Can the Earth's Magnetic Field Support a Space Railgun?

Fig. 1: Railgun diagram (HowStuffWorks)

In physics class, you’ve probably learned about railguns and how they use magnetic fields and current to generate a propulsive force. And if you’re anything like me, you’ve probably wondered if we can launch a person or spaceship with one, because why not? What would be even cooler is if we could do it with the earth’s ambient magnetic field. Well, let’s test this idea a little bit.

We know that the force generated by a magnetic field on a charged particle is

1.       F = d/dt(L)q x B = qv x B (L is displacement, q is charge, B is magnetic field, v is velocity)

An equivalent statement more applicable to railguns is 

2.       F = d/dt(q)L x B = IL x B          (I is current)

This change of derivative position is acceptable because it represents a change in reference frame, the displacement derivative in the frame of the charged particle and the charge derivative from a fixed external frame of reference, watching the charges flow by. In the setup of a railgun, B and I are orthogonal and the resultant vector of the cross product is in the direction of propulsion. Solving the cross product yields

3.       F = ILB 

We also know the induced current through the railgun circuit to be related to the generated emf, 

4.       Ɛ = - d/dt(Φ_B) = - d/dt(BA) = - Bd/dt(A)          (Φ_B is magnetic flux, A is area)

By geometric analysis, we know the change in area to be the fixed base length/bar length times the change in height, or

5.       d/dt(A) = L d/dt(h) = Lv          (h is height)

The full expression is

6.       Ɛ = IR = - BLv

7.       I = - BLv/R

Appending our force equation to account for induced current that works against the initial current, we get

8.       F = (I - BLv/R)LB = LBI – (L²B²/R)v

This equation is useful because by setting the magnetic force to zero we can solve for a terminal velocity,

9.       v = IR/LB

This is no good! If we’re to launch someone high into the sky, we need a constant force to produce a constant acceleration to exceed that of gravity. So what if we keep a constant effective current to counteract the current induced by propulsion? Looking at equation 8, we see that this is possible if we establish current as a time-dependent function given by

      10.   I(t) = I₀ + BLv/R                                                 (R is resistance)

      11.   I(t) = I₀ + BL(at)/R = I₀ + BL(Ft/m)/R               (from kinematic equations v=at and F=ma)

The F term here should be substituted with the constant force we predict to be generated by the current adjustment

12.   F = I₀LB

Substitution and regrouping produces

13.   I(t) = I₀(1 + (B²L²/Rm)t)          (m is mass)

This equation can be verified by substitution into equation 8. Now that our force is constant, let’s calculate what current would accelerate the average person [1] on a weightless conductive beam unaffected by air drag 1m in length enough to counteract gravity (a pretty low bar for testing viability of concept). Since the magnetic field generated by the circuit would probably diminish too close to the walls of a 1m launcher, an external field is necessary. Let’s try using the upper end of the earth’s magnetic field strength range [2].

                     14.   F = mg = I₀LB

                     15.   I₀ = mg/LB = (62kg)(9.8m/s²)/((1m)(65x10^-6 N/Am)) = 9.3x10⁶ A

That’s a lot of current, but perhaps possible? Adjusting for a spaceship weighing 1,000 times the average person 10m wide with an acceleration of 1m/s² [3], we get a figure on the scale of 1x10⁷ A. In 2014, the CERN Superconductors team was able to pass 20x10³ A through an MgB₂ superconducting wire, a world record at the time [4]. This record value is clearly magnitudes less than even the current necessary to lift a single person, let alone a spaceship. Looks like the earth's ambient field is a no-go. But! If the person were in a magnetic field of an achievable 1T even, then with currently achievable currents it should be possible to throw them at least into the sky if not into space. Hooray for space railguns!

If you liked this post or have any ideas for another, let me know by leaving a comment. Thanks for reading!