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

Calculate the **s****peed of the asteroid** and the **radius of its orbit**.

The aim of this article is to find the **speed** at which the **asteroid** is moving and the **radius** of its **orbital movement**.

The basic concept behind this article is **Keplerâ€™s Third Law for Orbital Time Period** and the expression for **Orbital Speed** of asteroid in terms of **Orbital Radius**.

**Keplerâ€™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:

\[v_o\ =\ \sqrt{\frac{G\ M_s}{r}}\]

## Expert Answer

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 **Keplerâ€™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 **Keplerâ€™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:

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

\[v_o\ =\ 16986.76\ \ \frac{m}{s} \]

\[v_o\ =\ 16.99\ \ \frac{km}{s} \]