Explain how Newton arrived at his law of gravitation from Kepler's third law.

III. Long Answer Questions · Q2

Q. Explain how Newton arrived at his law of gravitation from Kepler's third law.

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Concept understanding — Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle with a force directed along the line joining them. For two point masses $m_1$ and $m_2$ separated by a distance $r$ , the magnitude of the attraction is

F = \frac{G\,m_1 m_2}{r^{2}},

where $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$ is the universal gravitational constant. The force is attractive, acts along the line of centres, and obeys Newton's third law (equal and opposite on the two bodies). Its defining feature is the inverse-square dependence on separation.

Superposition. When several masses act on a body, the net gravitational force is the vector sum of the individual forces. For an extended body — a rod, ring or shell — the body is treated as a collection of point-mass elements and the contributions are integrated. For a uniform ring of mass $M$ and radius $R$ , the force on a point mass $m$ placed on its axis at distance $h$ from the centre is directed along the axis and has magnitude

F = \frac{G M m\, h}{\left(R^{2} + h^{2}\right)^{3/2}},

because the components perpendicular to the axis cancel by symmetry while the axial components add. This axial formula is not a simple $1/h^{2}$ law: as $h$ changes, the force scales as $h / (R^{2}+h^{2})^{3/2}$ , so comparing the force at two axial positions (say $h$ and $2h$ ) means substituting into this expression rather than using the point-mass inverse square directly. Only when $h \gg R$ does the ring behave like a point mass and $F \to GMm/h^{2}$ .

Newton's law of gravitation is one of the earliest topics in the NCERT Class 11 Physics chapter on Gravitation, and "universal law of gravitation formula and examples" or "gravitation class 11 important questions" are searched constantly by CBSE board and NEET/JEE Main aspirants. The extended-body superposition case shown here — finding the force due to a ring on an axial point mass — is a classic JEE Main gravitation numerical built directly on this NCERT Class 11 foundation.

Step 1. For a planet in a (nearly) circular orbit of radius $r$ , the centripetal acceleration is $a=v^2/r$ . Using orbital speed $v=2\pi r/T$ , this becomes $a=4\pi^2r/T^2$ .

Step 2. Newton's second law then gives the required central force as $F=ma=4\pi^2mr/T^2$ , where $m$ is the orbiting planet's mass.

Step 3. Kepler's third law states $r^3/T^2=k$ , a constant that is the same for every planet orbiting the Sun. Rearranging, $1/T^2=k/r^3$ , and substituting into Step 2's expression for $F$ gives

F=4\pi^2mr\cdot\frac{k}{r^3}=\frac{4\pi^2mk}{r^2}.

This is already an inverse-square force in $r$ .

Step 4. Newton then reasoned using his own third law: if the Sun attracts the planet, the planet must equally attract the Sun, so the Sun's mass $M$ should appear explicitly and symmetrically in the force expression too (mass $m$ already appears, but $M$ does not yet).

Step 5. He therefore identified the constant $4\pi^2k$ with $GM$ (where $G$ is a new universal constant), giving

F=\frac{GMm}{r^2},

the full law of universal gravitation, with both masses now appearing symmetrically.

✓Final answer

F\propto m/r^2

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Q.Explain how Newton arrived at his law of gravitation from Kepler's third law.

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