Is it possible to orbit inside a gas giant?

Like JDÅ‚ugosz wrote, what will cause problems in the scenario you describe isn't so much your orbit as the fact that you are within the gas giant's atmosphere.

I'm going to use Jupiter here to have some specific gas giant to use for examples. Feel free to look up the relevant data for any other gas giant, or come up with your own.

For the case we are interested in, a small mass orbiting a much larger mass where the radius of the orbit is equal to the larger body's radius (just dipping your toes into the Jovian atmosphere), orbital speed can be approximated as $$ v_o \approx \frac{v_e}{\sqrt{2}} $$

The escape velocity of Jupiter is approximately 59.5 km/s, so to dip our toes into the atmosphere we get an orbital velocity of approximately $$ v_o \approx \frac{59~500~\text{m/s}}{\sqrt{2}} \approx 42~100~\text{m/s} $$

To give an idea of how freakishly fast this is, it's equivalent to approximately 152,000 km/h or 94,200 miles per hour. It gets you between the Earth and the Moon in 2.5 hours. In mid-1976, an airplane managed to get to 3,530 km/h, which is about 1/43 of the orbital speed at the edge of Jupiter's atmosphere. The best we have managed on anything resembling a repeat basis is around 2,500 km/h, or 1/60 of what you would need.

For comparison, Jupiter's wind speeds peak in excess of 150 m/s. While quite a stiff gale, that's nowhere near orbital velocity; by the above estimate, about 1/280 (and that's assuming that top wind speeds occur in the uppermost layers of the atmosphere, which might not be the case). With such a large difference between orbital speeds and wind speeds, we can largely ignore wind speeds for the purposes of this question; even in a perfect situation, wind speed will contribute less than 0.36% of the required velocity. (Interestingly enough, according to the same source, Jupiter wind speeds have a peak very near the equator, which works well for us.)

Given that Jupiter has an equatorial diameter of 142,984 km and that the circumference of a circle is $\pi d$, 42.1 km/s gives an orbital period (if you can call it orbital) of $\frac{142984 \pi}{42.1} \approx 10~700~\text{seconds}$ or just under three hours. For comparison, Wikipedia gives Jupiter's sidereal rotation period ("day") of 9.925 hours (a shade over 9 hours 55 minutes).

For comparison, to get into a reasonably stable low Earth orbit you need a velocity of approximately 7.8 km/s (corresponding to an orbital period of about 90 minutes). To go to the Moon (which is pretty close to escape velocity), you need about 10.5 km/s relative to the Earth. Actual Earth escape velocity is 11,186 m/s. Compare Apollo by the numbers: Translunar Injection and look at particularly the Earth Fixed velocity figures for the various lunar missions.

Let's say you can somehow handwave the issue of absolute speed away. (After all, you got there somehow, and that already takes quite a bit of speed.) Let's also say that your craft is a very, very long, perfect cylinder with a forward cross section of 1 square meter, built to handle constant hurricane-level wind speeds. Every second, you are moving through 42,100 meters of atmosphere. That means that every second, your craft will need to push aside 42,100 cubic meters of atmospheric gases while maintaining its speed (at least if you plan on staying at that altitude). Wikipedia gives the composition of Jupiter's atmosphere as approximately $89.8 \pm 2.0 \% ~\text{H}_2$ and $10.2 \pm 2.0 \% ~\text{He}$. Despite the fact that these two gases are among the lightest known, and that the density is going to still be low at the altitude we are talking about, pushing aside over 40,000 cubic meters of gas per second is going to cause some massive drag.

And that, my friend, is what will cause your craft to heat up, lose speed very quickly and eventually descend into the atmosphere, ruining your day.

posted over 8 years ago

Canina‭

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