r}$ .
A system: add $-\dfrac{Gm_im_j}{r_{ij}}$ over EVERY pair. Four masses on a square have 4 sides and 2 diagonals; a fifth mass at the centre adds 4 more pairs.
Work done by an external agent: $W = U_f - U_i$ . Spreading masses out needs positive work.
Lifting m from the surface to height h: $\Delta U = \dfrac{GMmh}{R(R + h)} = \dfrac{mgh}{1 + h/R}$ .
Only for $h \ll R$ does this reduce to $mgh$ . At $h = R$ it is $\tfrac{1}{2}mgR$ ; at $h = 2R$ , $\tfrac{2}{3}mgR$ ; at $h = 3R$ , $\tfrac{3}{4}mgR$ .

Potential energy

U = -\sum_{\text{pairs}} \frac{Gm_im_j}{r_{ij}} \qquad \Delta U_{R \to R+h} = \frac{mgh}{1 + h/R}

Worked example

Masses of 1 kg, 2 kg and 3 kg sit at the corners of an equilateral triangle of side 2 m. They are moved to the corners of a triangle of side 4 m. How much work is needed? Take

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q20Moderate

Example 3 · Gravitation · Newton's Law, Gravitational Field and Potential

Three masses

200\text{ }kg,300\text{ }kg

and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m . They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process

\_\_\_\_

J. (Gravitational constant

G = 6.7 \times10^{- 11}\text{ }N{\text{ }m}^{2}/kg^{2}

)

Count the diagonals

A square of masses has six pairs, not four. Leaving out the two diagonal pairs is the most common wrong option.

mgh fails when h is comparable to R

Raising m to height 2R gains (2/3)mgR, not 2mgR. Use mgh/(1 + h/R) unless h is a small fraction of R.

Maximum potential energy usually means maximum size

Gravitational potential energy is negative. When a question asks when it is 'maximum', it means largest in magnitude, that is, most negative.

Drill 5 more on potential energy of a system and work to rearrange it

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Newton's law and adding the pulls of several masses
Newton's law of gravitation
$F = \frac{Gm_1m_2}{r^{2}} \qquad \vec F_{\text{net}} = \sum_i \vec F_i$
Field and potential of a sphere, a shell and a cavity
Uniform sphere of mass M and radius R
$r \ge R:\ E = \frac{GM}{r^{2}},\ V = -\frac{GM}{r} \qquad r < R:\ E = \frac{GMr}{R^{3}},\ V = -\frac{GM(3R^{2} - r^{2})}{2R^{3}}$
Potential energy of a system and work to rearrange it
Potential energy
$U = -\sum_{\text{pairs}} \frac{Gm_im_j}{r_{ij}} \qquad \Delta U_{R \to R+h} = \frac{mgh}{1 + h/R}$

Watch out for (9)

Opposite masses leave only their difference
→ Newton's law and adding the pulls of several masses
A semicircle does not cancel
→ Newton's law and adding the pulls of several masses
Moving mass between two bodies weakens the pull
→ Newton's law and adding the pulls of several masses
Zero field inside a shell, not zero potential
→ Field and potential of a sphere, a shell and a cavity
A cavity ratio is always more than 1
→ Field and potential of a sphere, a shell and a cavity
Measure the cavity's distance to its own centre
→ Field and potential of a sphere, a shell and a cavity
Count the diagonals
→ Potential energy of a system and work to rearrange it
mgh fails when h is comparable to R
→ Potential energy of a system and work to rearrange it
Maximum potential energy usually means maximum size
→ Potential energy of a system and work to rearrange it

Test yourself on Gravitation

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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