The distance of a planet from the sun is 5 times the distance between the earth and the sun. The Time period of the planet is

To find the time period of the planet, we can use Kepler's Third Law of Planetary Motion.

Kepler's Third Law states that the square of the orbital period ($T^2$) of a planet is directly proportional to the cube of the semi-major axis of its orbit ($R^3$), which is its average distance from the Sun.
Mathematically, this relation is expressed as:
$T^2\propto R^3$
Or, in ratio form comparing the planet to the Earth:
${\left(\frac{T_p}{T_e}\right)}^2={\left(\frac{R_p}{R_e}\right)}^3$

Let's define the given variables:
- Let $T_e$ be the time period of the Earth, which is 1 year ($T_e=1\text{year}$).
- Let $R_e$ be the distance of the Earth from the Sun.
- Let $R_p$ be the distance of the planet from the Sun. According to the question, $R_p=5R_e$.
- Let $T_p$ be the orbital time period of the planet.

Now, substitute the values into the equation:
${\left(\frac{T_p}{1}\right)}^2={\left(\frac{5R_e}{R_e}\right)}^3$
Simplifying the expression:
${T_p}^2=5^3$

To solve for $T_p$, we take the square root of both sides:
$T_p={\left(5^3\right)}^{\frac{1}{2}}$
Using the laws of exponents, ${\left(a^m\right)}^n=a^{m\times n}$:
$T_p=5^{\frac{3}{2}}$
Since $\frac{3}{2}=1.5$:
$T_p=5^{1.5}\text{years}$

Thus, the time period of the planet is 5^1.5 years.