Universal law of gravitation

Drop an apple and it falls. Watch the sky long enough and the Moon never falls away from Earth, circling it forever instead. Isaac Newton's great insight was that these are the same fact: the apple and the Moon are both being pulled by the Earth, by exactly the same kind of force, obeying exactly the same rule. He called it gravitation, and he showed that it is not special to Earth or apples or moons at all — every object in the universe, from a pebble to a planet, pulls on every other object. This article states that rule precisely as a formula, shows how the force changes as masses and distances change, and works through the numbers so you can calculate it for yourself. By the end you should be able to write the law, plug numbers into it confidently, and reason about how the force scales — skills the rest of the gravitation chapter (acceleration due to gravity, weight, floatation) all build on.

The core insight

Every pair of objects in the universe attracts each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres, F=Gm1m2/r2F = Gm_1m_2/r^2.

The core idea

One law, one force, everywhere. Before Newton, people treated "why things fall" and "why the Moon orbits" as two separate mysteries. Newton's leap was to unify them: the force pulling the apple down and the force bending the Moon's path into a circle are the same gravitational force, just acting over very different distances. This is why the law is called universal — it applies to every pair of masses in the cosmos, not to some special category of "heavenly" force.

Newton's Universal Law of Gravitation states that every object attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. For two objects of masses m1m_1 and m2m_2 with centres separated by a distance rr, the force of attraction between them is:

F=G m1m2r2F = \frac{G\,m_1 m_2}{r^2}

where: FF is the gravitational force (newton, N), m1m_1 and m2m_2 are the two masses (kg), rr is the distance between their centres (m), and GG is the universal gravitational constant:

G=6.674×10−11 N ⋅ m2/kg2G = 6.674 \times 10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2

GG has the same value everywhere in the universe — on Earth, on the Moon, between two galaxies. It does not depend on the masses involved, the distance between them, or the medium (air, water, vacuum) in between, which is exactly why the law is "universal" rather than tied to any one place or pair of objects.

How the force scales. Two proportionalities follow directly from the formula, and exam numericals almost always test one of them:

  • Force and mass. F∝m1F \propto m_1 and F∝m2F \propto m_2. If either mass doubles while rr stays fixed, the force doubles; if both masses double, the force becomes four times as strong.
  • Force and distance — the inverse-square law. F∝1/r2F \propto 1/r^2. Doubling the distance between the centres does not halve the force; it cuts the force to one-quarter. Halving the distance quadruples the force. This "square" behaviour, not a simple inverse, is the single most common thing students get wrong.

Because GG is so tiny (10−1110^{-11}), the gravitational pull between two everyday objects — say, two students sitting side by side — is far too small to feel. Gravity only becomes noticeable when at least one mass is astronomically large, like a planet.

Figure
Two circles labelled apple and Moon both pulled toward Earth by arrows of force, with the distance to the apple shown as one Earth radius and the distance to the Moon shown as sixty Earth radii
Same force, same law — near or far Earth apple r = 1 Earth radius Moon r = 60 Earth radii Both the apple and the Moon feel the same F = Gm1m2/r squared The far Moon feels a much weaker pull than the near apple — same law, different r
Derivation AScaling the force when masses and distance both change

The goal is to find, in one clean chain of algebra, exactly how the gravitational force between two objects changes when both of their masses and the distance between them are altered by given factors — a skill needed whenever a numerical asks "what happens to FF if..." rather than for a single number, and one that trips students who try to scale mass and distance separately instead of substituting directly into the formula.

Step 1. Write the original force. For masses m1m_1, m2m_2 separated by rr, F=Gm1m2r2F = \dfrac{Gm_1m_2}{r^2}.

Step 2. Suppose each mass is doubled and the distance is also doubled: m1′=2m1m_1' = 2m_1, m2′=2m2m_2' = 2m_2, r′=2rr' = 2r. Substitute these directly into the same formula rather than guessing: F′=G (2m1)(2m2)(2r)2F' = \dfrac{G\,(2m_1)(2m_2)}{(2r)^2}.

Step 3. Simplify the numerator and denominator separately. The numerator becomes G×4 m1m2G \times 4\,m_1m_2. The denominator becomes 4r24r^2. So F′=4Gm1m24r2=Gm1m2r2=FF' = \dfrac{4Gm_1m_2}{4r^2} = \dfrac{Gm_1m_2}{r^2} = F.

Step 4. Read off the result. Doubling both masses and doubling the distance leaves the force completely unchanged, because the mass factor of 44 in the numerator is exactly cancelled by the distance factor of 44 in the denominator (from squaring the doubled rr). This is the general method for any scaling question: substitute the new values into F=Gm1m2/r2F = Gm_1m_2/r^2 symbolically first, simplify, and only then compare to the original FF — never try to combine the mass and distance effects by intuition alone.

Worked example

Two objects of masses 40 kg and 60 kg are placed with their centres 5 m apart. (a) Find the gravitational force of attraction between them. (b) If the distance is now reduced to one-third of the original, find the new force in terms of the original.

Step 1. List the knowns for part (a): m1=40m_1 = 40 kg, m2=60m_2 = 60 kg, r=5r = 5 m, G=6.674×10−11 N ⋅ m2/kg2G = 6.674\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2.

Step 2. Substitute into the law: F=Gm1m2r2=6.674×10−11×40×6052=6.674×10−11×240025F = \dfrac{Gm_1m_2}{r^2} = \dfrac{6.674\times10^{-11} \times 40 \times 60}{5^2} = \dfrac{6.674\times10^{-11}\times 2400}{25}.

Step 3. Simplify: F=1.6018×10−725=6.41×10−9 NF = \dfrac{1.6018\times10^{-7}}{25} = 6.41\times10^{-9}\ \text{N}. This is about six-billionths of a newton — far too small to feel, which is exactly why we never notice everyday objects attracting each other.

Step 4. For part (b), use the inverse-square scaling directly rather than recomputing from scratch. New distance r′=r/3r' = r/3, so F′=Gm1m2(r/3)2=Gm1m2×9r2=9F=9×6.41×10−9=5.77×10−8 NF' = \dfrac{Gm_1m_2}{(r/3)^2} = \dfrac{Gm_1m_2 \times 9}{r^2} = 9F = 9 \times 6.41\times10^{-9} = 5.77\times10^{-8}\ \text{N}.

The gravitational force is 6.41 × 10⁻⁹ N at 5 m, and it grows to 9 times that value, 5.77 × 10⁻⁸ N, once the distance is cut to one-third.

Common traps

  • Writing F∝1/rF \propto 1/r instead of 1/r21/r^2. Doubling the distance cuts the force to one-quarter, not one-half — the "square" in inverse-square law is not optional.
  • Forgetting the unit of GG. G=6.674×10−11 N ⋅ m2/kg2G = 6.674\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2; dropping or misremembering this unit produces dimensionally wrong answers even when the number is right.
  • Confusing GG with gg. GG is the universal gravitational constant, the same everywhere in the universe; gg is the acceleration due to gravity, which varies from planet to planet. They are related but never interchangeable.
  • Scaling masses and distance separately by intuition. When both change at once (as in the derivation above), substitute the new values into F=Gm1m2/r2F = Gm_1m_2/r^2 symbolically and simplify — do not try to add or guess the combined effect.

Ask the tutor: Want to try a few more scaling numericals, or see how this same law is used to work out the value of gg on Earth in the next article — just ask.

Why don't I feel the gravitational pull of the person sitting next to me?

You are being pulled, just by an amount too tiny to sense. Using F=Gm1m2/r2F = Gm_1m_2/r^2 with two people of ordinary mass a metre or so apart, the force works out to a few times 10−910^{-9} N — roughly what you calculated for the 40 kg and 60 kg objects in the worked example. Human muscles and nerves simply cannot detect a force that small. Gravity only becomes noticeable when one of the masses is enormous, like the Earth, because FF is directly proportional to mass, and Earth's mass is about 6×10246\times10^{24} kg.

Why is it called the universal law if we only ever seem to test it with Earth in the problems?

Earth shows up in most numericals simply because its huge mass makes the force large enough to be interesting, and because it is the planet you live on. But the formula itself, F=Gm1m2/r2F = Gm_1m_2/r^2, contains no reference to Earth at all — it is written purely in terms of any two masses and the distance between them. The same equation correctly predicts the pull between two stars, two planets, or two marbles on a table. "Universal" means the rule and the constant GG never change no matter which two objects you plug in.

Does the gravitational force depend on what the objects are made of, like whether they are iron or wood?

No. The formula depends only on the mass of each object and the distance between their centres — not on shape, material, colour, or temperature. Two objects with the same mass, one made of iron and one made of wood, exert exactly the same gravitational force at the same distance. This is different from, say, magnetic force, which does depend strongly on what material is involved; gravity is far simpler and universal in that sense too.

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