# How fast would one have to move to achieve permanent lunar day by perpetually moving along with the solar terminator?

This comment got posted and I did not understand it.

for those interested, one rod per moonhour is about a quarter furlong per fortnight.

If I googled correctly and jumped to the right conclusions...

The terminator on the Moon moves at about 9.3 miles per hour (14.9 $$\frac{\text{km}}{\text{h}}$$).

Which means you could easily drive around the Moon at the equator and always stay in day or night (your choice). At lunar noon, drive 20mph for 4 hours, stop for 4 hours, and it would be lunar noon again.

How fast, on average, would you need to advance at the lunar equator to stay in eternal daylight?

• See the short story "A Walk in the Sun" by Geoffrey Landis: en.wikipedia.org/wiki/A_Walk_in_the_Sun_(short_story) Commented Nov 12, 2019 at 19:59
• @MarkFoskey walking on the Moon in a space suit with 15.4 km/h for 30 days continously is impossible anyway. Far away from the lunar equator the walking speed may be possible for some hours but not for many days.
– Uwe
Commented Nov 12, 2019 at 21:27
• A single circumnavigation beginning at local sunrise and arriving at local sunset could go slower. A lot of rough terrain - better have a course well mapped out. Commented Nov 12, 2019 at 22:21
• @Uwe Landis did a good job of making it seem believable. Ken Fabian's observation I think is key, along with an assumption of good luck on the part of the character. But it was too relevant a story not to mention. Commented Nov 14, 2019 at 5:49
• So the question is "if night moves at 14.9 kph, at which speed should I walk to stay in the night?" ?? Commented Jul 14, 2020 at 12:25

## 2 Answers

Using the diameter 3476 km and the Moon solar day of 29.53 Earth days I calculated a speed of 15.4 km/h for the day night line at the Moon equator.

$$\frac{3476 km*π} {29.53*24 h} = 15.4 km/h = 4.28 m/s$$

That is the necessary average speed at the lunar equator to stay in eternal day light.

• A rod is 5.0292 m, so a rod per moon-hour is 0.0473 mm/s
• A furlong is 201.168 m, so a furlong per fortnight is 0.598 m/h or 0.1663 mm/s
• The ratio is then 0.0473/0.1663 or about 0.284

A more generic answer could be:

Angular speed (constant in any point on surface) is $$2\pi$$ radians every $$29.53 \cdot 24$$ hours, which gives $$0.00443 \frac{\text{rad}}{\text{h}}$$.

Linear speed depend on local distance from rotation axis, which is $$r \cdot cos(latitude)$$.

Hence linear speed in any point on Moon surface is:

$$v = 0.00443 \cdot r \cdot cos(latitude)\ \left[\frac{\text{rad} \cdot \text{km}}{\text{h}}\right]$$

Of course for $$latitude=0°$$ it becomes just $$0.0443 \cdot r = 0.00443 \cdot 3476 = 15.4 \frac{\text{km}}{\text{h}}$$.

The "standard walk speed" is around 3 km/h:

$$3 = 0.00443 \cdot 3476 \cdot cos(latitude)$$

$$\implies latitude = arccos\left(\frac{3}{0.00443 \cdot 3476}\right) = 78.8°$$

But Sun would be vety low over the horizon, hence may be covered by craters' rims.