Worksheet 3, Problem 1: Martian Solar Day

What is the difference between sidereal and solar day on Mars if Mars has the same rotation period and orbits at 1.5 AU?

\(a_{Mars}=1.5AU\)
\(a_{Earth}=1AU\)
\(p_{Mars}=?\)
\(p_{Earth}=366.25 sidereal\text{ } days\)
\(day_{Mars, Solar}=?\)
\(day_{Earth, Solar}=24h\)
\(day_{Earth, Sidereal}=23h,56'\)
\(day_{Mars, Sidereal}=23h,56'\)

\(\theta=?\)

First let's review the background concepts.  A sidereal day is the time that a planet takes to rotate \(360^{\circ}\) relative to fixed distant stars.  A solar day is how long a planet takes to have the sun appear at the zenith at a given longitude.  Due the the orbit of the planet around the sun, a solar day will result in more than \(360^{\circ}\) of rotation and is thus longer than a sidereal day.  The difference between these two times will be dependent upon the angle through which the planet moves each day.

To begin the calculation, we should find the Mars' Period.
Kepler's Third Law of Planetary motion states:
\[p_{planet}^2\propto a_{planet}^3\]
We can use Earth as a reference point to find the period of Mars.
\[\frac{p_{Mars}^2}{a_{Mars}^3}=\frac{p_{Earth}^2}{a_{Earth}^3}\]
\[p_{Mars}=\left(\frac{p_{Earth}^2}{a_{Earth}^3}\times a_{Mars}^3\right)^{\frac{1}{2}}\]
\[p_{Mars}=\left(\frac{(366.25sidereal\text{ }days)^2}{(1AU)^3}\times (1.5AU)^3\right)^{\frac{1}{2}}\]
\[\boxed{p_{Mars}=673sidereal\text{ }days}\]

Next to find \(\theta\) we divide the degrees in a circle by the time it takes Mars to travel around the circle.
\[\theta =\frac{360^{\circ}}{p_{Mars}}=\frac{360^{\circ}}{673sidereal\text{ }days}=\boxed{.54^{\circ}/sidereal\text{ }day}\]
Converting from a degrees system to an hours system gives us:
\[.54^{\circ}/sidereal\text{ }day\times \frac{1440min}{360^{\circ}}=\boxed{2.1\frac{minutes}{day}}\]
This gives us the difference between a Martian Sidereal Day and a Martian Solar Day.  Our calculations show that a Martian Solar Day is 2.1 minutes longer than a Martian Sidereal Day.
This makes sense, because Marts moves through a lesser angle of orbit per day than Earth, so its time difference should also be less.

Calculated in collaboration with G. Grell, J. Herrmann, and H. Sigurslid.