# Earth is 1.5 x 10^11 m from the sun and has a period of 365.2 days. Mars is an average of 2.28 x...

## Question:

Earth is {eq}1.5 \times 10^{11} {/eq} m from the sun and has a period of 365.2 days. Mars is an average of {eq}2.28 \times 10^{11} {/eq} m from the sun. What is the orbital period of mars in earth years?

## Third Law of Kepler:

The mathematical expression that obtained form the Kepler's third law is shown below:

{eq}\displaystyle T^2=\frac{4\pi^2}{GM}R^3 {/eq}, where,

• T is the orbital period of a planet.
• G is known as universal gravitational constant.
• M is the mass of a planet.
• R is known as the semi-major axis or distance.

Given information:

• The distance between Sum and Earth is {eq}R_E=\rm 1.5 \times 10^{11}\ m {/eq}.
• The distance between Sum and Mars is {eq}R_M=\rm 2.28 \times 10^{11}\ m {/eq}.
• The value of the orbital period of Earth is {eq}T_E=\rm 365.2\ days {/eq}.

From the mathematical expression {eq}\displaystyle T^2=\frac{4\pi^2}{GM}R^3 {/eq}, we have:

{eq}\displaystyle T^2\propto R^3 {/eq}, where,

• {eq}\frac{4\pi^2}{GM} {/eq} is constant.

The relation between orbital period and distance of the given planets using the above expression is:

{eq}\displaystyle \frac{T_E^2}{T_M^2}=\frac{R_E^3}{R_M^3} {/eq}

Plugging all the given values in the above expression and simplifying it, we get:

{eq}\begin{align*} \displaystyle \frac{(\rm 365.2\ days)^2}{T_M^2}&=\frac{(\rm 1.5 \times 10^{11}\ m)^3}{(\rm 2.28 \times 10^{11}\ m)^3}\\ \displaystyle \frac{\rm 133371.04\ days^2}{T_M^2}&=\frac{\rm 3.375 \times 10^{33}\ m^3}{\rm 11.852352 \times 10^{33}\ m^3}\\ \displaystyle \frac{\rm 133371.04\ days^2}{T_M^2}&=\frac{3.375}{11.852352}\\ \displaystyle T_M^2&=11.852352\times \frac{\rm 133371.04\ days^2}{3.375}\\ \displaystyle T_M^2&=\rm 468373.4852\ days^2\\ \displaystyle T_M&=\sqrt{\rm 468373.4852\ days^2}\\ \displaystyle T_M&\approx \boxed{\rm 684.4\ days}\\ \end{align*} {/eq}

Thus, the orbital period of Mars is {eq}684.4 {/eq} days.