I am writing a story in which a planet has a moon that orbits it about once a minute. In the story, the moon is pretty bright too, so the night sky has a little bit of a slow strobe light effect: 30 seconds really bright, 30 seconds dark, etc. The planet itself rotates pretty slowly, so the nighttime is about 48 hours long. The point is that the long nighttime is pretty annoying.

My questions:

  • Is there any reason why this wouldn't be possible?
  • What would the effects of such a quickly orbiting moon be on this planet? (Another potentially relevant fact: the planet is about half the size of earth, with about half the gravity. It does have water, not clear yet exactly how much).
  • What would make the moon extra bright? An extra bright sun? Or could it be something else?
  • 3
    $\begingroup$ In order for something to orbit a planet faster, it must orbit the planet closer. There is a thing called the Roche Limit that is extremely relevant to this question. Basically, if you get too close to a planet, you risk being torn apart by tidal forces (if you're a moon-like thing). As a result, there is a lower bound on how close you can get, and therefore an upper bound on orbital speeds. $\endgroup$ – MozerShmozer Jun 28 '17 at 22:41
  • 2
    $\begingroup$ On top of the problem with the Roche Limit, there is another physical limit with regard to orbital radius: actual size of the planet. For a planet to have half the gravity of the Earth, it will probably have to be somewhere between the size of Mars and Earth. The fastest you can orbit the Earth without having to counteract significant atmospheric drag is about 90 minutes. I don't know about Mars off the top of my head, but I doubt it's hugely different (though it is small and has less atmosphere, it also has less gravity, so orbits tend to be slower). $\endgroup$ – MozerShmozer Jun 28 '17 at 22:48
  • $\begingroup$ So, with the moon, is that 9,492km? Or something like a 5-6 hour orbit? $\endgroup$ – MichaelHouse Jun 28 '17 at 22:54
  • $\begingroup$ Ok cool, thanks. If I have the masses and radii of the planet and the moon is there a way to calculate the maximum orbital speed outside of the roche limit? $\endgroup$ – Mike Miller Jun 28 '17 at 23:01
  • $\begingroup$ One other thing to note is that since the moon causes the tides, you would have crazy wave patterns (like having the tide change drastically every minute or two) $\endgroup$ – The Mattbat999 Jun 29 '17 at 2:07

The formula for orbital speed is

$$v = \sqrt{\frac{G\cdot m_P}{r}}$$

Rearranging, the formula for orbital height is

$$h = \frac{G \cdot m_P}{v^2} - R_P$$

Where $G$ is a constant. The mass of the planet is $m_P$. The orbital velocity in $\frac{m}{s}$ is $v$. The radius of the planet is $R_P$. And that gives you the height above ground as $h$.

But what you actually want here is to calculate the orbital height purely in terms of the orbital period, not the velocity.

$$h = \sqrt[3]{\frac{G\cdot m_P \cdot t^2}{4\pi^2}} - R_P$$

You describe the planet as half the mass and radius of earth. You give the orbital period as sixty seconds. So that gives us

$$h = -1.6\cdot 10^6 m$$

So this would be about 1600 kilometers below ground.

I think that we can safely say that this is not feasible.

  • $\begingroup$ So it cannot occur naturally. Could a planet capture a high-speed transiting object, even temporarily? Probably not at the OP's velocity... It would just fly past the planet, wouldn't it? $\endgroup$ – CaM Jun 29 '17 at 14:57
  • 2
    $\begingroup$ @CM_Dayton Natural or not has nothing to do with it, it's simply how gravity works. Spacecraft obey the exact same orbital laws as moons, although, being strong they don't need to worry about the Roche limit. $\endgroup$ – Loren Pechtel Jun 29 '17 at 17:48
  • $\begingroup$ @Loren being strong has little to do with it. Mainly they are safe because they are small. $\endgroup$ – Mołot Jun 29 '17 at 20:32
  • 1
    $\begingroup$ @Mołot Being small lowers the Roche limit but doesn't protect you. The thing is the Roche limit is for bodies held together only by gravity. A satellite is held together by it's own structural strength, if part of the satellite is outside it's gravitational area it doesn't wander off. $\endgroup$ – Loren Pechtel Jun 29 '17 at 20:56

As other people have pointed out, you cannot have a moon orbiting that quickly. Nor is there a good reason for a super-bright moon if you think about it as a normal moon.

However, you can get a similar effect. Imagine a slowly orbiting moon that was also a parabolic mirror focused on the surface of your planet; it would aim a very bright dot at a very small area of the planet. Now, if that moon wobbled, you would have a very bright dot that moved around on the planet, and you would still have your strobe effect.


Binary White Dwarf Several Light Weeks Away

I can get you a light brighter than the moon that turns off for a minute once every five minutes, but it won't move in the sky. Put your planet around a normal star that is in a distant orbit with a pair of white dwarfs that orbit very close to each other, such as HM Cancri. Make one of the white dwarfs much older, and thus dimmer, than the other. Whenever the old white dwarf eclipses the young one (i.e., every five minutes) the majority of the light will be blocked. Note that for half the year, this blinking light will be in the daylight sky, and so less obnoxious.

You could make the period shorter by putting the dwarfs closer together, probably down to about 2 minutes.

You can make the light brighter by bringing the binary star closer to the star hosting your planet, but the X-ray radiation may begin to become a problem.

Note that this arrangement puts a time limit on the existence of whatever civilization you are writing about: The blinking will get faster and faster, and in about 300,000 years the stars will merge and wipe out everyone on your planet.


Rearranging Brythan's answer delves more into this problem:

$$h = \sqrt[3]{\frac{G\cdot m_P \cdot t^2}{4\pi^2}} - R_P$$ can be changed into:

$$\frac{h}{R_P} = \sqrt[3]{\frac{G\cdot \rho \cdot t^2}{3\pi}} - 1$$

So long as $$\sqrt[3]{\frac{G\cdot \rho \cdot t^2}{3\pi}} \ge 1$$ an object can have an orbit at this period. This implies:

$$\rho \cdot t^2 \ge \frac{3\cdot\pi}{G} = 1.41214639 × 10^{11} \frac{kg s^2}{m^3}$$

A planet of pure Osmium, which would have the highest density and have a surface humans could live on, would have a density of $22,590 \frac{kg}{m^3}$ This leaves you with a minimum of:

$$t \ge 2500 s$$

If you want a denser planet, your planet's going to be radioactive. Handwaving to PTU is fun, but Hassnium in the island of stability will still keep your orbit to about half an hour.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.