Tuesday, February 28, 2017

Some moon math

So "Brianna Wu" (famous for gamergate) is trending, and because I love punishment, I clicked on it to see why. Apparently she tweeted that Elon Musk's plan to go to the moon is bad, because once there he can drop rocks on the Earth with the power of 100s of nuclear bombs. People are mocking her for the stupidity of this.

But the math checks out.

First of all, she probably got the idea from Heinlein's book The Moon is a Harsh Mistress where the rebel moon colonists do just that. I doubt she did her own math, and relied upon Heinlein to do it for her. But let's do the math ourselves.

Let's say that we want to stand at the height of the moon and drop a rock. How big a rock do we need to equal the energy of an atomic bomb? To make things simple, let's assume the size of bombs we want is that of the one dropped on Hiroshima.

As we know from high school physics, the energy of a dropped object (ignoring air) is:
energy = 0.5 * mass * velocity * velocity
Solving for mass (the size of the rock), the equation is:
mass = 2 * energy/(velocity * velocity)
We choose "energy" as that of an atomic bomb, but what is "velocity" in this equation, the speed of something dropped from the height of the moon?

The answer is something close to the escape velocity, which is defined as the speed of something dropped infinitely far away from the Earth. The moon isn't infinitely far away (only 250,000 miles away), but it's close.

How close? Well, let's use the formula for escape velocity from Wikipedia [*]:

where G is the "gravitational constant", M is the "mass of Earth", and r is the radius. Plugging in "radius of earth" and we get an escape velocity from the surface of the Earth of 11.18 km/s, which matches what Google tells us. Plugging in the radius of the moon's orbit, we get 1.44 km/s [*]. Thus, we get the following as the speed of an object dropped from the height of the moon to the surface of the earth, barring air resistance [*]:
9.74 km/s
Plugging these numbers in gets the following result:

So the answer for the mass of the rock, dropped from the moon, to equal a Hiroshima blast, is 1.3 billion grams, or 1.3 million kilograms, or 1.3 thousand metric tons.

Well, that's a fine number and all, but what does that equal? Is that the size of Rhode Island? or just a big truck?

The answer is: nearly the same mass as the Space Shuttle during launch (2.03 million kilograms [*]). Or, a rock about 24 feet on a side.

That's big rock, but not so big that it's impractical, especially since things weigh 1/6th as on Earth. In Heinlein's books, instead of shooting rocks via rockets, it shot them into space using a railgun, magnetic rings. Since the moon doesn't have an atmosphere, you don't need to shoot things straight up. Instead, you can accelerate them horizontally across the moon's surface, to an escape velocity of 5,000 mph (escape velocity from moon's surface). As the moon's surface curves away, they'll head out into space (or toward Earth)

Thus, Elon Musk would need to:
  • go the moon
  • setup a colony, underground
  • mine iron ore
  • build a magnetic launch gun
  • build fields full of solar panels for energy
  • mine some rock
  • cover it in iron (for magnet gun to hold onto)
  • bomb earth
At that point, he could drop hundreds of "nukes" on top of us. I, for one, would welcome our Lunar overlords. Free Luna!

Update: I've made a number of short cuts, but I don't think they'll affect the math much.

We don't need escape velocity for the moon as a whole, just enough to reach the point where Earth's gravity takes over. On the other hand, we need to kill the speed of the Moons's orbit (2,000 miles per hour) in order to get down to Earth, or we just end up orbiting the Earth. I just assume the two roughly cancel each other out and ignore it.

I also ignore the atmosphere. Meteors from outer space hitting the earth of this size tend to disintegrate or blow up before reaching the surface. The Chelyabinsk meteor, the one in all those dashcam videos from 2013, was roughly 5 times the size of our moon rocks, and blew up in the atmosphere, high above the surface, with about 5 times the energy of a Hiroshima bomb. Presumably, we want our moon rocks to reach the surface, so they'll need some protection. Probably make them longer and thinner, and put an ablative heat shield up from, and wrap them in something strong like iron.

I don't know how much this will slow down the rock. Presumably, if coming straight down, it won't slow down by much, but if coming in at a steep angle (as meteors do), then it could slow down quite a lot.

Update: First version of this post used "height of moon", which Wolfram Alfa interpreted as "diameter of moon". This error was found by . The current version of this post changes this to the correct value "radius of moon's orbit".

Update: I made a stupid error about Earth's gravitational strength at the height of the Moon's orbit. I've changed the equations to fix this.


newt0311 said...

I'd call this bullshit but I'm not used to bullshit being this obviously bullshit-ty...

For starters, the radius of the earth is ~4k miles. The distance from the Earth to the moon is ~240k miles. I'm not sure where you got the idea that the Earth's gravity on the moon was the same as Earth's gravity on Earth's surface but it is complete nonsense by a factor of ~3600x.

I don't have any stake in Brianna's statements or your statements or Elon Musk's supposed plans for world domination and normally I wouldn't normally comment on such things but your post blows through a rather startling number of basic common-sense checks.

1. If the Earth's gravity was roughly the same on the moon as it is on the earth's surface then Neil Armstrong wouldn't have been able to walk on the moon! After all, the pull towards Earth would have been six times as strong as the pull toward's the moon's surface.

2. Can we get a space-shuttle sized object to the moon? Probably though it might be expensive. Can we get an object 1/100 the size of the space-shuttle to the moon? Almost certainly yes though again it might be expensive. Do we have any rockets in space that can put out the same amount of energy as a hundred nuclear warheads? One nuclear warhead? As much energy as the Little Boy? Hell no. Especially when you start accounting for the oxidizer volume. Since getting to the moon takes more energy than falling back, that itself tells us that dropping a space shuttle from the moon to earth will not result in a 100x nuclear explosion.

3. The moon's gravity is 1/6th of Earth's, not say 1/1000th. Anything that you drop on the Earth will need to be lifted from the moon which implies a linear relationship between the amount of energy spent moon-side and the amount of energy released Earth-side (I think...). This point is a bit sketchy because I haven't done the math on this but I'm fairly sure it is correct. Of course the moon's gravity is 1/6th of Earth so there's some benefit but it isn't more than an order of magnitude or two. The real benefit is that the moon rocks apparatus can spend months building up the energy on the moon and it will be released in an instant when the rock hits earth... But you know, that can also be done on Earth (kinetic warheads in LEO for example). All going to the moon gets you is the large but mostly unusable store of potential energy in the moon's sheer mass. The massive logistical issues inherent in such an enterprise are probably not worth it.

I strongly suspect it would be a lot easier for Elon to just buy some third-world country and some PMCs and setup Musk-topia there. Certainly recruitment would be a lot easier when it doesn't involve a literal trip to the moon.

Finally if somebody really wanted to setup an orbital bombardment system I think the place they'd want to look at is the asteroid belt. Each individual asteroid is a lot smaller and already it's own piece so there's no need for expensive mining equipment, just some propellor stages (perhaps the Orion project has contributions to make). The asteroids could be brought into Earth orbit for ostensibly legitimate purposes and then forced into a decaying orbit whenever the controlling entity felt like it. Feels like we should be more worried about the Deep Space Mining company instead.

Anupam Kapoor said...

heinlien proposed this in 'moon is a harsh mistress' where the protagonist (mike the sentient computer) claims: 'But we can throw rocks at Earth, Man. We will.'

newt0311 said...

Um... not quite?

So let's see...

We can model this as two objects, the earth and the moon. Let's for the sake of this discussion say that they are stationary and the atmosphere, etc... is not relevant. Our hypothetical weapon starts with a rock on the surface of the moon, gives it just enough energy to get to the gravitational equilibrium point between the earth and moon and then Earth's gravity takes over from there. Let x be the distance from earth's center to the moon's center. So x varies from 0 to ~250k miles (we'll call the latter number d_me).

Then for any given x, the potential energy of the rock is*:

U_r(x) = U_e(x) + U_m(x) = -G m_r (m_e / x + m_m / (d_me - x))

U_r is the potential energy of the rock.

U_e, U_m are the potential energy contributions of the earth and the moon (resp.)

G is the gravitational constant.

m_r, m_e, m_m are the mass of the rock, earth, and moon (resp.)

Now we need to find the equilibrium point x_e. For this, we just need the force from earth's and moon's gravity to be equal:

G m_e m_r / x_e^2 = G m_m m_r / (d_me - x_e)^2

This simplifies to:

x_e = d_me * sqrt(m_e) / (sqrt(m_e) +/- sqrt(m_m))

It is clear that we want the (+) solution because the other one is past both the earth and the moon. Note also that this is the point where U_r is maximized which makes intuitive sense (it takes energy input to get the rock to this point). I get x_e = 3.332e8 meters which corresponds to about 0.9 * d_me.

So at the equilibrium point, all the energy of the rock is the potential energy U_r. Then when this rock actually hits, x = r_e (radius of earth) and U_r is much smaller and the rest is kinetic energy. So E_e, the energy of the explosion is:

E_e = U_r(x_e) - U_r(r_e)

After some math, we get:

E_e = G m_r(m_e / r_e + m_m / (d_me - r_e) - m_e / x_e - m_m / (d_me - x_e)) = m_r * 6.125e7 J.

Note that the implied velocity is ~11 km/s so that part of your post is roughly right.

But recall point #3 I made in my original comment: the weapon still needs to accelerate the rock to get it to the equilibrium point! The energy for that is:

E_m = U_r(x_e) - U_r(d_me - r_m)

By now it should be pretty obvious that all of this is linear in the mass of the rock. If we look at E_e / E_m, we get 23.78 independent of the mass of the rock in question.

So it may be possible to cause a 100x nuclear explosion on earth but it would take 4.2 nukes worth of energy on the moon to do it. Assuming unrealistically ideal conditions.

Maybe asteroid bombardment is still feasible?

Side note: apparently I was wrong about rocket thrusts and whether we can get a space-shuttle sized object to the moon. Ah well...

* This website has a good explanation for gravitational potential energy: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

Sys-Error said...

Thank you for nerding out for us. Breaking out my slide rule for this one!

Unknown said...

BRB, making a Minecraft mod