No, it will not, no matter the changes done while keeping it a F-22

Wikipedia gives the data for the F-22 as empty weight 19,700 kg and a maximum take-off weight (MTOW) of 38,000 kg. That gives a mass ratio of about 0.52 if it lands on fuel vapors in the tank. While exact specifications will obviously vary, I suspect that this is relatively representative of this class of aircraft.

We can approximate the F-22 in this case as a single stage to orbit (SSTO), because it lacks any meaningful staging capabilities. (No, ditching extra tanks doesn't really count for much, because the mass of those tanks when empty is likely negligible.)

The rocket equation describes how a single stage rocket can change its velocity (delta-v or $\Delta v$) as a function of the mass ratio and exhaust velocity ($v_e$):

$$ \Delta v = v_e \times \ln\left(\frac{m_i}{m_f}\right) $$

where $m_i$ is the initial mass and $m_f$ is the final mass. Plugging in the above numbers, and assuming an exhaust velocity of 2,000 m/s (this is probably way higher than the actual F-22, so the calculation becomes very optimistic), and ignoring that a rocket needs to bring its own oxidizer as well as fuel and propellant, we get

$$ \Delta v = 2\,000~\text{m/s} \times \ln\left(\frac{38\,000~\text{kg}}{19\,700~\text{kg}}\right) \approx 1\,300~\text{m/s} $$

The escape velocity of Mars is 5.03 km/s. We can approximate the mean orbital speed as $$ v_o \approx \frac{v_e}{\sqrt{2}} \approx \frac{5\,030~\text{m/s}}{\sqrt{2}} \approx 3\,560~\text{m/s} $$

None of this is exact, but that doesn't really matter because our F-22 comes up about 2/3 short based on publicly available data, optimistic guesses and a highly improbable scenario (not needing to bring any oxidizer or propellant, only fuel). Because of the exponential nature of the rocket equation, the remaining two thirds are far harder than they would appear to be simply looking at the velocity change figures.

I suspect that, in order to be able to attain orbit around Mars, an F-22 which has been modified to work in that environment would need a mass ratio of less than 0.1 (bringing about ten times its own final mass in fuel). Keeping one of the values fixed, this corresponds to a MTOW of around 200,000 kg or an empty weight of less than 4,000 kg. If the exhaust velocity is lower (which is probable), this becomes even worse. Given that it can only carry its own weight in fuel (a mass ratio of 0.5 is pushing it), this is clearly not feasible.

And of course, the person flying it might want to get back down, too, which having attained a stable (even short-term) orbit requires either waiting for orbital decay (for example due to atmospheric drag, such as that experienced by the ISS), or a deorbit burn. While the deorbit burn can require shedding only a small fraction of the orbital velocity, you still need the fuel for that.

posted over 7 years ago

Canina‭

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