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I am trying to figure out how fast an object will have to travel (at a height around 10,600 m) to never be in the presence of the moon. Meaning the object will always be on the opposite side of the moon to the Earth.

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    $\begingroup$ Awww... I liked the concept of "Cursing" altitude. $\endgroup$ – Carl Witthoft Jan 15 at 20:02
  • $\begingroup$ Cross-posted to physics.stackexchange.com/questions/454464/… $\endgroup$ – PM 2Ring Jan 15 at 20:55
  • $\begingroup$ Do you mean that the object must stay at the point directly opposite the moon, or is a location anywhere where it cannot see the moon ok? $\endgroup$ – antispinwards Jan 15 at 22:49
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Other answers already note that the average period of the Moon in the sky (in hours) is about

$$\frac{1}{\frac{1}{1 \ \text{day}} - \frac{1}{27.32 \ \text{days}}} \approx 1.038 \ \text{days, or } 24.91 \ \text{hours}. $$

At the Earth's equator, where the circumference is about 6378.137 km plus the OP's 10.6 km alitude = 6388.7 km, that would be about 1611 km/hour.

However, you can prorate that by $\cos(\text{latitude})$ so at 60 degrees north or south that would be only half, or only about 806 km/s and that's only about 0.76 the speed of sound at cruise altitude.

But what's really interesting is that if you go far enough towards one of the poles, you may be able to hide for the moon for almost two weeks without moving at all!

According to this answer which makes for very interesting reading, the Moon can remain above the horizon at one pole for two weeks while being below the horizon and therefore invisible at the other for those two weeks.

That can get you ready for your flight from one pole to the other, in a sort-of spiral shape to stay hidden from the Moon just long enough to reach the the other pole. Then you've got close to two more weeks before you make your return flight.

You will have quite a challenge finding a place to land near the North pole, and global warming means that's even more challenging for more months a year, but who knows maybe after a few decades of warming there will be floating airports in the arctic ocean and hotels and...

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    $\begingroup$ I was going to mention that near-polar solution, but wanted to stick with the observer being diametrically opposite the moon. Good point, though. $\endgroup$ – Carl Witthoft Jan 16 at 20:06
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Well, as a first approximation, you know the Earth rotates once every 24 hours while the moon moves "only a little" in that time. This means your object must appear to an observer on the Earth' surface to rise and fall 12 hours opposite to the moon. First thing, then, is you better be moving East to West, just as the moon appears to do.

Using the Earth's radius as approximately 6400 km, your object is essentially at the Earth's surface to 2 significant digits. The orbital circumference is thus $2*\pi*6400$ km. You want to cover that distance in 24 hours, or 1676 km/hr .

If you want to get very accurate results, you'll need to take into account the moon's eccentricity with respect to the equator (so that you are always on the opposite end of a diameter thru the Earth's center from the moon) as well as the fact that it moves approximately 1/28 of a revolution per day.

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There are 24.84 hours before the moon appears in the same position in the sky. So if you wanted to keep the moon apparently fixed on the opposite side of Earth, you would need to travel the circumference at 10,000m of the Earth (40,100km) in 24.84 hours, or 1610 km/hr.

At that altitude the speed of sound is about 1090 km/hr, so you would need to travel substantially above the speed of sound

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