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  3. Newton's Law of Universal Gravitation

Gravitation · JEE & NEET Physics

Newton's Law of Universal Gravitation: notes and previous year questions

One pull for apples and moons: F = Gm₁m₂/r², the inverse-square law, the constant G, centre-to-centre distances, pairs of forces and adding pulls as vectors.

Newton's Law of Universal Gravitation in short

  • The pull that drops an apple also holds the Moon in orbit: the Moon is always falling, and always missing.
  • Every mass attracts every other mass with F = Gm₁m₂/r².
  • It is an inverse-square law because the pull spreads over an area that grows as r².
  • G = 6.67 × 10⁻¹¹ N m²/kg² is universal and tiny; g is local and changes.

1The big idea: the Moon is falling

An apple falls because the Earth pulls it. Newton's bold idea was that the same pull reaches the Moon and holds it in its orbit.

Picture a cannon on a very tall mountain. A slow ball falls nearby. A faster one lands farther away, because the ground curves away beneath it. At about 8 km/s the ball falls exactly as fast as the ground curves away, so it goes all the way round and never lands: it is in orbit.

2The law

Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them:

F=G m1m2r2F = G\,\frac{m_1 m_2}{r^2}r is the distance between the centres. G is the universal gravitational constant.
SymbolMeaningUnit
FFgravitational forceN
GGuniversal gravitational constant, 6.67×10−116.67 \times 10^{-11}N m²/kg²
m1,m2m_1, m_2the two masseskg
rrdistance between the centresm

Double one mass and the force doubles. Double both and it becomes 4 times as big. Double the distance and it falls to a quarter.

Why the square? The pull spreads out in all directions. At twice the distance it is spread over four times the area (the area of a sphere is 4πr24\pi r^2), so each bit gets a quarter. At three times the distance, one ninth. This is the inverse-square law; light and sound spread out the same way.

DistanceForce
rrFF
2r2rF/4F/4
3r3rF/9F/9
10r10rF/100F/100
r/2r/24F4F

3Big G

In 1798 Henry Cavendish hung a light rod with two small lead balls from a thin wire and brought two big lead balls close to them. The tiny pull twisted the wire; from the twist he found the force, and so GG.

G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2From G = Fr²/(m₁m₂). In base units m³ kg⁻¹ s⁻²; dimensions [M⁻¹ L³ T⁻²].

It is tiny. Two 50 kg students 1 m apart: F=6.67×10−11×50×50÷12≈1.7×10−7F = 6.67 \times 10^{-11} \times 50 \times 50 \div 1^2 \approx 1.7 \times 10^{-7} N, far too small to feel. Gravity is only strong when one of the masses is huge, like a planet.

4Centre to centre

A 50 kg student on the Earth (M=6×1024M = 6 \times 10^{24} kg, R=6.4×106R = 6.4 \times 10^6 m):

F=6.67×10−11×6×1024×50(6.4×106)2≈488 NF = \frac{6.67\times10^{-11} \times 6\times10^{24} \times 50}{(6.4\times10^6)^2} \approx 488\ \text{N}The same as the weight mg = 50 × 9.8 = 490 N.

So mg=GMm/R2mg = GMm/R^2. The body's mass cancels, leaving g=GM/R2g = GM/R^2: every body falls with the same gg. (More in the next lesson.)

For a body at a height hh, the distance in the law is r=R+hr = R + h. At h=Rh = R, r=2Rr = 2R and the pull is a quarter of its value on the ground. At h=2Rh = 2R, r=3Rr = 3R: one ninth.

Half the surface pull: 1/r2=1/(2R2)1/r^2 = 1/(2R^2), so r=2R≈1.41Rr = \sqrt2 R \approx 1.41R, a height of about 0.41R0.41R.

5Equal and opposite; the vector form

The Earth pulls a 100 g apple with about 1 N, and the apple pulls the Earth with 1 N too (Newton's third law). The apple accelerates at 10 m/s²; the Earth at only 1÷(6×1024)≈1.7×10−251 \div (6\times10^{24}) \approx 1.7 \times 10^{-25} m/s².

F⃗12=−G m1m2r2 r^12\vec F_{12} = -G\,\frac{m_1 m_2}{r^2}\,\hat r_{12}F₁₂: the force on m₂ due to m₁. r̂₁₂: unit vector from m₁ to m₂; the minus sign points F₁₂ back towards m₁.
F⃗21=−F⃗12\vec F_{21} = -\vec F_{12}
  • Always attractive: gravity never pushes.
  • Central: it acts along the line joining the centres.
  • Long range: it reaches to infinity, getting weaker as 1/r21/r^2.
  • Nothing placed between the masses blocks it.
  • It is the weakest force in nature, but huge masses make it win.

6Many masses: add as vectors

Each mass pulls on its own, as if the others were not there. The total is the vector sum (the principle of superposition):

F⃗net=F⃗1+F⃗2+F⃗3+…\vec F_{net} = \vec F_1 + \vec F_2 + \vec F_3 + \dots

Two equal pulls FF at right angles give F2+F2=F2\sqrt{F^2 + F^2} = F\sqrt2, not 2F2F. Two 3 N pulls at 90° give 32≈4.23\sqrt2 \approx 4.2 N.

Summary

Key ideas

  • The pull that drops an apple also holds the Moon in orbit: the Moon is always falling, and always missing.
  • Every mass attracts every other mass with F = Gm₁m₂/r².
  • It is an inverse-square law because the pull spreads over an area that grows as r².
  • G = 6.67 × 10⁻¹¹ N m²/kg² is universal and tiny; g is local and changes.
  • r is always measured from centre to centre: r = R + h above the Earth.
  • The pulls come in equal and opposite pairs; the lighter body accelerates far more.
  • Gravity is always attractive, central, long-range and cannot be blocked.
  • Pulls from several masses add as vectors.
  • mg = GMm/R² gives g = GM/R².

Every equation

Newton's law
F=Gm1m2/r2F = Gm_1m_2/r^2
The constant
G=6.67×10−11 N m2/kg2G = 6.67\times10^{-11}\ \text{N m}^2/\text{kg}^2
Dimensions of G
[G]=[M−1L3T−2][G] = [M^{-1}L^3T^{-2}]
Vector form
F⃗12=−Gm1m2r2r^12\vec F_{12} = -\frac{Gm_1m_2}{r^2}\hat r_{12}
Third law
F⃗21=−F⃗12\vec F_{21} = -\vec F_{12}
Distance above the Earth
r=R+hr = R + h
Superposition
F⃗net=F⃗1+F⃗2+…\vec F_{net} = \vec F_1 + \vec F_2 + \dots
At the surface
g=GM/R2g = GM/R^2
Half the surface pull
r=2 Rr = \sqrt2\,R
Square of masses
F=(2+12) Gm2/a2F = (\sqrt2 + \tfrac12)\,Gm^2/a^2
Earth–Moon null point
x=0.9 dx = 0.9\,d

Previous year questions with solutions

Real JEE and NEET questions on newton's law of universal gravitation. Try each one before you open the solution.

Q1JEE Main 2024One correct option

Four identical particles of mass mm are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is (22+132)Gm2L2(\frac{2\sqrt{2}+1}{32})\frac{{\mathrm{Gm}}^{2}}{L^{2}}, the length of the sides of the square is

  1. A4L
  2. B3L
  3. C2L
  4. DL2\frac{L}{2}
Show answer and solution

Answer: Option A

Let the side of the square be ss. The net force on a corner is (2+12)Gm2s2=22+12×Gm2s2\left(\sqrt{2} + \dfrac{1}{2}\right)\dfrac{Gm^{2}}{s^{2}} = \dfrac{2\sqrt{2} + 1}{2} \times \dfrac{Gm^{2}}{s^{2}}: two neighbours at right angles, plus the far corner at 2 s\sqrt{2}\,s, all along the diagonal. Setting it equal to the given force,

22+12s2=22+132L2\dfrac{2\sqrt{2} + 1}{2 s^{2}} = \dfrac{2\sqrt{2} + 1}{32 L^{2}}, so s2=16L2s^{2} = 16L^{2} and s=4Ls = 4L.

The slip to avoid is stopping at s2=16L2s^{2} = 16L^{2}: take the square root. Check the others by putting them back: s=2Ls = 2L gives a force four times too big, s=3Ls = 3L one 169\dfrac{16}{9} times too big, and s=L2s = \dfrac{L}{2} one 6464 times too big.

Q2NEET 2017One correct option

Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will

  1. Amove towards each other.
  2. Bmove away from each other.
  3. Cwill become stationary.
  4. Dkeep floating at the same distance between them.
Show answer and solution

Answer: Option A

Far from every planet and star, the only force on each astronaut is the other's gravitational pull, F=Gm1m2r2F = \dfrac{G m_{1} m_{2}}{r^{2}}. It always attracts, so each accelerates towards the other, and they drift together. The pull is tiny (for two 100 kg100\ \mathrm{kg} astronauts 1 m1\ \mathrm{m} apart it is 6.67×10−7 N6.67 \times 10^{-7}\ \mathrm{N}), but it is not zero.

The trap is D: a small force is still a force, and an unbalanced force changes the motion. B would need gravity to push, which it never does. C, becoming stationary, would need some force to bring them to rest, and nothing supplies one.

Q3JEE Main 2024One correct option

A metal wire of uniform mass density having length LL and mass MM is bent to form a semicircular arc and a particle of mass mm is placed at the centre of the arc. The gravitational force on the particle by the wire is :

  1. AGmMπ2L2\frac{\mathrm{GmM}{\pi}^{2}}{L^{2}}
  2. BGMmπ2L2\frac{\mathrm{GMm}\pi }{2L^{2}}
  3. C0
  4. D2GmMπL2\frac{2\mathrm{GmM}\pi }{L^{2}}
Show answer and solution

Answer: Option D

A semicircle of length LL has radius r=Lπr = \dfrac{L}{\pi}, and its mass per unit length is λ=ML\lambda = \dfrac{M}{L}. It spans 180∘180^{\circ}, so α=π2\alpha = \dfrac{\pi}{2} and sin⁡α=1\sin\alpha = 1:

F=2Gmλr=2Gm×ML×πL=2πGMmL2F = \dfrac{2Gm\lambda}{r} = 2Gm \times \dfrac{M}{L} \times \dfrac{\pi}{L} = \dfrac{2\pi GMm}{L^{2}}

The trap is C, zero: only a full ring surrounds the particle symmetrically. A semicircle's pulls all have a part towards the middle of the arc, and those never cancel. A, π2GMmL2\dfrac{\pi^{2}GMm}{L^{2}}, is GMmr2\dfrac{GMm}{r^{2}}, as if every piece pulled in the same direction. B is a quarter of the right value.

Practice questions, easy to hard

Three questions from the newton's law of universal gravitation practice ladder: one easy, one medium, one hard.

Q4One or more correct options

The gravitational pull has a few fixed features.

It is mutual: m1m_{1} pulls m2m_{2} exactly as hard as m2m_{2} pulls m1m_{1}, in opposite directions. The two pulls are an action–reaction pair, as Newton's third law requires.

It acts along the line joining the two particles.

It always attracts. Gravity never pushes.

It does not depend on what lies between the bodies. Air, water or a wall in between changes nothing: GG is the same, and nothing screens gravity off.

It is by far the weakest of nature's fundamental forces. We notice it only because the Earth's mass is so enormous.

Which statements are correct?

  1. AThe Earth pulls a falling apple with exactly as large a force as the apple pulls the Earth
  2. BA thick steel sheet placed between two bodies weakens the gravitational pull between them
  3. CThe pull on each of two particles points along the line joining them, towards the other
  4. DGravity is the strongest of the fundamental forces, which is why it holds the planets in their orbits
Show answer and solution

Answer: Options A, C

A is the mutual pull: the apple and the Earth pull each other equally hard, in opposite directions. C is the direction of the force, along the line joining the two, and towards the other body because gravity attracts.

B is the trap: nothing between two bodies changes their gravitational pull, since F=Gm1m2r2F = \dfrac{G m_{1} m_{2}}{r^{2}} has only the masses and the distance in it. D has it backwards. Gravity is the weakest of the fundamental forces; it rules the planets only because their masses are huge.

Q5One correct option

When the pulls are at an angle, it often helps to split each into components and let symmetry cancel some of them.

Two particles, each of mass MM, are fixed a distance 2d2d apart. A particle of mass mm sits on the perpendicular bisector of the line joining them, a distance dd from its midpoint. What is the net gravitational force on it?

  1. AGMmd2\dfrac{GMm}{d^{2}}
  2. Bzero
  3. CGMm2 d2\dfrac{GMm}{\sqrt{2}\,d^{2}}
  4. D2 GMmd2\dfrac{\sqrt{2}\,GMm}{d^{2}}
Show answer and solution

Answer: Option C

Each MM is d2+d2=2 d\sqrt{d^{2} + d^{2}} = \sqrt{2}\,d from mm, so each pulls with GMm2d2\dfrac{GMm}{2d^{2}}, at 45∘45^{\circ} to the bisector. The parts along the line joining the two masses cancel. The parts along the bisector add: 2×GMm2d2×cos⁡45∘=GMm2 d22 \times \dfrac{GMm}{2d^{2}} \times \cos 45^{\circ} = \dfrac{GMm}{\sqrt{2}\,d^{2}}, towards the midpoint.

The trap is A, GMmd2\dfrac{GMm}{d^{2}}, adding the two pulls as numbers without the cos⁡45∘\cos 45^{\circ}. B, zero, cancels everything, but only the sideways parts cancel. D, 2 GMmd2\dfrac{\sqrt{2}\,GMm}{d^{2}}, takes each mass as dd away instead of 2 d\sqrt{2}\,d.

Q6One correct option

A wire bent into an arc of a circle of radius rr, with mass per unit length λ\lambda, pulls a particle mm at the centre of that circle. Every piece is rr from mm. The piece at angle θ\theta from the arc's line of symmetry has length r dθr\,d\theta and mass λr dθ\lambda r\,d\theta, and pulls with Gmλr dθr2=Gmλr dθ\dfrac{Gm\lambda r\,d\theta}{r^{2}} = \dfrac{Gm\lambda}{r}\,d\theta. Across the line of symmetry the parts cancel in pairs; along it they add, each giving Gmλrcos⁡θ dθ\dfrac{Gm\lambda}{r}\cos\theta\,d\theta. An arc running from θ=−α\theta = -\alpha to θ=+α\theta = +\alpha gives

F=Gmλr∫−ααcos⁡θ dθ=2Gmλrsin⁡αF = \dfrac{Gm\lambda}{r}\displaystyle\int_{-\alpha}^{\alpha}\cos\theta\,d\theta = \dfrac{2Gm\lambda}{r}\sin\alpha

A full ring, α=π\alpha = \pi, gives zero, as symmetry demands.

A uniform wire of mass MM is bent into a quarter of a circle of radius rr. What is its pull on a particle mm at the centre of that circle?

  1. A22 GMmπr2\dfrac{2\sqrt{2}\,GMm}{\pi r^{2}}
  2. BGMmr2\dfrac{GMm}{r^{2}}
  3. C2 GMmπr2\dfrac{\sqrt{2}\,GMm}{\pi r^{2}}
  4. D4GMmπr2\dfrac{4GMm}{\pi r^{2}}
Show answer and solution

Answer: Option A

A quarter circle has length πr2\dfrac{\pi r}{2}, so λ=2Mπr\lambda = \dfrac{2M}{\pi r}, and it spans 90∘90^{\circ}, so α=π4\alpha = \dfrac{\pi}{4}. Then F=2Gmr×2Mπr×sin⁡π4=4GMmπr2×12=22 GMmπr2F = \dfrac{2Gm}{r} \times \dfrac{2M}{\pi r} \times \sin\dfrac{\pi}{4} = \dfrac{4GMm}{\pi r^{2}} \times \dfrac{1}{\sqrt{2}} = \dfrac{2\sqrt{2}\,GMm}{\pi r^{2}}.

The trap is D, 4GMmπr2\dfrac{4GMm}{\pi r^{2}}, which takes α\alpha as the whole π2\dfrac{\pi}{2} the arc spans instead of half of it. B, GMmr2\dfrac{GMm}{r^{2}}, lets every piece pull in the same direction, ignoring the cancelling parts. C uses the length of a semicircle, πr\pi r, for λ\lambda.