# The region of the solar system between Mars and Jupiter contains many asteroids that orbit the...

## Question:

The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consider an asteroid in a circular orbit of radius 4.3 10{eq}^{11} {/eq} m. Find the period of the orbit.

## Kepler's Third Law

Kepler's third law or law of periods states that the square of the orbital period **T** is directly proportional to the cube of the distance of the object or radius of the orbit **r**, Mathematically

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

## Answer and Explanation:

**Data Given**

- Radius of the orbit of asteroids {eq}r = 4.3 \times 10^{11} \ m {/eq}

- Mass of the sun {eq}M = 2.0 \times 10^{30} \ kg {/eq}

We know that time period is given by

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

{eq}\begin{align} T = \sqrt{\frac{4 \pi^2}{6.67 \times 10^{-11} \ N.m^2 kg^{-2} \times 2.0 \times 10^{30} \ kg} \times (4.3 \times 10^{11} \ m)^3} \end{align} {/eq}

{eq}\begin{align} T = 8.6543\times 10^7 \ s \end{align} {/eq}

{eq}\begin{align} T = \frac{8.6543\times 10^7 \ s}{365.24 \times 24 \times 60 \times 60 \ s/\text{year}} \end{align} {/eq}

{eq}\begin{align} \color{green}{\boxed{T = 2.74 \ \text{year}}} \end{align} {/eq}

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