if it takes a planet 2.8×10^8 s to orbit a star with a mass of 6.2×10^30 kg, what is the average distance…

Answer:

B. $9.36\times 10^{11}\text{ m}$

Explanation:

Step1: Recall the formula

The formula for the period of a planet orbiting a star is $T^{2}=\frac{4\pi^{2}r^{3}}{GM}$, where $T$ is the period, $r$ is the average - distance between the planet and the star, $G = 6.67\times10^{- 11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$ is the gravitational constant, and $M$ is the mass of the star.

Step2: Rearrange the formula for $r$

We can rewrite the formula as $r^{3}=\frac{GMT^{2}}{4\pi^{2}}$.

Step3: Substitute the given values

Given $T = 2.8\times10^{8}\text{ s}$, $M = 6.2\times10^{30}\text{ kg}$, and $G=6.67\times 10^{-11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$. [ \begin{align*} r^{3}&=\frac{6.67\times 10^{-11}\times6.2\times10^{30}\times(2.8\times10^{8})^{2}}{4\pi^{2}}\ &=\frac{6.67\times 10^{-11}\times6.2\times10^{30}\times7.84\times10^{16}}{4\pi^{2}}\ &=\frac{6.67\times6.2\times7.84\times10^{-11 + 30+16}}{4\pi^{2}}\ &=\frac{6.67\times6.2\times7.84\times10^{35}}{4\pi^{2}}\ \end{align*} ] [ \begin{align*} r^{3}&=\frac{6.67\times6.2\times7.84\times10^{35}}{4\times3.14^{2}}\ &=\frac{6.67\times6.2\times7.84\times10^{35}}{4\times9.86}\ &=\frac{6.67\times6.2\times7.84\times10^{35}}{39.44}\ &=\frac{321.9\times10^{35}}{39.44}\ &\approx8.16\times10^{36} \end{align*} ]

Step4: Solve for $r$

Take the cube - root of both sides: $r=\sqrt[3]{8.16\times10^{36}}\approx9.36\times10^{11}\text{ m}$.