# The asteroid belt circles the sun between the orbits of Mars and Jupiter.

The period of the asteroid is assumed to be $5$ Earth Years.

Calculate the speed of the asteroid and the radius of its orbit.

Kepler’s Third Law explains that the time period $T$ for a planetary body to orbit a star increases as the radius of its orbit increases. It is expressed as follows:

$T^2\ =\ \frac{4\pi^2r^3}{GM_s}$

Where:

$T\ =$ Asteroid Period in Second

$G\ =$ Universal Gravitational Constant $=\ 6.67\ \times\ {10}^{-11}\ \dfrac{Nm^2}{{\rm kg}^2}$

$M_s\ =$ The Mass of the star around which the asteroid is moving

$r\ =$ The radius of the orbit in which the asteroid is moving

The orbital speed $v_o$ of an asteroid is represented in terms of its orbital radius $r$ as follows:

$v_o\ =\ \sqrt{\frac{G\ M_s}{r}}$

Given that:

Time Period of Asteroid $T\ =\ 5\ Years$

Converting the time into seconds:

$T\ =\ 5\ \times\ 365\ \times\ 24\ \times\ 60\ \times\ 60\ =\ 1.5768\times{10}^8\ s$

We know that the Mass of Sun $M_s\ =\ 1.99\times{10}^{30}\ kg$.

Using the Kepler’s Third Law:

$T^2\ =\ \frac{4\pi^2r^3}{G\ M_s}$

By rearranging the equation, we get:

$r\ =\ \left[\frac{T^2\ G\ M_s}{4\pi^2}\right]^\frac{1}{3}$

We will substitute the given values in the above equation:

$r\ =\ \left[\frac{\left(1.5768\times{\ 10}^8s\right)^2\times\left(6.67\ \times\ {10}^{-11}\ \dfrac{Nm^2}{{\rm kg}^2}\right)\times\left(1.99\times{\ 10}^{30}kg\right)}{4\pi^2}\right]^\frac{1}{3}$

$r\ =\ 4.38\ \times\ {10}^{11}\ m$

$r\ =\ 4.38\ \times\ {10}^8\ km$

Now using the concept for orbital speed $v_o$, we know that:

$v_o\ =\ \sqrt{\frac{G\ M_s}{r}}$

We will substitute the given and calculated values in the above equation:

$v_o\ =\ \sqrt{\frac{\left(6.67\ \times\ {10}^{-11}\ \dfrac{Nm^2}{{\rm kg}^2}\right)\times\left(1.99\times{10}^{30}kg\right)}{4.38\ \times\ {10}^{11}\ m}}$

$v_o\ =\ 17408.14\ \ \frac{m}{s}$

$v_o\ =\ 17.408\ \ \frac{km}{s}$

## Numerical Result

The Radius $r$ of the Orbit of the asteroid is:

$r\ =\ 4.38\ \times\ {10}^8\ km$

The Orbital Speed $v_o$ of the asteroid is:

$v_o\ =\ 17.408\ \ \frac{km}{s}$

## Example

A planetary body circles around the sun for a period of $5.4$ Earth Years.

Calculate the speed of the planet and the radius of its orbit.

Solution

Given that:

Time Period of Asteroid $T\ =\ 5.4\ Years$

Converting the time into seconds:

$T\ =\ 5.4\ \times\ 365\ \times\ 24\ \times\ 60\ \times\ 60\ =\ 1.702944\times{10}^8\ s$

We know that the Mass of Sun $M_s\ =\ 1.99\times{10}^{30}\ kg$.

Using the Kepler’s Third Law:

$T^2\ =\ \frac{4\pi^2r^3}{G\ M_s}$

$r\ =\ \left[\frac{T^2\ G\ M_s}{4\pi^2}\right]^\frac{1}{3}$

We will substitute the given values in the above equation:

$r\ =\ \left[\frac{\left(1.702944\times{\ 10}^8s\right)^2\times\left(6.67\ \times\ {10}^{-11}\ \dfrac{Nm^2}{{\rm kg}^2}\right)\times\left(1.99\times{\ 10}^{30}kg\right)}{4\pi^2}\right]^\frac{1}{3}$

$r\ =\ 4.6\ \times\ {10}^{11}\ m$

$r\ =\ 4.6\ \times\ {10}^8\ km$

Now using the concept for orbital speed $v_o$, we know that:

$v_o\ =\ \sqrt{\frac{G\ M_s}{r}}$

We will substitute the given and calculated values in the above equation:

$v_o\ =\ \sqrt{\frac{\left(6.67\ \times\ {10}^{-11}\ \dfrac{Nm^2}{{\rm kg}^2}\right)\times\left(1.99\times{10}^{30}kg\right)}{4.6\ \times\ {10}^{11}\ m}}$

$v_o\ =\ 16986.76\ \ \frac{m}{s}$

$v_o\ =\ 16.99\ \ \frac{km}{s}$