# How do you calculate the mass of the sun, M_"sun", using Kepler's third law (T^2=(4 pi^2 r^3)/(G M_"sun"))?

## Assume the period of the Earth is $T = 3.156 \times {10}^{7}$ seconds and the Earth's distance from the Sun is $1.496 \times {10}^{11}$ meters.

May 13, 2017

Plug the given values into the given equation. Answer: $\approx 1.98955 \times {10}^{30}$ $\text{kg}$

#### Explanation:

Since we are given the equation:
${T}^{2} = \frac{4 {\pi}^{2} {r}^{3}}{G {M}_{s u n}}$
where $T = 3.156 \times {10}^{7}$ seconds, $r = 1.496 \times {10}^{11}$ meters, $\pi \approx 3.14$ is the mathematical constant, and $G = 6.67 \times {10}^{-} 11$ as the gravitational constant, we can first solve for ${M}_{s u n}$ with variables then substitute the given values to find ${M}_{s u n}$:

First, we will use variables to solve for ${M}_{s u n}$ to avoid the amount of numbers in the equation:
${T}^{2} = \frac{4 {\pi}^{2} {r}^{3}}{G {M}_{s u n}}$
${T}^{2} \left(G {M}_{s u n}\right) = 4 {\pi}^{2} {r}^{3}$
${M}_{s u n} = \frac{4 {\pi}^{2} {r}^{3}}{{T}^{2} G}$

Now, we can substitute our given values:
${M}_{s u n} = \frac{4 {\pi}^{2} {\left(1.496 \times {10}^{11}\right)}^{3}}{{\left(3.156 \times {10}^{7}\right)}^{2} \left(6.67 \times {10}^{-} 11\right)}$

$\approx 1.98955 \times {10}^{30}$ $\text{kg}$ rounded to 5 decimals