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 * velocitySolving 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/sPlugging 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
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 @hiergiltdiestfu. 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.