In one of the early chapters of my upcoming novel, my hero drives a rover along a service road to repair a faulty railroad switch on Mars. I carefully describe the Hellas Planitia terrain, Mars’ dusty atmospheric conditions, even the midday summer temperature. The train roars by, its slip stream kicking up a cloud of dust. All of it to give you a sense of being there.
But one thing I didn’t discuss is what it would be like to drive down a gravel road on Mars. As I noodled on that, I realized driving on Mars will be quite different from here on Earth. So much so that the roads will have to be constructed differently than they are here on Earth.
I suspected that if I took my Toyota RAV4 to Mars, I’d likely skid off the road at the first curve I came to. Why? Because Mars’s force of gravity is only about one third of Earth gravity. Less gravity means my car would weigh less. Less weight means less friction between the tires and the gravel road. I’ll slide off the road as if it were paved with ice.
If I want to avoid visiting any roadside craters, I either need to slow down significantly, or design a more gradual (i.e., larger radius) curve. To confirm my hunch, I dusted off my old physics text book and consulted a number of videos on the internet. Here’s what I found. (For those whose eyes glaze over when you see math, just skip to my conclusion.)
The equation to calculate the maximum velocity my car can travel on a curve is pretty simple:
v = √(µ · g · r), where:
v = velocity in meters per second
µ = the Greek letter mu, the coefficient of friction, a dimensionless number. For a gravel road µ = 0.5.
g = the gravity constant. Earth’s gravity constant G, is 9.8 meters per second per second. Mars’s g =3.72m/sec^2
r = the radius of the curve, in meters. For this exercise, I’ve chosen 100m. If I extend the curve indefinitely, it becomes a circle 200 meters across. For simplicity, I’ve assumed the curve is flat, not banked.
The maximum speed I can travel on a flat curve on a gravel road here on Earth is:
v = √(0.5 · 9.81 m/sec^2· 100 m)
v = √490 m^2/sec^2 =22 m/sec =
80 km/hr, or about 50 mph Now let’s look at how fast I could drive this same curve on Mars:
v = √(0.5 · 3.72 m/sec^2· 100 m)
v = √186 m^2/sec^2= 13.6 m/sec =
49 km/hr, or about 30 mph Bottom line, I will have to drive slower on Mars by half to stay on my 100-meter radius curve. How much larger must my curve be for me to be able to drive the same speed (80 km/hr)? If I substitute in my value for velocity and solve for r:
80 m/sec = √(0.5 · 3.72 m/sec^2· r)
490 m2/sec^2= 0.5 · 3.72 m/sec^2 · r
r = (490m^2/sec^2)/(0.5 · 3.72m/sec^2) =
263.4m The radius of my curve will be over twice as large!
Looking down the road (pun intended), what does this imply for construction on Mars? It means having to rewrite engineering manuals where any equations rely on the gravity constant. That includes any formulas that depend on weight. As I’ve just shown, roads will have to be designed differently, or speeds lowered considerably. For example, any systems relying on the gravity flow of water will have to upsize pipes—like in a hydroponic system—to get the same flow rates.
This speaks to the importance of the Artemis program, where many of the apparatuses that will be used on Mars will be tested in the low grav lunar environment. Better to find out the engineers didn’t account for gravity in a critical system when the fix is three days away, instead of nine months.
It also means I’ll have to pay more attention to the effect of low gravity on the operations of Ep City!
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For further viewinghttps://www.youtube.com/watch?v=CNx0IBBicUQ
