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There are some concepts of providing propulsion (or at least preventing orbital decay) by driving current through extended rods from a satellite.

AFAIK currently they are used only to provide torque (magnetorquers) to stabilize spin of satellites, but some works - like the non-rotating skyhook foresee using them for acceleration - boosting the orbit of the vessel.

I know current moving parallel to lines of a magnetic field doesn't exert any force on the wire it travels through, so certainly there is a "dead direction" where a magnetic rod would provide no propulsion/force whatsoever. I'm not sure about shape of Earth's magnetic field though, and as result what direction it is.

So - if we take a satellite in equatorial, or other not heavily inclined orbit, what translation maneuvers can we perform using just magnetic rods? How do the rods need to be oriented for them?

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    $\begingroup$ To double check; does 'magnetic rod' mean (for example) a rod of ferrite with a coil of wire wrapped around it, limited to the size of the spacecraft? In a (locally) uniform field, a rod or a loop only provides torque, assuming your circuit is closed (complete loop). No matter what the shape, the total force (propulsive) adds up to zero. But if you want to push on the gradient of the field, you can get a very very very tiny force, basically proportional to the very small change in the Earth's field from one side of the spacecraft to the other. $\endgroup$ – uhoh Apr 2 '16 at 10:22
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    $\begingroup$ @uhoh: AFAIR this is achieved by a rod where circuit is not closed directly - it has ion emitters at both ends, so the circuit is "closed" through vacuum; "magnetic rod" would be a kind of misnomer, it's just a rod through which electric current travels in one direction, positive ions discarded on one end, electrons on the other. $\endgroup$ – SF. Apr 2 '16 at 12:15
  • $\begingroup$ OK got it - I'll give it a go. $\endgroup$ – uhoh Apr 3 '16 at 0:10
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The earth's magnetic field is roughly that of a dipole, tilted from the rotational axis by about 10 degrees at the moment, but it's actually somewhat lumpy/bumpy. I plotted the three components (x, y, z = North, East, Down) as a function of longitude, for a LEO at 300km above the equator using the World Magnetic Model WMM2015 from NOAA > NESDIS > NCEI (formerly NGDC) > Geomagnetism using the menu settings and script shown below. The field is roughly 0.3 Gauss pointing North, but there are "bumps" available in the other two directions of about +/- 0.1 Gauss. The full description of the model is here. I also plotted the fields from a dipole approximation.

You can read more about the Earth's magnetic field here for example.

Earth's lumpy magnetic field

The force on a current in a field follows the Laplace right hand rule (I never knew he had one!) so for a satellite in orbit, using North, East, and Down as the three axes, to boost an equatorial LEO satellite higher, I think we need an East-West force (accelerate the speed), which means either an Up-Down rod using the strongest North-South field, or - as you mention in the question - a North-South rod using the weaker Up-Down field.

The magnitude of the force (Newtons) is equal to the length of the rod (meters) times the field strength (Tesla). So 1 Amp in 0.3 gauss gives 3E-5 Newtons. I'm not sure how the ends of the rod work - but if you assume a 1.5 eV work function for the electron and ion guns (and otherwise no losses), that's 1E-05 Newtons per Watt, or 100kW per Newton - very roughly the same scale as plasma thrusters.

So in order to carry a current, the rods must emit negative charge (electrons or ions) from one end and positive charge (ions) from the other. But I have no idea where an Amp of ions can come from! If this technology is supposed to take advantage of gas that it encounters in its orbit, it will probably be microamps or nanoamps.

I have a sneaking suspicion that the drag from a given gas pressure may be worse than the thrust from a rod carrying a current produced by ionizing the gas at that pressure, but that's a different question.

If instead it carries it's own gas to ionize, then I wonder if maybe a conventional ion thruster is a better choice?

Let me know if this is helpful or if I've missed the point.

Earth magnetic field 300km above equator

Right Hand Rule of Laplace

Field coordinates used in WMM2015

Input settings

Some scrappy Python to give an idea what I did:

import numpy as np
import matplotlib.pyplot as plt

fnames     = ["equator_X.csv", "equator_Y.csv", "equator_Z.csv"]
directions = ["North",         "East",          "Down"]

# http://www.ngdc.noaa.gov/geomag-web/#igrfgrid

fields = []
for fname in fnames:
    with open(fname, "r") as infile:
        lines = [line.rstrip('\n') for line in infile]
    print lines[13]

    lon, hkm, nT = np.array([[float(x) for x in line.split(',')]
                             for line in lines if "#" not in line]).T[2:5]
    fields.append(nT * 1E-09 * 1E+04) # nT to T to gauss

fields = np.vstack(tuple(fields))
total  = np.sqrt((fields**2).sum(axis=0))

fields = np.vstack((fields, total))
directions += ["Total"]

lon = np.array(lon)
zlon = np.zeros_like(lon)


# Now get simple dipole approximation
# Geomagnetic poles - axis of "best dipole approximation" to earth's real field

gmlat, gmlon = 80.3, -72.6

rads, degs = np.pi/180., 180./np.pi

mlon = lon - gmlon - 90
sinmlat = -np.sin(rads*mlon)*np.sin(rads*(90-gmlat))
mlat = np.arcsin(sinmlat)

plt.figure()
plt.plot(lon, degs*mlat)
plt.show()

def earth_dipole(r_km, maglat):
    """from: https://en.wikipedia.org/wiki/Dipole_model_of_the_Earth%27s_magnetic_field"""

    a_km = 6371. # km
    B0 = 3.12E-5 # T

    R = r_km / a_km  # in earth radii

    Br  = -2. * (B0 / R**3) * np.sin(maglat)
    Bth =       (B0 / R**3) * np.cos(maglat)

    return Br, Bth

r_km = (6371. + 300)*np.ones_like(mlat)

Bz_dip, Bx_dip = earth_dipole(r_km, mlat)

fields = np.vstack((fields, 1E+04*Bx_dip, -1E+04*Bz_dip)) # Tesla to Gauss
directions += ["North(dipole)", "Down(dipole)"]


plt.figure()

for field, label in zip(fields, directions):
    plt.plot(lon, field, label=label)

plt.plot(lon, zlon, '-k')
plt.legend(loc='center right', shadow=False, fontsize = 'large')
plt.xlim(lon.min(), lon.max())
plt.title("Magnetic field 300km above Earth Equator", fontsize=18)

plt.show()
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  • $\begingroup$ there's roughly 100,000 coulombs worth of charge (x2; +&-) in a mole of any substance on the 1st oxidation level. It's primarily meant for massive, long structures like that skyhook, where the current travels for hundreds of kilometers. $\endgroup$ – SF. Apr 4 '16 at 8:06
  • $\begingroup$ 100,000 coulombs will give you 1 Newton per meter of length, for 3 seconds. Roughly what is the mass of this structure per meter of length? $\endgroup$ – uhoh Apr 4 '16 at 12:45
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    $\begingroup$ we won't know until we have the buckytube technology up to snuff, but a meter of buckytube rope isn't all that excessive, probably would be well below a kilogram. Also, deliveries of cargo/"fuel" would be significantly cheaper (plain suborbital or even just hypersonic) and it would be needed primarily to prevent orbital decay. Plus that 100,000 coulombs would provide the same 1 newton per meter to every meter of all the 600 kilometers or so of it! Just try to do something like that with any other propulsion! 600KN for 3s out of 1 mole of propellant? $\endgroup$ – SF. Apr 4 '16 at 20:16
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    $\begingroup$ I doubt so, but we're operating with very high voltages (~10kV?) and nanotubes are a fair conductor (I had trouble finding the actual value though) - plus the rope is going to be pretty thick. Plus we definitely don't need 600kN at a time. I suppose 1kN sustained would be a more plausible ballpark. $\endgroup$ – SF. Apr 5 '16 at 7:33
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    $\begingroup$ Sooner or later it'll need all the physics (mass, drag, conductivity, magnetic "thrust") all parameterized in one place. In power transmission over a pair ow wires, you can lower current (and therefore I^2R losses) by raising voltage. But here it's the current that is necessary - the thrust is proportional to current times length, the ohmic losses proportional to current squared times length. Good luck! $\endgroup$ – uhoh Apr 5 '16 at 13:16

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