# Mass of Sun from Gravitational Constant

Jump to navigation
Jump to search

## Theorem

Let the 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:

- $T^2 = \left({\dfrac {4 \pi^2} {G M} }\right) a^3$

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 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:

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

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

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 $1$