Worked exercise: the mass of the Sun from Kepler

Problem

The Earth orbits the Sun at distance $d = 1{,}5\cdot 10^{11}\ \text{m}$ with period $T = 3{,}15\cdot 10^7\ \text{s}$ (one year). Assuming a circular orbit, calculate the mass of the Sun.

Solution

Setup. The Sun’s gravitational force on the Earth provides the centripetal acceleration of uniform circular motion: $\frac{GM_\star m_T}{d^2} = m_T\,\omega^2\,d \;\Rightarrow\; M_\star = \frac{\omega^2 d^3}{G} = \frac{(2\pi/T)^2 d^3}{G}$

The Earth’s mass $m_T$ cancels out: it is not needed.

Angular velocity. $\omega = \frac{2\pi}{T} = \frac{2\pi}{3{,}15\cdot 10^7} = 1{,}99\cdot 10^{-7}\ \text{rad/s}$

Numerical substitution. $M_\star = \frac{(1{,}99\cdot 10^{-7})^2\cdot(1{,}5\cdot 10^{11})^3}{6{,}67\cdot 10^{-11}}$

$\ev{M_\star \approx 2{,}0\cdot 10^{30}\ \text{kg}}$

About $330\,000$ times the mass of the Earth. By measuring the period and radius of an orbit, one can thus “weigh” the central body.

Links

Topics: Gravitation Concepts: Kepler’s laws · Law of universal gravitation · Centripetal force · Uniform circular motion Skills: Symbolic setup