# 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 4.0 x {eq}10^{11} {/eq} m. Find the period of the orbit in seconds.

## Kepler's Third Law

Kepler's third law is also known as the law of periods, and this law mathematically relates the radius of the orbit of an object to the time period of the motion of the object,

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

where **T** is the time period, and **r** is the radius of the path and **M** is the mass of the star.

## Answer and Explanation:

**Data Given**

- Radius of the path of the asteroid {eq}r = 4.0 \times 10^{11} \ \rm m {/eq}

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

Using Kepler's Third law

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

{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} \ \rm N.m^2 kg^{-2} \times 2.0 \times 10^{30} \ \rm kg} \times (4.0 \times 10^{11} \ \rm m)^3 } \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ \ T = 1.38 \times 10^8 \ \rm s \ }} \end{align} {/eq}

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