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?

  • 1
    $\begingroup$ See the short story "A Walk in the Sun" by Geoffrey Landis: en.wikipedia.org/wiki/A_Walk_in_the_Sun_(short_story) $\endgroup$ Commented Nov 12, 2019 at 19:59
  • $\begingroup$ @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. $\endgroup$
    – Uwe
    Commented Nov 12, 2019 at 21:27
  • $\begingroup$ 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. $\endgroup$
    – Ken Fabian
    Commented Nov 12, 2019 at 22:21
  • $\begingroup$ @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. $\endgroup$ Commented Nov 14, 2019 at 5:49
  • $\begingroup$ So the question is "if night moves at 14.9 kph, at which speed should I walk to stay in the night?" ?? $\endgroup$
    – jumpjack
    Commented Jul 14, 2020 at 12:25

2 Answers 2


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.


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