8
$\begingroup$

After reading about Ingenuity, whose main challenge I heard was dealing with the relatively thin atmosphere of Mars, I wondered if the thicker atmospheres of gas giants could allow for powered flight similar to Ingenuity?

Obviously this would exclude Jupiter and Neptune, the former due to its highly powerful magnetic field and the latter due to its extremely fast winds. For the remaining two gas giants, could anything achieved powered flight there, perhaps in the upper atmospheres so that the magnetic field isn't as strong? I recall reading Uranus for instance has magnetic fields ranging from 0.1 to 1 Gauss which should be somewhat bearable for any unmanned craft.

Could such a hypothetical flyer also fly along the direction of the wind so that they're not blown to pieces as well?

$\endgroup$
8
  • $\begingroup$ Are you asking about the helicopter "Ingenuity"? $\endgroup$ May 12 at 20:06
  • 1
    $\begingroup$ @OrganicMarble Very sorry that is what I had meant. Thank you for pointing the error out $\endgroup$
    – Hash
    May 12 at 20:19
  • 1
    $\begingroup$ A helicopter needs a lot of energy. A glider or even an aerostat would probably be the better choice. $\endgroup$
    – user40414
    May 13 at 13:26
  • 1
    $\begingroup$ "Fast" winds don't matter, as airspeed is all that an airplane cares about (unless of course you intend to "land" somewhere, but again if you have a floating city moving with the wind, that's easy). If there's huge wind shear then yeah you have problems. Other than that, just compare atmospheric densities with what aircraft on Earth can handle (or the Mars 'copter) $\endgroup$ May 13 at 15:12
  • 1
    $\begingroup$ Obligatory xkcd: what-if.xkcd.com/30 $\endgroup$
    – kgutwin
    May 14 at 15:52
5
$\begingroup$

tl;drs:

  1. Don't worry about the magnetic fields.
  2. Jupiter certainly has an altitude range with the same pressures at which we fly on Earth, just develop your aircraft aerodynamically to fly on Earth with 2.5 times as much weight as you'll fly there (simulating the 2.5 x higher surface gravity), and fly it there instead but good luck finding a power source! At 5.5 AU from the Sun sunlight is only 3% what it is here, so don't imagine a Solar Impulse-styled solar powered aircraft! Instead think more along the lines of a kilopower small space-rated nuclear reactor currently under development, or put a humungous solar collector farm in Jupiter orbit and beam kilowatts of power to your aircraft using a laser or microwave beam while they are in line-of-sight contact, and then let it operate as a glider (if possible, though at 2.5 g it's going to be rough) during the time when they are out of sight. Batteries are probably going to be too heavy in this gravity, I think you will need a nuclear reactor or solar farm one way or the other.

We are used to handling small magnets where the gradients are large, the field near each pole surface is much larger than just a few cm away, so ferromagnetic metal objects are strongly attracted to the poles.

In a nearly uniform magnetic field, a piece of steel or iron is not pushed or pulled at all. There is no gradient, so there's no decrease or increase in energy moving in any direction.

There certainly can be torques on the object about its center of mass, but those can be zeroed out by distributing any ferromagnetic materials (or internally generated fields due to permanent magnets or electromagnets) through some careful engineering by balancing their distribution, adding some extra electromagnets to null the torque (magnetotorquers) or just plain internal shielding or flux closing, like the yolks around motors that have permanent magnets inside.

The magnetic part (2nd term) of the Lorentz force

$$\mathbf{F} = q \mathbf{E} + q \mathbf{v} \times \mathbf{B}$$

says that there is a force on a charge that's moving in a magnetic field.

If the field is uniform and the charge is in a conducting loop like a circuit or a metal satellite frame, then this will induce no net current in the loop, so don't worry about eddy currents producing drag and loss of altitude (at least to first order).

What will happen though is that the Lorentz force will tend to push charges to one side of the spacecraft; there could build up a small static dipole charge across the craft pushed there by the Lorenz force until their electric field were just strong enough to cancel the Lorentz force.

$$F_E = qE = qvB$$

or just

$$E = vB$$

We can calculate the size as follows. The equatorial field strength of Jupiter at the surface is about 4.2 Gauss, or 4.2$\times 10^{-4}$ Tesla. The standard gravitational parameter of Jupiter is 1.27$\times 10^{17}$ m3/s2 so the orbital velocity at Jupiter's equatorial radius of 7.15$\times 10^{5}$ meters we can calculate the orbital velocity of a Karman plane to be 42,000 m/s (82,000 knotts) using the vis-viva equation

So at orbital speed in our Karman plane which hopefully you won't be doing, the field built up across a 1 meter wide spacecraft at the 1 atmospheric pressure point (1 bar) of Jupiter's atmosphere will be 17.7 volts, using

$$\Delta V = dE = dvB$$

where the voltage difference $\Delta V$ is equal to the distance across the spacecraft $d$ times the electric field $E$.

At a nice Piper Cub cruising speed of 34 m/s it's only 14 millivolts, so not much to worry about.

Basically, just build your craft to work in 10 gauss and you can fly pole to equator (where it's double the strength at the equator and pointing up/down) without worries.


$\endgroup$
2
  • $\begingroup$ 2) For fuel issues would it be possible to fly in Uranus and use the methane there as a fuel source? If we have endless quantities of methane in its atmosphere isn't it basically an infinite fuel source for us. Also when you had stated "the field a few cm away from each pole is much larger than near the pole surface", why is this so? Being near the pole surface means much closer to the surface so wouldn't the magnetic field logically be higher as you're closer to it? $\endgroup$
    – Hash
    May 14 at 6:21
  • 3
    $\begingroup$ @Hash You need an oxidizer too. If you had both a useful fuel and oxidizer combination in the atmosphere, I guess the planet would have caught fire a long time ago. Yes that sentence is wackbards; I'll fix it, good catch! $\endgroup$
    – uhoh
    May 14 at 11:36
4
$\begingroup$

This rough graph I made gives a sense of the (aerodynamic) difficulty of flight on other planetary bodies: Planetary Atmospheres

Consider pressure analogous to density (and thus analogous to lift). The 'equivalence' curves are equal gravity-to-pressure ratios (1:1 for sea level, 5:1 for airline heights [think 1g at 0.2 atm pressure]). They are a sort of 'contours of (aerodynamic) difficulty' measure. The 'float' region is approximately where buoyancy starts to play a non-negligible role.

Interestingly, flight on Titan is about as big a leap towards 'float' as Ingenuity was towards 'flight'.

$\endgroup$
4
  • $\begingroup$ So essentially flight on Jupiter will be the most difficult of them all? $\endgroup$
    – Hash
    May 18 at 17:52
  • $\begingroup$ @Hash depends on what pressure (altitude) you go to. Go to the far right and it's easy, go to the far left and it's very difficult. $\endgroup$ May 18 at 17:55
  • $\begingroup$ What makes Jupiter have more or less the same difficulty across all pressures then? $\endgroup$
    – Hash
    May 19 at 10:10
  • $\begingroup$ @Hash it doesn't, it just has roughly the same gravity across all pressures (altitudes). The difficulty comes from the ratio of gravity (weight) to pressure (density / lift), represented by the 'equivalence' lines. At the high gravities of Jupiter these 'equivalence' lines are nearly perpendicular to the Jupiter line (implying changing difficulty). $\endgroup$ May 19 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.