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Gravitation · JEE & NEET Physics

Gravitational Field and Potential: notes and previous year questions

The field E = GM/r², shells and spheres, potential energy U = −GMm/r, potential V = −GM/r, E = −dV/dr, adding potentials, and solving with energy.

Gravitational Field and Potential in short

  • The field is the pull per kilogram: E = GM/r², in N/kg; near the Earth it equals g.
  • Inside a hollow shell E = 0; outside any sphere, treat it as a point mass.
  • U = −GMm/r is zero at infinity and negative nearer: the masses are bound.
  • mgh is the small-height case of −GMm/r.

1The gravitational field

Every mass changes the space around it. The gravitational field at a point is the pull on each kilogram placed there:

E⃗=F⃗m=−GMr2 r^\vec E = \frac{\vec F}{m} = -\frac{GM}{r^2}\,\hat rSize GM/r², pointing towards the mass. Unit N/kg, which is the same as m/s².

Force per mass is an acceleration, so near the Earth the field is just gg. If the field at a planet's surface is 10 N/kg, it is 2.5 N/kg at r=2Rr = 2R, anywhere on that circle.

  • Field lines point towards the mass.
  • They crowd together where the field is strong.
  • They never cross (a point has only one field direction).
  • They reach out forever.

2Shells and spheres

BodyInside ($r < R$)Outside ($r \ge R$)
Thin hollow shellE=0E = 0E=GM/r2E = GM/r^2
Uniform solid sphereE=GMr/R3E = GMr/R^3E=GM/r2E = GM/r^2

Inside a shell, the pulls from all sides cancel exactly. Inside a solid sphere, only the mass nearer the centre pulls, so the field grows in a straight line and is largest at the surface. Outside any sphere, treat it as a point mass at its centre; that is why rr is always measured from the centre.

3Potential energy

We take the potential energy to be zero when the masses are infinitely far apart, where they no longer pull. As they come together, gravity does positive work, so UU drops below zero:

U=−GMmrU = -\frac{GMm}{r}Zero at infinity, negative everywhere else. At the Earth's surface U = −GMm/R = −mgR.

The minus sign means the masses are bound: to pull mm away to infinity from the surface you must supply GMm/RGMm/R.

Lifting from the surface to a height hh:

ΔU=GMm(1R−1R+h)=GMmhR(R+h)\Delta U = GMm\Big(\frac1R - \frac1{R+h}\Big) = \frac{GMmh}{R(R+h)}

When h≪Rh \ll R, R+h≈RR + h \approx R and ΔU≈GMmh/R2=mgh\Delta U \approx GMmh/R^2 = mgh. So mghmgh is a small-height special case; for big heights use −GMm/r-GMm/r.

4Potential

The gravitational potential is the potential energy per kilogram, a property of the place itself:

V=Um=−GMrV = \frac{U}{m} = -\frac{GM}{r}Unit J/kg (= m²/s²). A scalar: a number with a sign, no direction.

It is the work done per kilogram to bring a mass in from infinity. Points at the same distance share the same VV: they form circles (spheres in space) around the planet. To move mm from point 1 to point 2, the work needed is W=m(V2−V1)W = m(V_2 - V_1).

At the Earth's surface: V=−GM/R=−6.67×10−11×6×1024÷6.4×106≈−6.25×107V = -GM/R = -6.67\times10^{-11} \times 6\times10^{24} \div 6.4\times10^6 \approx -6.25\times10^7 J/kg. Check: −gR=−9.8×6.4×106≈−6.3×107-gR = -9.8 \times 6.4\times10^6 \approx -6.3\times10^7 J/kg.

5Field from potential

E=−dVdr,F=−dUdrE = -\frac{dV}{dr},\qquad F = -\frac{dU}{dr}

On the VV–rr graph, the field is the slope (with a minus sign). ddr(−GM/r)=GM/r2\frac{d}{dr}(-GM/r) = GM/r^2, so E=−GM/r2E = -GM/r^2: inward. Near the planet the curve is steep and the field strong; far away it is almost flat. The field always points downhill, towards lower potential. Going back: V=−∫E drV = -\int E\,dr.

At the surface the slope is gg; it is g/9g/9 at r=3Rr = 3R. A field of 5×10−65\times10^{-6} N/kg means VV drops by 5×10−65\times10^{-6} J/kg for each metre moved along the field (J/kg per metre is the same as N/kg).

6Adding potentials

V=V1+V2+⋯=−G(M1r1+M2r2+…)V = V_1 + V_2 + \dots = -G\Big(\frac{M_1}{r_1} + \frac{M_2}{r_2} + \dots\Big)Potentials add as plain numbers. Fields and forces are vectors and add tip to tail.

Midway between two equal masses mm a distance rr apart, the fields cancel (E=0E = 0), but each potential is −Gm/(r/2)-Gm/(r/2), so V=−4Gm/rV = -4Gm/r. Zero field does not mean zero potential. At the centre of a square with a mass mm at each corner, a distance dd away: V=−4Gm/dV = -4Gm/d.

The potential energy of a system adds over every pair. Three masses mm on a triangle of side aa: U=−3Gm2/aU = -3Gm^2/a. At side 2a2a, U=−3Gm2/2aU = -3Gm^2/2a, so pulling them apart needs W=3Gm2/2aW = 3Gm^2/2a.

7The energy way

With only gravity acting, KE+UKE + U stays the same. Energy is a scalar and only the start and the end matter, so use it for speeds, heights and work.

To just reach h=Rh = R: 12v2=GM/R−GM/2R\tfrac12 v^2 = GM/R - GM/2R, so v2=GM/R=gRv^2 = GM/R = gR; with g=10g = 10 m/s² and R=6400R = 6400 km, v=8v = 8 km/s.

The work to move a mass between two points does not depend on the path: gravity is a conservative force.

Summary

Key ideas

  • The field is the pull per kilogram: E = GM/r², in N/kg; near the Earth it equals g.
  • Inside a hollow shell E = 0; outside any sphere, treat it as a point mass.
  • U = −GMm/r is zero at infinity and negative nearer: the masses are bound.
  • mgh is the small-height case of −GMm/r.
  • Potential is energy per kilogram, V = −GM/r, a scalar in J/kg.
  • The field is minus the slope of the V–r graph and points towards lower potential.
  • Potentials add as plain numbers; zero field does not mean zero potential.
  • The potential energy of a system adds over every pair of masses.
  • With only gravity acting, KE + U is conserved: solve with energy.

Every equation

Field
E=F/m=GM/r2E = F/m = GM/r^2
Shell inside
E=0E = 0
Solid sphere inside
E=GMr/R3E = GMr/R^3
Potential energy
U=−GMm/rU = -GMm/r
At the surface
U=−mgRU = -mgR
Lifting to height h
ΔU=GMmh/(R(R+h))\Delta U = GMmh/(R(R+h))
Small heights
ΔU≈mgh\Delta U \approx mgh
Potential
V=U/m=−GM/rV = U/m = -GM/r
Surface potential
V=−gRV = -gR
Work
W=m(V2−V1)W = m(V_2 - V_1)
Work between distances
W=GMm(1/r1−1/r2)W = GMm(1/r_1 - 1/r_2)
Field from potential
E=−dV/drE = -dV/dr
Force from energy
F=−dU/drF = -dU/dr
Adding potentials
V=−G∑Mi/riV = -G\sum M_i/r_i
Three masses
U=−3Gm2/aU = -3Gm^2/a
Self-energy of a sphere
U=−3GM2/5RU = -3GM^2/5R

Previous year questions with solutions

Real JEE and NEET questions on gravitational field and potential. Try each one before you open the solution.

Q1JEE Main 2026One correct option

Three masses 200kg,300kg200 \mathrm{kg},300 \mathrm{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=6.7×10−11Nm2/kg2G=6.7\times {10}^{-11} Nm^{2}/{\mathrm{kg}}^{2} )

  1. A4.77×10−74.77\times {10}^{-7}
  2. B1.74×10−71.74\times {10}^{-7}
  3. C9.86×10−69.86\times {10}^{-6}
  4. D2.85×10−72.85\times {10}^{-7}
Show answer and solution

Answer: Option B

Every pair's distance goes from 20 m20\ \mathrm{m} to 25 m25\ \mathrm{m}, so

W=Uf−Ui=G(m1m2+m1m3+m2m3)(120−125)W = U_{f} - U_{i} = G(m_{1}m_{2} + m_{1}m_{3} + m_{2}m_{3})\left(\dfrac{1}{20} - \dfrac{1}{25}\right)

The pairs: 200×300+200×400+300×400=60 000+80 000+120 000=2.6×105 kg2200 \times 300 + 200 \times 400 + 300 \times 400 = 60\,000 + 80\,000 + 120\,000 = 2.6 \times 10^{5}\ \mathrm{kg^{2}}, and 120−125=0.01 m−1\dfrac{1}{20} - \dfrac{1}{25} = 0.01\ \mathrm{m^{-1}}. So

W=6.7×10−11×2.6×105×0.01≈1.74×10−7 JW = 6.7 \times 10^{-11} \times 2.6 \times 10^{5} \times 0.01 \approx 1.74 \times 10^{-7}\ \mathrm{J}

It is positive because the masses are pulled further apart. Keeping the same centre changes nothing: only the distances between the masses count.

The traps are to leave out a pair, to use the total mass in place of the three pair products, or to take the reciprocals the wrong way round, which makes the work negative. None of A, C or D comes from the full sum of pairs times (120−125)\left(\dfrac{1}{20} - \dfrac{1}{25}\right).

Q2NEET 2026One correct option

The amount of work done to raise a mass ' mm ' from the surface of the Earth to a height equal to the radius of the Earth ' RR ' will be

  1. A2mgR2mgR
  2. BmgR4mg\frac{R}{4}
  3. CmgRmgR
  4. DmgR2mg\frac{R}{2}
Show answer and solution

Answer: Option D

With h=Rh = R, ΔU=mgh1+h/R=mgR2\Delta U = \dfrac{mgh}{1 + h/R} = \dfrac{mgR}{2}. Or straight from the potential energies: Ui=−GMmRU_{i} = -\dfrac{GMm}{R} and Uf=−GMm2RU_{f} = -\dfrac{GMm}{2R}, so W=Uf−Ui=GMm2R=mgR2W = U_{f} - U_{i} = \dfrac{GMm}{2R} = \dfrac{mgR}{2}, using GM=gR2GM = gR^{2}.

The trap is C, mgRmgR, which is mghmgh used over a climb on which gg falls to a quarter of its surface value. A, 2mgR2mgR, doubles where it should halve, and B, mgR4\dfrac{mgR}{4}, uses the weakened gg at the top for the whole climb.

Q3JEE Main 2026One correct option

A body of mass mm is taken from the surface of earth to a height equal to twice the radius of earth (Re)(R_{e}). The increase in potential energy will be ____\_\_\_\_ .

( gg is acceleration due to gravity at the surface of earth)

  1. A12mgRe\frac{1}{2}mgR_{e}
  2. B34mgRe\frac{3}{4}mgR_{e}
  3. C14mgRe\frac{1}{4}mgR_{e}
  4. D23mgRe\frac{2}{3}mgR_{e}
Show answer and solution

Answer: Option D

A height of 2Re2R_{e} above the surface is r=3Rer = 3R_{e} from the centre. So

ΔU=GMm(1Re−13Re)=23GMmRe=23mgRe\Delta U = GMm\left(\dfrac{1}{R_{e}} - \dfrac{1}{3R_{e}}\right) = \dfrac{2}{3}\dfrac{GMm}{R_{e}} = \dfrac{2}{3}mgR_{e}

using GM=gRe2GM = gR_{e}^{2}. The same comes from mgh1+h/Re\dfrac{mgh}{1 + h/R_{e}} with h=2Reh = 2R_{e}: 2mgRe3\dfrac{2mgR_{e}}{3}.

The trap is A, 12mgRe\dfrac{1}{2}mgR_{e}, which takes r=2Rer = 2R_{e}: the height is measured from the surface, not the centre. B, 34mgRe\dfrac{3}{4}mgR_{e}, is the answer for a height of 3Re3R_{e}. C, 14mgRe\dfrac{1}{4}mgR_{e}, is less than even the first ReR_{e} of the climb costs.

Practice questions, easy to hard

Three questions from the gravitational field and potential practice ladder: one easy, one medium, one hard.

Q4One or more correct options

Fields add as vectors, just like the forces they come from. To find the field at a point due to several masses, work out each one's Gmr2\dfrac{Gm}{r^{2}}, give it its direction (towards that mass), and add the arrows.

Four equal masses mm sit at the corners of a square of side aa. Each corner is a2\dfrac{a}{\sqrt{2}} from the centre. Which statements are correct?

  1. AThe field at the centre of the square is zero
  2. BIf one mass is removed, the field at the centre is 2Gma2\dfrac{2Gm}{a^{2}}, pointing towards the corner opposite the empty one
  3. CIf one mass is removed, the field at the centre points towards the empty corner
  4. DThe field at the centre is 8Gma2\dfrac{8Gm}{a^{2}}, the four pulls of 2Gma2\dfrac{2Gm}{a^{2}} added together
Show answer and solution

Answer: Options A, B

A: each mass pulls towards its own corner with Gm(a/2)2=2Gma2\dfrac{Gm}{(a/\sqrt{2})^{2}} = \dfrac{2Gm}{a^{2}}, and opposite corners pull in opposite directions, so the four arrows cancel in pairs. B: take one mass away and its partner across the diagonal is left unbalanced. The field is that one pull, 2Gma2\dfrac{2Gm}{a^{2}}, towards the corner opposite the gap.

C is the trap: it is the mass across the diagonal that is left unbalanced, and it pulls towards itself, away from the gap. D adds the four pulls as plain numbers, forgetting that the field is a vector.

Q5One correct option

Because potential is a scalar, the potentials of several masses simply add as numbers, all of them negative, with no directions to worry about. That makes it very different from the field: at a point where the fields of several masses cancel, their potentials do not cancel; they pile up.

Four equal masses mm sit at the corners of a square of side aa, each a2\dfrac{a}{\sqrt{2}} from the centre. What is the gravitational potential at the centre?

  1. AZero, because the four pulls cancel there
  2. B−22Gma-\dfrac{2\sqrt{2}Gm}{a}
  3. C−42Gma-\dfrac{4\sqrt{2}Gm}{a}
  4. D−4Gma-\dfrac{4Gm}{a}
Show answer and solution

Answer: Option C

Each mass gives −Gma/2=−2Gma-\dfrac{Gm}{a/\sqrt{2}} = -\dfrac{\sqrt{2}Gm}{a}, and four of them add to −42Gma-\dfrac{4\sqrt{2}Gm}{a}.

The trap is A: the field at the centre is zero, but the potential is a sum of negative numbers and cannot cancel. D uses the side aa as the distance. B counts only two of the masses, as if opposite corners cancelled; that is how fields combine, not potentials.

Q6One or more correct options

When two bodies pull each other and nothing else acts, neither stays put: both move. Two laws together settle their speeds. Momentum: they start at rest, so the total stays zero, and m1v1=m2v2m_{1}v_{1} = m_{2}v_{2}. Energy: the kinetic energy they share equals the fall in their mutual potential energy,

12m1v12+12m2v22=Gm1m2(1r−1r0)\dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2} = Gm_{1}m_{2}\left(\dfrac{1}{r} - \dfrac{1}{r_{0}}\right)

when they have come from r0r_{0} apart to rr apart.

Two small spheres of masses mm and 3m3m are released from rest a distance dd apart, far out in space. Which statements are correct when they are d2\dfrac{d}{2} apart?

  1. ATheir momenta are equal in size and opposite in direction
  2. BThe lighter sphere moves three times as fast as the heavier
  3. CThe heavier sphere's speed is Gm2d\sqrt{\dfrac{Gm}{2d}}
  4. DThey share the kinetic energy equally
Show answer and solution

Answer: Options A, B, C

A is momentum conservation from rest. B follows: mv1=3mv2mv_{1} = 3mv_{2}, so v1=3v2v_{1} = 3v_{2}. C: the potential energy falls by 3Gm2(2d−1d)=3Gm2d3Gm^{2}\left(\dfrac{2}{d} - \dfrac{1}{d}\right) = \dfrac{3Gm^{2}}{d}, and the kinetic energy is 12m(3v2)2+12(3m)v22=6mv22\dfrac{1}{2}m(3v_{2})^{2} + \dfrac{1}{2}(3m)v_{2}^{2} = 6mv_{2}^{2}. Setting them equal, v22=Gm2dv_{2}^{2} = \dfrac{Gm}{2d}.

D is the trap. Equal momenta do not mean equal energies: with K=p22mK = \dfrac{p^{2}}{2m} and the same pp, the lighter sphere gets three times the kinetic energy of the heavier, 34\dfrac{3}{4} of the total against 14\dfrac{1}{4}.