# Mass of Sun from Universal Gravitational Constant

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## Theorem

Let the universal gravitational constant be known.

Let the mean distance from the Earth to the sun be known.

Then it is possible to calculate the mass of the sun.

## Proof

From Kepler's Third Law of Planetary Motion:

- $T^2 = \paren {\dfrac {4 \pi^2} {G M} } a^3$

where:

- $T$ is the orbital period of the planet in question (in this case, the Earth)
- $a$ is the distance from the planet (in this case, the Earth) to the sun
- $M$ is the mass of the sun
- $G$ is the universal gravitational constant

In MKS units:

- $T = 60 \times 60 \times 24 \times 365.24219 \, \mathrm s$ by definition of year
- $a = 149 \, 597 \, 870 \, 700 \, \mathrm m$ by definition of astronomical unit
- $G = 6.674 \times 10^{-11} \, \mathrm N \, \mathrm m^2 \, \mathrm{kg}^{-2}$ by measurement, as by hypothesis.

- $M$ will be the result in kilograms.

Thus:

\(\ds M\) | \(=\) | \(\ds \dfrac {4 \pi^2 a^3} {T^2 G}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {4 \times \paren {3.14159}^2 \times \paren {149.60 \times 10^9}^3} {\paren {31.557 \times 10^6} \times 6.674 \times 10^{-11} }\) |

The calculation has been left as an exercise for anyone who has the patience.

$\blacksquare$

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation: Problem $1$