The period of the Earth's orbit around the Sun is T = 365…

The period of the Earth's orbit around the Sun is $T = 365$ days and the Earth–Sun distance is $d = 1.5 \cdot 10^{11}\ \mathrm{m}$. Calculate the mass of the Sun, and show that $M_S = 4\pi^2d^3/(\gamma T^2)$.

Step 1: Idea

We cannot weigh the Sun, but we can watch how the Earth moves around it. The stronger the Sun's pull, the faster the Earth must move to stay on its orbit. Treating the orbit as a circle, gravity supplies the centripetal force.

Step 2: Derivation

For uniform circular motion with period $T$ the centripetal acceleration is $\omega^2d = (2\pi/T)^2d$. Newton's second law with the gravitational force: $\gamma\frac{M_Sm}{d^2} = m\left(\frac{2\pi}{T}\right)^2d$.

The Earth's mass $m$ cancels. Solving for $M_S$: $M_S = \frac{4\pi^2d^3}{\gamma T^2}$.

This is Kepler's third law ($T^2 \propto d^3$) solved for the mass of the central body.

Step 3: Numbers

The period must be in seconds: $T = 365 \cdot 86\,400 = 3.154 \cdot 10^7\ \mathrm{s}$.

\[ M_S = \frac{4\pi^2 \cdot (1.5 \cdot 10^{11})^3}{6.67 \cdot 10^{-11} \cdot (3.154 \cdot 10^7)^2}, \]

\[ M_S = \frac{1.332 \cdot 10^{35}}{6.635 \cdot 10^4}. \]

\[M_S \approx 2.0 \cdot 10^{30}\ \mathrm{kg}\]

That is about 330 000 times the mass of the Earth.

Check

Units: $ \mathrm{m^3}/(\mathrm{N\,m^2\,kg^{-2}} \cdot \mathrm{s^2}) $ $= \mathrm{kg}\ \checkmark $

Accepted value $1.99 \cdot 10^{30}\ \mathrm{kg}\ \checkmark$