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Answer: Kepler's laws describe planetary orbits; using his third law (\(T^2 \propto r^3\)) with the centripetal force, Newton showed that the gravitational force \(F \propto \tfrac{1}{r^2}\).

Kepler's three laws

  1. First law: The orbit of a planet is an ellipse with the Sun at one of the foci.
  2. Second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
  3. Third law: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun, i.e. \(T^2 \propto r^3\) or \(\tfrac{T^2}{r^3} = K\) (constant).

How they helped Newton

Newton assumed, for simplicity, that a planet moves in a circular orbit of radius \(r\) around the Sun with uniform speed \(v\). The force needed to keep it moving in a circle is the centripetal force:

\(F = \dfrac{mv^2}{r}\) ... (1)

In one revolution the planet covers the circumference \(2\pi r\) in time \(T\), so \(v = \dfrac{2\pi r}{T}\). Putting this in (1):

\(F = \dfrac{m}{r}\cdot\dfrac{4\pi^2 r^2}{T^2} = \dfrac{4\pi^2 m r}{T^2}\) ... (2)

By Kepler's third law, \(T^2 = K r^3\). Substituting in (2):

\(F = \dfrac{4\pi^2 m r}{K r^3} = \dfrac{4\pi^2 m}{K}\cdot\dfrac{1}{r^2}\)

Since \(4\pi^2\), \(m\) and \(K\) are constants for a given planet, \(F \propto \dfrac{1}{r^2}\).

Thus Kepler's third law showed Newton that the force of attraction between the Sun and a planet must be inversely proportional to the square of the distance between them. This is the inverse square law of gravitation.

Doubtly · 3 May 2021 Report
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Answer: Kepler's three laws describe planetary motion; combining his third law with the formula for centripetal force shows that the force on a planet varies as \(1/r^2\).

Kepler's laws:

  1. The orbit of a planet is an ellipse with the Sun at one of the foci.
  2. The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
  3. The square of a planet's period of revolution around the Sun is directly proportional to the cube of its mean distance from the Sun: \(T^2 \propto r^3\), i.e. \(\dfrac{T^2}{r^3} = K\) (constant).

How they helped Newton: Take the orbit as nearly circular, of radius \(r\). A planet of mass \(m\) moving with speed \(v\) needs a centripetal force

\[F = \frac{mv^2}{r}, \qquad v = \frac{2\pi r}{T}\]

\[F = \frac{m}{r}\cdot\frac{4\pi^2 r^2}{T^2} = \frac{4\pi^2 m r}{T^2}\]

From Kepler's third law, \(T^2 = K r^3\). Substituting,

\[F = \frac{4\pi^2 m r}{K r^3} = \frac{4\pi^2 m}{K}\cdot\frac{1}{r^2}\]

So \(F \propto \dfrac{1}{r^2}\). Newton concluded that this centripetal force is provided by the Sun's gravitational attraction, which must therefore be inversely proportional to the square of the distance. This is the inverse square law of gravitation.

Doubtly AI · 23 Oct 2026 AI-verified · Oct 2026 Report
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