Question

The planet Mars takes \[1.88\] years to complete one revolution around the Sun.The mean distance of the Earth from the Sun is \[1.5 \times {10^8}\,km\] . Calculate that of planet Mars?

Answer 1

Hint: From Sun, Mars is the fourth planet, a dusty, cold, desert world with a very thin atmosphere. Seasons, polar ice caps, canyons, extinct volcanoes, and evidence that Mars was once much more active make Mars a complex world.

Complete step by step answer:
The third Kepler's law of planetary motion can be used to calculate the average distance between Mars and the Sun. The square of a planet's orbital length is proportional to the cube of the semi-major axis, which is the mean size of its orbit:
\[\dfrac{{P_M^2}}{{a_M^3}} = \dfrac{{P_E^2}}{{a_E^3}}\]
Here, \[{P_M}\] denotes Mars' orbital period, \[{P_E}\] denotes Earth's orbital period, \[{a_M}\] denotes Mars' mean distance from the Sun, and \[{a_E}\] denotes Earth's mean distance from the Sun.

Then, using this formula, we can calculate the average distance between Mars and the Sun.
That is,
\[a_M^3 = \dfrac{{P_M^2}}{{P_E^2}}a_E^3\]
\[ \Rightarrow {a_M} = \sqrt[3]{{\dfrac{{P_M^2}}{{P_E^2}}a_E^3}}\]
It is given that, \[{P_M} = 1.88\,year\] , \[{P_E} = 1.0\,year\] , and \[{a_E} = 1.5 \times {10^8}\,km\] .
Now we are going to substitute the given values in the formula to find out \[{a_M}\] .
\[ \Rightarrow {a_M} = \sqrt[3]{{\dfrac{{{{\left( {1.88\,year} \right)}^2}}}{{{{\left( {1.0\,year} \right)}^2}}} \times {{\left( {1.5 \times {{10}^8}km} \right)}^3}}}\]
\[ \therefore {a_M} = 2.28 \times {10^8}km\]

So, the answer is \[2.28 \times {10^8}\,km\].

Note: Remember the third Kepler's law of planetary motion. The third law states that when a planet gets farther away from the Sun, its orbital speed slows down, and vice versa.Between \[1609\] and \[1619\] , Johannes Kepler published his laws of planetary motion, which explain the orbits of planets around the Sun. The laws changed Nicolaus Copernicus' heliocentric theory, describing how planetary velocities differ by replacing circular orbits and epicycles with elliptical trajectories. As a result of his own laws of motion and universal gravitation, Isaac Newton demonstrated in \[1687\] that relationships like Kepler's would extend to the Solar System to a good approximation.