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In the story I'm writing, humanity has cracked FTL travel by compressing a chunk of matter outside the craft into a short lived, gravitationally weak black hole that leads to a higher dimension (known simply as a rift). Although they're usually stable enough for a ship to travel through, there is one MAJOR catch. If directed matter (such as air or debris) fell into a rift, the rift will grow larger and last longer. And, if enough was to fall in, the mass would destabilize the rift into a classical, high gravity black hole.

In a huge battle near a space station (which is big enough to be destroyed the Earth's Roche limit if it came too close) that's positioned around the Earth-Moon L4 point, one of the factions decides to weaponize one of these FTL drives to disable and clear away much of the enemy fleet. However, after the ship is suddenly shot like Swiss cheese, the drive fired almost directly at the space station, consuming both it's mass and almost all the mass of both fleets.

If the black hole tidally tugged on the Moon for a day and a half, would it be enough to de-orbit it? (or at least lower it's perigee to where it could aerobrake against the Earth's atmosphere)

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    $\begingroup$ The gravitational effect the black hole would have on the moon would actually be smaller than the effect the station and fleets had before being consumed. $\endgroup$ – Ville Niemi Jul 7 '16 at 5:48
  • $\begingroup$ Compare Black Hole Moon. $\endgroup$ – a CVn Jul 7 '16 at 7:24
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    $\begingroup$ Also, it looks to me like your actual question is at best peripherally related to the title. In principle, all you'd need in terms of delta-v to give people on Earth a bad day is whatever is enough to decircularize the Moon's orbit such that its orbit becomes sufficiently elliptical that it gets close enough to Earth at some point that Earth's gravity can do the rest, which probably isn't a whole lot, but delta-v is traditionally a measure used with propulsive maneuvers, not gravitational attraction. $\endgroup$ – a CVn Jul 7 '16 at 7:27
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    $\begingroup$ And of course, black holes aren't magic things. They have exactly the same gravitational attraction as any other object the same mass -- they are just a lot smaller for any given mass. Think of a black hole as simply a gravity point source instead of a large object with the same gravitational accelleration, and you get pretty close. $\endgroup$ – a CVn Jul 7 '16 at 7:29
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    $\begingroup$ As @MichaelKjörling said, you might want to rethink your title. I came in here ready to throw a huge heap of math at you regarding how much force it would take to cause the moon to impact Earth within a given time period... $\endgroup$ – guildsbounty Jul 7 '16 at 13:45
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Only thing that matters is mass (unless your tech actually increases mass, in other words creates matter from the rift or whatever).

So if the mass of the fleets and the space station wasn't enough to destabilize Moon's orbit, then converting that mass to a black hole will not do it either. Nothing really changed, as far as gravity from a moderate distance of a few kilometers (for that small a total mass) is concerned.

If you get too close to a black hole, tidal forces will start to have an effect, but for a black hole of that tiny size, that would probably be in the order of centimeters from the event horizon.

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    $\begingroup$ I think centimeters is overly far. Compare Black Hole Moon. $\endgroup$ – a CVn Jul 7 '16 at 7:24
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    $\begingroup$ Turns out the event horizon of the Earth, compressed to a black hole, is about 9.0 mm. $\endgroup$ – a CVn Jul 7 '16 at 15:53
  • $\begingroup$ @MichaelKjörling Note that region where tidal forces could cause havoc is much larger than event horizon (or the unstable region of space around event horizon, size 3r I believe). $\endgroup$ – hyde Jul 8 '16 at 4:05
  • $\begingroup$ Yes, of course; that's the Roche limit rather than the event horizon or Schwartzschild radius. It's still a similar magnitude, however. (Note: I'm not arguing against your answer, and even upvoted it.) $\endgroup$ – a CVn Jul 8 '16 at 7:45
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What you're looking for actually isn't delta-V. dV is a measure of changed velocity, so to find the dV for making Earth-based life less alive, all you really need to do is find the dV to put the Moon's periapsis under the Earth's surface (or honestly just kinda close would suffice).

What you're really looking for is how much energy it would take to get the moon to that point of crashing into the Earth. Using some well defined physics and orbital-mechanics equations, you could easily calculate the amount of force exerted on the moon by the black hole you've created, then find out how much energy that yields after sustained exertion for a day and a half. This amount of energy, applied retrograde (or not quite retrograde if you're looking for max realism) may or may not de-orbit the Moon. If not, consider relocating your black hole, changing the mass of it (space stations really don't have all that much mass once we get to BH levels), or changing the duration.

Also worth noting is that once a black hole evaporates away, a tremendous explosion takes place, so this might actually be a better propulsive force for do-orbit of the Moon.

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MozerShmozer suggested I actually supply the equations I mentioned, and they're completely right! Here is the first Google result for searching "Orbital Mechanics Equations", and it's actually surprisingly thorough. What you're looking for is the relationship between orbital velocity and altitude.

v = sqrt((GM)/r) 

Where G is the gravitational constant (a short Google search away), M is the mass of the Earth, and r is the altitude of your orbiting body.

What you need to do is find the velocity of the Moon at its current altitude, then find the velocity at its much less desirable altitude (note it will not be 0, that makes for super not fun math as it divides by 0. Enter the distance from the center of the Earth to the surface), and find the difference of the two. This is your dV (originally sought answer), and now you need to find the amount of energy needed to accelerate an object the mass of the Moon that much.

What this really means, is how much kinetic energy do we need to exert in the opposite direction of the Moon's current trajectory (also known as retrograde)? Luckily for us, someone figured that equation out a long time ago. It's

(m(v^2))/2

And we already know both mass of the Moon and both velocities the Moon will be travelling at. So plug all that in to find your dKe and now we have a (astronomically (pun) huge number) that describes how much energy (joules) we need to exert retrograde on the Moon. All that remains is to calculate how much force your black hole exerts on the Moon, and for how long, and then you'll come up with an amount of work done (in joules) on the Moon by the black hole. This should meet or exceed your dKe requirement (also in joules) for a Moon-Earth-impact, and thus the un-aliving of all things currently alive.

Edit 1: Spelling and some clarification. Edit 2: Followed MozerShmozer's advice of a fuller explanation.

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  • $\begingroup$ for making Earth-based life less alive - classic! On a serious note, you mention some well defined orbital mechanics equations and physics and ask OP to use them. Why not use them yourself and more completely answer his question? $\endgroup$ – MozerShmozer Jul 7 '16 at 15:54
  • $\begingroup$ I like that idea you mentioned of using a hawking radiation boom to mess with the Moon's orbit. Turns out somebody already did the math for it on the physics stack exchange for the total energy released (physics.stackexchange.com/questions/243650/…), but would be enough to substantially change the Moon's orbit, or would it just wind up blowing it to pieces? $\endgroup$ – Mattias Jan 25 '18 at 23:49

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