A Walk in the Sun

I recently read the short story “A Walk in the Sun” by Geoffrey A. Landis. It was based on the fact that on the Moon, if you walk towards the Sun, you can keep up with it.

I absolutely needed to check it!

The calculation is trivial. The circumference of the Moon at the equator is 10,921 km. Its orbital period around Earth is 27.322 days. Then, the surface speed at the equator is 10,921 / 27.322 / 24 = 16.65 km/h (or 10.35 miles/h). Considering that the Moon gravity is about 1/6g, it should be possible to sustain 16.65 km/h for quite a while.

In the story, an astronaut stranded on the Moon, in order to survive, needs to keep the solar panels of her life-support unit in sunlight. She has no problem in keeping up with Sun, at least on the near side of the Moon, which is less rugged than the far side. In fact, at times, she can easily gain ground.

The pilot crash-lands her shuttle near Mare Smythii, which is close to the equator. Therefore, Landis is right in saying that she needs to run at about 10 miles/h. But, in my opinion, he made a mistake: instead of running directly towards the Sun, considering that she could actually outrun it and that the Sun was still quite high, the pilot should have also moved towards one of the poles (e.g., the North Pole).

This is because what the pilot needed was to move towards the Sun 360 / 27.322 / 24 = 0.55 degrees of longitude per hour, not to run at 10 miles/h around the equator. The closer to a pole, the slower she would have needed to move. At a pole, by climbing to the top of a hill or of a crater, she might have not needed to move at all, and just adjust the orientation of the solar array every now and then.

Assuming that the pilot moved along a major circle of constant inclination with respect to the equator, to calculate how she should have moved, we have to look at spherical triangles. We can first of all define the following:

O: The point of origin, where the shuttle crashed, assumed to be on the equator. We can also use this symbol to indicate the angle of inclination of the pilot’s path.

t: The current time.

P: The point reached by the pilot at time t while moving on a great circle with inclination O and speed V.

E: The point on the equator on the same meridian of P.

d: The distance between O and P measured on the Moon surface.

R: The Moon radius (average = 1,737.10 km).

lat: The latitude of P. That is, the angle subtended by the radii through P and E.

lon: The difference in longitude between O and P (or between O and E, as E, by definition, has the same longitude as P). That is, the angle subtended by the radii through O and E.

W: The speed of 16.65 km/h necessary to keep up with the Sun when running on the equator.

The spherical triangle OPE is right in E, because a meridian always intersect the equator at 90 degrees. Therefore, from a simple (!) formula of spherical trigonometry for right triangles, we can write:

(1): sin(lat) = sin(O) * sin(d/R)

In case you are interested, the formula can be expressed in words as follows: “In every right spherical triangle, the sine of one of the sides adjacent to the right angle equals the sine of the opposite angle multiplied by the sine of the hypotenuse”. Note that in spherical geometry, instead of saying “the angle formed by the radii passing through the ends of a segment of great circle”, you can simply say “the sine of the segment”.

Another formula of spherical trigonometry let us calculate the difference in longitude between O and P:

tg(lon) = tg(d/R) * cos(O)

If we impose to keep pace with the Sun, tg(lon) becomes tg(W*t/R).

(2): tg(W*t/R) = tg(d/R) * cos(O)

This lets us resolve (2) in d:

(3): d(t) = R * arctg(tg(W*t/R) / cos(O))

When the pilot starts her run, she needs to maintain a speed given by W/cos(O). For example, if she travels with an inclination of 30 degrees, she needs to go just a bit faster than 19.2 km/h. With an inclination of 45 degrees, her initial speed needs to be around 23.5 km/h.

As she progresses, the length of the parallel she is on becomes shorter. That means that she can walk slower and still maintain the 0.55 degrees/hour necessary to keep up with the Sun.

There is a problem, though: when t passes the value 1.57*R/V (i.e., 163.8 h), W*t/R becomes greater than 90 degrees (which means that the pilot has completed 1/4 of the journey around the Moon), and tg(W*t/R) goes through a point of singularity before becoming negative. It turns out (but I am too lazy to study the configuration and understand why) that the singularity always occurs when the latitude equals the inclination (i.e., when lat equals O).

The solution is to move O to the coordinates of P when the tangent diverges, and keep going from there. To do so, we need to replace W with the speed needed to keep up with the Sun when travelling at the latitude of P, which is given by W_p = W * cos(lat_p). But in tg(W*t/R), we also need to replace R with R * cos(lat_p). This means that we can leave (3) as it is. And it makes sense, because we are only interested in the angle around the Moon, not in the actual distance travelled.

Well, I have spent quite a bit of effort on this problem, and it’s time to stop. The conclusion is that, if the pilot had run at an inclination of 30 degrees, she would have reached a latitude of 60 degrees in less than two weeks, progressively slowing down from the initial speed of 19.2 km/h to a more sedated 9.6 km/h. Then, she could have followed the 60 degree parallel and kept up with the Sun by walking at 8.3 km/h.

If the pilot had maintained a more ambitious inclination of 45 degrees, after two weeks, she would have reached the pole. The initial speed would have needed to be a more brisk 23.5 km/h. But only initially, because already after six days, the speed to keep up with the Sun would have gone down to 16.8 km/h.

The bottom line is: if you remain stranded on the Moon and need to keep pace with the Sun, don’t just run towards the Sun. Go as quickly as possible towards the nearest pole.