Where is Mars Now?

Where is Mars now?

If you haven't found out about it previously, please find out where Mars is in this excellent web program that gives you the current state of the solar system, and all NASA missions. Check it out here and click on the solar system icon at the top of the page.

Due to our ability to harness electricity and create light pollution, where I now live on the east coast of the United States, it's well nigh impossible to pick out where Mars is in the night sky. I know, you can download a free app that shows you the position of the planets and the stars and thus you can see where Mars is right now, but it's not the same.

In the Australian outback, it's easy to see Mars as an angry red dot on the sky. No doubt the various Australian aboriginal tribes who lived beneath those skies for 65,000 years were familiar with the red planet (Hamacher, 2017).

'Planet' is derived from the Greek word planētēs (πλανήτης) which literally means 'wanderer'. To ancient peoples observing the sky and the steady transits of the stars across the skies, Mars was a red anomaly, which not only didn't match the cycle of the stars, but even travelled in a retrograde motion and became brighter at the same time. Many European cultures associated this intruder with the god of war, as did the Greeks.

The Orbit of Mars

When Galileo finally turned his telescope upon the moon in 1609 and discovered the four Galilean moons in 1610, he only had 8x magnification and really terrible color aberration by today's standards. Half a century later, in 1666, another Italian, Giovanni Cassini, was able to use an improved telescope to observe surface features on Mars, and he used these to determine the length of the Martian day as 24 hours and 40 minutes long. Today we know the Martian day to be 24 hours 37 minutes 22.66 seconds long. In 1672, Cassini and a colleague to French Guiana so they could conduct simultaneous measurements and make a parallax measurement of Mars in both locations to measure its differing position on the fixed stars when viewed from Paris and French Guiana. He was thus able to establish the distance from Sun to Mars - Mars is 1.524 times further away from the Sun than the Earth is. Because we live on Earth, we call it's distance from the Sun 1 astronomical unit (or a.u. for short).

While Galileo conducted observations with his telescope, Johannes Kepler was busy improving Copernicus heliocentric model of the Solar system. In 1609, as Galileo reported his findings of mountains on the Moon, Kepler eventually hit upon the idea of using ellipses to fit the locations of Mars using his mentor Tycho Brahe's measurements (Kepler, 1609). We won't discuss too much the fact that Brahe was famously incorrigible and Kepler had to wait until after he died to obtain full access to observational results that he had taken on the Danish island of Uraninborg. Using insights from these observations of Mars, Kepler was later able to come up with his third law which relates the time it takes for a planet to go around a star to the distance of that planet from the star (Kepler, 1619). Kepler's final work of 1619 is called "Harmonices Mundi" or the Harmony of the Worlds. Inspired by the idea of a harmony existing between geometry of polygons and the regularity and harmony of the orbits of the planets, Kepler was able to deduce a relationship between the distance of a planet to its star, and the amount of time it took to orbit the star:

t^2 = d^3 <eqn 1>

d is the distance from Mars to the sun, and if we insert that into the equation, we get:

t^2 = (1.524)^3 = 3.54

t = sqrt(3.54) = 1.881

Therefore the Martian year is 1.88 times longer than Earth. Given that Earth has 365.25 days in a year, the Martian year is 687 Earth days long.

Seasons on Mars

In 1800, one hundred and fifty years after Cassini discovered how far Mars was from the Earth, a German born British astronomer, William Herschel, who can rightly be called the first (Father?) spectroscopist. Herschel recognised the phenomenon of infrared radiation by carrying out a very cool heat experiment. Herschel used a prism to split the light from the sun into different colours "of the rainbow". He then used a thermometer to measure the amount of heat delivered across the spectrum from blue to red. He then moved his detector into the region beyond the red colours, and found that an even greater amount of energy was being deposited in this "infrared" region by invisible light. In carrying out this first infrared (IR) experiment, he had built the first spectroscope. As we shall discuss below, the story of how this instrument has been improved over the past 200 years is key to understanding how our knowledge of the seasons on Mars.

Not only did Father Spectroscopy discover the IR, he was also an avid telescope developer. In fact, his IR discovery was the result of him chasing down sources of glare in his telescope. He also became the first discoverer of a planet since antiquity when he discovered Uranus. But he is part of this Martian story because he improved Cassini's rotation period of Mars to 24 hours, 37 minutes and 8 seconds long, very close to today's accepted value. He also measured the inclination of Mars to the ecliptic (angle of rotation relative to the Sun-Mars line) as 24 degrees, and recognised it as similar to that of the Earth. In the course of these observations, he was able to observe the white polar caps of Mars and see that they varied over time. He speculated that the polar caps of Mars might consist of water ice, by analogy with the Earth. He also speculated that Mars may, as a result, be inhabited. However, he also carried out an observation of stars traveling behind Mars and noted that they did not 'twinkle' for a large period as they were obscured by Mars, which led him to infer (correctly) that today's Martian atmosphere must be relatively thin.

Insolation of the Surface of Mars

Insolation of a surface refers to the amount of sunlight reaching the surface. Following Cross (1971) we can work out the amount of light hitting Mars at any particular time for any given longitude and latitude.

Let's start with the assumptions we will make to make this task tractable, while at the same time getting results that are useful for the task at hand.

1. We are assuming Mars is a 3D spherical ball (it's not - like all planets in fact it's it's an ellipsoid, but lets ignore that).

2. We are assuming the Martian atmosphere (clouds and Rayleigh scattering due to small particles) plays no role in our calculation - effectively we are deriving the 'Top Of the Atmosphere' (TOA) measurement of the solar insolation.

3. In line with (2), we are assuming no topography is shadowing our observation point.

With these assumptions in mind, let us think about what we need to specify a particular point on a 3D ball. On Earth, we are familiar with latitude and longitude, and the same concept is relevant for Mars. Let's call our Martian longitude x and the declination of the Sun as seen from our point on Mars we will call y. Under these conditions, if we think of the local surface as just a plane, the amount of light hitting a person is going to be at an angle given by the sin of these two quantities.

Sin(x) is shown in the Figure below. This shows a function that starts at 0, peaks in the middle, and shrinks to 0 at the end. We can use this function to project the solar insolation onto the globe of Mars, with the North pole being at the origin (0 deg), the equator in the middle (90 deg) and the south pole at the right (180deg).

In addition, the solar insolation is also reduce according to the local time of day. If we again use the sin function to map this, it starts at 0, that means the start of the day, peaks in the middle of the day, and shrinks to 0 at the end of the day as the sun rises. Let's call the angle representing the day the "latitude of the subsolar point" following Briggs 1974 or the "solar co-latitude" according to Cross 1971. Then we can take the product of the two sin functions to take both the latitude and local time of day into account. Then we get the following plot.

Looking at this plot, and mentally projecting it onto a ball, we can get the amount of light at any time of the day for any latitude. For example, if we use a latitude of theta = 80 deg in the north pole region, we need to map this correctly into the space for our sin function. If you give theta = 0 (the equator), we want pi/2, if you give theta = 90, we want to give pi. If you give -90 (south pole) we want pi.

A function that does what we want is:

plot sin(-1((abs(y)-90)*pi/180)), y=-90..90

and the result is given below, courtesy of Wolfram Alpha. This map has as its domain (input) the values -90 to 90 and gives back the sin of that value after converting to radians. This gives us the 'standard' latitude coordinates used by planetary scientists.

So we now have a way to go from a latitude value in degrees to the amount of light that we receive at that latitude from the Sun. At the equator, we get the full amount (x1). At 90 or -90 we get nothing. At 45 degrees, half way in degrees to the pole, we get 1/sqrt(2), or 0.707107... and at 80 degrees, 10 degrees of the north pole, we get 0.1736...

Now in fact the same relationship holds for the local time of day. At local midday, when the sun is as high as it gets over a certain location, the amount of light will be maximal, and as the sun rises, the angle would be -90, and the output would be 0. As the sun sets below the horizon, the day angle is 90, and the output is again 0. Lets call this day angle delta. Then, we can get the combined effects of day angle and latitude by multiplying them together like this:

sin(theta)sin(delta).

This gives us the instantaneous amount of light on an imagined point on the surface of the planet.

So far we have assumed Mars is sitting next to the sun. If we now want to factor on how much light is received when the Mars is moved out to its position in the solar system, we multiply the solar output (I_0)by (r_0/r)^2. That is due to the 1/r^2 law that governs Solar irradiation around the solar system.

Keplers first Law is

Following Murray et al (1973) we can use the following equation for the average yearly insolation <I>:

• <I> =S.sin(delta)/(pi.(1-e^2)^(1/2)

this