Worksheet 3, Problem 1: The Sidereal Day vs. Solar Day on Mars

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

We would like to solve for \(\phi\) since this how much more Mars must rotate after a sidereal day to complete a solar day. Since a sidereal day is relative to the fixed stars which have an arbitrary and large distance from Mars, we can assume that they are infinitely far away, which makes the lines extending from Mars to the stars horizontal. Therefore, we can approximate that \(\phi\approx\theta\). To solve for \(\theta\), which is the angle subtended by one rotation of Mars about the Sun (given to be 24 hours), we must first find the period of Mars' orbit around the Sun. We can do this using Kepler's Third Law.

\(P^2=\dfrac{4\pi ^2d^3}{GM_{tot}}\)

We were not given the total mass of the system, however, so we must solve for the period of Mars using a ratio to the known period of Earth and the assumption that the total mass of the system is approximately equal for both the Sun and the Earth and the Sun and Mars. Letting \(P_E\) and \(P_M\) be the period of Earth and Mars respectively, \(d_E\) and \(d_M\) be the orbital radii, and \(M_E\) and \(M_M\) be the total mass of their respective systems, we have:

\(\big(\dfrac{P_M}{P_E}\big)^2=\dfrac{\frac{4\pi ^2d_M^3}{GM_M}}{\frac{4\pi ^2d_E^3}{GM_E}}\)

\(\big(\dfrac{P_M}{P_E}\big)^2=\dfrac{d_M^3}{d_E^3}\)

\(P_M=P_E\sqrt{\dfrac{d_M^3}{d_E^3}}\)

Plugging in the values \(P_E=365\) days, \(d_E=1\) A.U., and \(d_M=1.5\) A.U., we get:

\(P_M=(365\text{ days})\sqrt{\dfrac{(1.5\text{ A.U.})^3}{(1\text{ A.U.})^3}}=670\) days

Now, we must find \(\theta\), the number of degrees Mars rotates about the Sun in one day. Assuming circular motion:

\(\theta = \dfrac{360^{\circ}}{670\text{ days}}\approx 0.54^{\circ}\)

Since we determined that \(\phi\approx\theta\), we know that it takes 0.54 degrees longer to complete a solar day. So to find the difference between the sidereal and solar day, we just need to find how long it takes to rotate \(0.54\) degrees.

\(\dfrac{24\text{ hours}}{360^{\circ}}=\dfrac{t}{0.54^{\circ}}\)

\(t=0.036\text{ days}\approx 2\text{ mins}\)

Therefore, the solar day is approximately 2 minutes longer than the sidereal day on Mars.

I would like to thank Charles Law, Daniel Chen, and Shai Szulanski for their help in solving this problem.