Planet Jupiter revolves around the Sun in 12 years ( 3.79 X 10 8 sec). What is its mean distance...

Question:

Planet Jupiter revolves around the Sun in 12 years ({eq}3.79 X 10^8 {/eq} sec). What is its mean distance from the center of Sun?

The mass of the sun is 1.99 x 10^30 kg.

Kepler's Third Law

Kepler's third law states that the square of the time period of a planet is directly proportional to the cube of the mean distance from the star, this law is also known as law of periods. Mathematically

{eq}\begin{align} T^2 \ \propto \ r^3 \end{align} {/eq}

Where T is the time period and r is the radius of the orbit of the planet.

Data Given

• Period of the Jupiter {eq}T = 12 \ \rm years = 3.79 \times 10^8 \ \rm s {/eq}
• Mass of the sun {eq}M = 1.99 \times 10^{30} \ \rm kg {/eq}

Now using Kepler's third law

{eq}\begin{align} T^2 \ \propto \ r^3 \end{align} {/eq}

{eq}\begin{align} T^2 = \frac{4 \pi^2}{GM} r^3 \end{align} {/eq}

{eq}\begin{align} r \ = \large \sqrt[3]{\frac{GMT^2}{4\pi^2}} \end{align} {/eq}

{eq}\begin{align} r \ = \large \sqrt[3]{\frac{6.67\times 10^{-11} \ \rm N.m^2kg^{-2} \times 1.99 \times 10^{30} \ \rm kg \times (3.79 \times 10^8 \ \rm s)^2}{4\pi^2}} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{r \ = 7.85 \times 10^{11} \ \rm m}} \end{align} {/eq}