Building a Space Launch Catapult

Ray Patrick

Fact or Fiction?

Lunar space catapult
Lunar space catapult

In Robert Heinlein’s novel The Moon is a Harsh Mistress, the inhabitants of a lunar prison colony stage a revolt, declaring themselves a nation independent from Earth. To make their revolution successful, the “Loonies” repurpose their electromagnetic space-launch catapult into a weapon. It was originally designed to place shipping containers into cislunar space, but they use it to hurl rocks down to Earth. After being accelerated to lunar escape velocity (about 2.38 km/s) the rocks fall down Earth’s steep gravity well, impacting at speeds closer to 11 km/s. The effect is equivalent to a nuclear attack.

Lunar revolution aside, the peaceful applications of this machine could be wonderful. Could such a catapult really be built? Similar machines, such as maglev trains, already exist, although they’re much smaller. Catapulting something to space would take a lot more energy than moving a commuter train a few miles.

The energy requirement grows with the square of the velocity. For an Earth-based catapult, the energy budget would be enormous. Besides the energy required to achieve orbital speed and altitude, even more would be needed to shove Earth’s thick atmosphere out of the way during the climb to space. By contrast, building one on the small, airless Moon would be much more doable - maybe even desirable. Imagine being able to place ships into orbit with infrastructure barely more complicated than a railroad! Zero to 5,342 mph with no moving parts! Could such an optimistic vision really come to pass? Let’s find out. We’ll begin by mathematically describing the problem:

Doing the Math

Choosing to build our catapult on the Moon is great for the reasons I outlined above, but also because it simplifies our mathematical analysis. There’s no wind resistance on the moon, and its 655-hour day means we can neglect its rotation when accounting for the forces acting on the projectile.1 For a given projectile mass, desired exit velocity, and acceleration, we can compute the required catapult length, operation time, and power consumption using just a few simple formulas. The simplest is operation time: for a known, constant acceleration and a desired exit velocity, the required operation time is given by:

t = v/a

If the projectile starts from rest, the required length of the catapult is given by:

L = ½at2

Since the projectile’s mass doesn’t change, we know from Newton’s second law that the force exerted on it by the catapult is equal to the product of its mass and its acceleration. We can combine this with the definition of work (force times displacement) to find the energy the catapult must impart to the projectile. Dividing the imparted energy by the operating time then yields the power required to launch the projectile (neglecting any losses).

E = m·a·L

P = E/t = (m·a·L)/t

Case 1: Lunar Escape Velocity

Basic Facts

Despite the high-tech subject, our analytical model is pretty simple. We can easily work some examples by hand. (Or, since we’re examining a piece of technology from a Heinlein novel, you could use your trusty slide rule for maximum authenticity.) Let’s assume we want to attain lunar escape velocity, 2.38 km/s, with a maximum acceleration of 3g (1 g being 9.81 m/s2, the mean acceleration due to gravity as felt at the Earth’s surface.) This works out to an operation time of 80.8 s (about 1 minute, 21 seconds) and a catapult length of 96.24 km.

Power Budgeting

How much power would be needed? It depends on the mass of the projectile being launched. To get a quantity that we can usefully compare between cases, we can instead compute the specific power, or power per unit mass. That is given by:

Pspecific = P/m = (a·L)/t

Using the figures we’ve got so far, this comes out to 35 kW/kg. In other words, every kilogram of projectile mass would cost 35 kW of power to launch to lunar escape velocity.

Electrical vs. Mechanical Power

But how much electricity would it take to use this thing? The “power” we just calculated is the true mechanical power it would take to accelerate a projectile at the given rate until it reached the desired velocity. The true cost in electrical power would be higher due to various losses such as heat and mechanical stress.

So would it be feasible to launch something off the Moon with this thing? Let’s say our projectile weighs as much as an Apollo command & service module (28,800 kg). This is a reasonable size for a spacecraft that could travel from the Moon to the Earth. Multiplying our specific power of 35 kW/kg by 28,800 kg reveals that, to accelerate a CSM from rest to lunar escape speed would take about 1 GW of mechanical power. Neglecting losses, let’s assume we need 1 GW of electrical power to do this. 1 GW (1000 MW) is about the typical output for a single-unit nuclear or coal-fired power plant. Since you can’t burn coal on the Moon, it looks like we’d need a decent-sized nuke plant to provide the needed power.

Alternate Energy Sources

A solar farm on the Moon
A solar farm on the Moon

Could we do it with solar power instead? After all, it never gets cloudy on the Moon. It seems like the perfect place for a giant solar farm. If we provide the needed power via solar panels, we’ll have a space launch complex with close to zero moving parts.

The bleeding edge of solar panel technology currently boasts about a 23% efficiency rate. That is, one of these panels can convert 23% of the sunlight it absorbs into electricity. The mean solar irradiance for Earth (and therefore the Moon) is 1362 W/m2. At 23% efficiency, solar panels could deliver 313.26 W/m2 continuously. To provide 1 GW of electrical power, we would need about 3.2 square kilometers of solar panels.2 We’d need to build a 1000m × 780m strip of solar panels. If we arranged them in a disk, it would have a 1-km radius.

Seems that launching from the Moon is a realistic goal as far as infrastructure requirements go. What about using a catapult to launch from Earth?

Case 2: Pikes Peak Catapult

Photo of Pikes Peak
Pikes Peak, elevation 4302 m (14,114 ft)

As you can probably tell, Heinlein got me thinking about this stuff to begin with. This example comes from Space Cadet (1948). After spending some shore leave on Earth, a military cadet returns to his orbiting school ship via a space launch catapult on Pikes Peak:

His position gave him an excellent view of the co-pilot and mate, waiting at the airplane-type controls. If the rocket motor failed to fire, after catapulting, it would be the mate’s business to fight the ship into level flight and bring her down to a deadstick landing on the Colorado prairie.

The captain manned the rocket-type controls. He spoke to the catapult control room, then sounded the siren. Shortly thereafter the ship mounted up the face of the mountain, at a bone-clamping six gravities. The acceleration lasted only ten seconds; then the ship was flung straight up at the sky, leaving the catapult at 1300 miles per hour.

They were in free-fall and climbing. The captain appeared to be taking his time about cutting in the jet; for a moment Matt held to the excited hope that an emergency landing was going to be necessary. But the jet roared on time.

The math checks out: six gees for ten seconds gives you:

v = a×t = (6×9.81 m/s2)×10 s
~= 589 m/s
~= 1317 mph

While this is only about 8.4% of the speed required to stay in Earth orbit, it could definitely give a rocket-powered craft ample initial velocity. As in the excerpt from Space Cadet, a winged vehicle could generate lift while coasting upwards, then ignite a rocket engine to go the rest of the way. This allows a significant savings on fuel compared to making the spacecraft take off from a conventional runway. Northrop Grumman’s Pegasus rocket and the North American X-15 manned rocketplane were deployed from an aircraft at high altitude for similar reasons.


  1. ^ The Moon’s equatorial circumference is 10,921 km. This means that, due to its rotation, a point on the Moon’s equator travels at merely 17 km/hr (about 10 mph.)
  2. ^ 789 acres, or a little over one square mile.