Can you use an aircraft to simulate Mars gravity on Earth, similar to the way weightlessness is simulated by fling in a parabolic arc? What would the flight profile have to look like to achieve mars gravity? I know the acceleration required to achieve weightlessness, 9.8 m/s2.
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2$\begingroup$ First: this SE is not, on the whole, about simulating reduced gravity in airplane flights, so it's really unclear from the way the question is worded what you're asking for. Second: you know Earth's gravity, Mars's gravity is easy to look up; all you have to do is subtract. $\endgroup$– Russell BorogoveCommented Oct 19, 2017 at 19:52
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3 Answers
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The short answer is "Yes, it's possible"
The typical technique to fly a zero-G parabola is to put the plane into a steep climb and when the airspeed is low enough, push forward on the yoke until the G-meter reads "0".
Mars gravity is roughly (3.71 / 9.8 = 0.38G). Instead of pushing to 0, the pilot just pushes to 0.38.
This would result in a somewhat flatter arc and the experience would be a little bit longer than a comparable zero-G flight.
Source: personal experience. I'm an aerobatic pilot and have flown similar profiles (but in a single seat plane, not a jet)
answered Oct 20, 2017 at 15:52
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$\begingroup$ I've just asked Could an aircraft ever simulate Martian gravity perpendicular to the aircraft's floor? $\endgroup$– uhohCommented Apr 17, 2018 at 2:05
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Yes, and it is done in such parabolic flights as explained and showed in this video, (in French but with English closed-captions available) especially the cockpit scene between about 24:30 and 26:30 where you can watch the instruments and exterior view of the plane. Basically, you adjust the trajectory and speed of the aircraft to fit the required acceleration.
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$\begingroup$ In theory, it is possible is the trajectory is carefully managed, but I don't see that 'lower gravity' anywhere in the video. Of course, when entering (and exiting) the parabolic trajectory, all the objects inside the plane are subjected to acceleration, and have 'extra weight' due to it. But during the phase of free fall, the apparent weight of those objects is zero... $\endgroup$– xxavierCommented Oct 22, 2017 at 10:20
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$\begingroup$ @xxavier I may have blend memories from several videos including this one as they were taken at the same time, buy I'm quite sure one of those explain they simulate several micro gravity settings (zero-G, moon gravity and marsian gravity) $\endgroup$– Manu HCommented Oct 22, 2017 at 21:32
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$\begingroup$ @Manu H It may be as you say, but it's not the case in the French video... $\endgroup$– xxavierCommented Oct 23, 2017 at 9:20
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$\begingroup$ I've just asked Could an aircraft ever simulate Martian gravity perpendicular to the aircraft's floor? $\endgroup$– uhohCommented Apr 17, 2018 at 2:05
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Suppose at the start of the parabola the aircraft has initial altitude $H$ meters, speed $V_o$ m/s, and is moving upward at an angle to the horizontal $M ^{\circ}$.
If the desired gravitational force is $G m/s^2$ then the aircraft must accelerate vertically at $(9.81-G) m/s^2$. The horizontal component of the speed is constant at $V_x(t) = V_o\times \cos M$, so that the horizontal co-ordinate after $t$ seconds is $X(t) = V_o\times \cos M\times t$.
Initially the vertical component of the speed is $V_y(t=0) = V_o\times\sin M$. The vertical component of the speed is $V_y(t) = V_o\times\sin M -(9.81-G)\times t$. At the peak vertical speed is zero, and occurs at $t_{\text{max h}} = V_o\times\sin M/(9.81-G) $.
The vertical co-ordinate is $$Y(t) = H + V_o\times\sin M\times t - {(9.81-G)\times(t^2)\over 2}$$ The maximum height attained is $$Y(t_{\text{max h}}) = H+V_o\times\sin M\times[V_o\times\sin M/(9.81-G)] - (9.81-G)\times{[V_o\times\sin M/(9.81-G)]^2\over 2}$$ By symmetry it takes $t_{\text{max h}}$ seconds to go from the peak to the base altitude $H$.
Greatest altitude and time at reduced gravity occur if the initial angle is $90 ^{\circ}$, but steep angles introduce other problems. For a given initial airspeed and initial angle, a larger perceived gravity (as of Mars) has longer reduced gravity time than a smaller perceived gravity (Luna or freefall).
There must be sufficient airspace below the base height to accelerate and maneuver into the parabola, and recover from the parabola. At the peak there must be sufficient air density and indicated air speed for the control surfaces to function. The indicated air speed may be significantly less than the true air speed, and the pilot must compensate for winds that vary with altitude.
I am not sure that the perceived gravity will be perpendicular to the aircraft floor. For free fall that would not matter.
Example: $$\begin{array}{lll} H & 6000 m & \text{base height} \\ V_o & 200 m/s & \text{initial speed} \\ M & 30 ^{\circ} & \text{initial flight angle} \\ & 3.71 m/s^2 & \text{desired perceived gravity} \\ & 6.10 m/s^2 & \text{needed vertical acceleration} \\ X(t) & 173 m/s & \text{horizontal component of speed} \\ t_{\text{max h}} & 16.4 s & \text{time to max height} \\ Y(t_{\text{max h}}) & 6820 m & \text{max height} \end{array}$$
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$\begingroup$ You raise a really good point about the direction of the gravity. Using a trajectory with constant ground speed (x) and a parabolic altitude profile (y) the simulated reduced gravity would continue to point to the center of the Earth, rather than be perpendicular to the floor. Trying to get the simulated gravity to be floor-normal makes the math problem much more interesting! And so I've just asked Could an aircraft ever simulate Martian gravity perpendicular to the aircraft's floor? $\endgroup$– uhohCommented Apr 17, 2018 at 2:04
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$\begingroup$ In comments here there is interest in you undeleting your answer. Two votes so far. Seeing as how other answers are branching out and being well-received, I think yours would be a great addition. Consider undeleting? $\endgroup$– uhohCommented Nov 6, 2018 at 14:23