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What would be someone's terminal velocity based on the gravity of Europa (.1335G) but given the drag coefficient and air density of Earth? I have a protagonist who is currently plummeting about four hundred feet, but I want to know how long his one hundred pound body will be aiborne and if his landing will pulverize him into protagonist soup. I may need to tweak things based on your answer. Thanks, A

PS--couldn't make a terminal velocity tag because I don't have enough reputation points.

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    $\begingroup$ Is this really about Space Exploration? You could make a case that this is planetary science, but I'm on the fenced. $\endgroup$ – HDE 226868 Dec 17 '14 at 0:15
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    $\begingroup$ In 400 ft your protagonist won't reach terminal velocity at 0.1335 g. If at 1 g $V_t = 122 \text{ mph}$ then at 0.1335 g (1.3092345 m/s) $V_t = 44.6 \text{ mph}$. He/she will be about 4.6 mph short and the fall will take about 13.65 seconds (1.6 seconds less than reaching terminal velocity). I'd however argue that this is more of a question for Physics than Space Exploration. Please don't cross-post tho, let the community decide if it's indeed off-topic here and if that'll be the case it'll be migrated for you. ;) $\endgroup$ – TildalWave Dec 17 '14 at 1:14
  • $\begingroup$ One other thought since you're writing a novel, that at 400 ft your protagonist could achieve significant horizontal component, all else being equal to Earth, and the fall would resemble more jumping off a train than falling off a tall building. This might be the difference between a folded and a rolled pancake ... $\endgroup$ – TildalWave Dec 17 '14 at 1:32
  • $\begingroup$ His "takeoff" is definitely horizontal (a leap from one level of a structure to the next, with a significant gap between them), so rolled pancake/fall of train analogies apply. Something intervenes and slows him down before impact, but I have a better idea of the nature and necessity of that intervention now. $\endgroup$ – adamholtwrites Dec 17 '14 at 1:50
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Terminal velocity is proportional to the square root of gravitational acceleration, if other factors are held equal. So it will be about 120 mph $\times \sqrt{0.1335} \approx $ 44 mph.

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There are two formulas for "resistive forces" that may be of use.

#1: Velocity is linearly related to force (Slow speeds).

Here, the resistive force $R$ is defined as $$R=-bv$$ where $v$ is velocity and $b$ is a constant that depends on the object and the medium. Using Newton's second law, $$\sum F=ma$$ The other force to take into account is gravity: $$\sum F=mg-bv=ma$$ $$\frac{dv}{dt}=\frac{mg}{m}-\frac{bv}{m}$$ $$\frac{dv}{dt}=g-\frac{bv}{m}$$ When an object reaches terminal velocity, $\frac{dv}{dt}=0$. so $$0=g-\frac{bv}{m}$$ $$g=\frac{bv}{m}$$ $$v=v_T=\frac{mg}{b}$$ $b$ will most likely have to be determined experimentally.

#2: Velocity is non-linearly related to force (High speeds).

The other equation (and the one I find is more common) is $$R=\frac{1}{2}D \rho Av^2$$ where $D$ is a constant, $\rho$ is the density of the medium, and $A$ is the cross-sectional area of the object. Again, using Newton's second law, $$\sum F=ma=mg-\frac{1}{2}D \rho Av^2=ma$$ $$\frac{dv}{dt}=\frac{mg}{m}-\frac{D \rho Av^2}{2m}$$ Setting $\frac{dv}{dt}$ to $0$, we have $$g=\frac{D \rho Av^2}{2m}$$ $$\frac{2mg}{D \rho A}=v^2$$ $$v=v_T=\sqrt{\frac{2mg}{D \rho A}}$$ Once again, $D$, like $b$ before, must be determined experimentally.


So without knowing the exact conditions and modeling them - or conducting them in an experiment on Europa - the precise velocities can not be determined.

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  • $\begingroup$ Thanks. The first question I posted involved interplanetary travel using constant acceleration drives, so I kept this one under Space Exploration as well. What are your thoughts on pericynthion's answer? $\endgroup$ – adamholtwrites Dec 17 '14 at 1:16
  • $\begingroup$ @adamholtwrites It's accurate, as far as I can tell, though the answer doesn't correct for $b$. But it's still accurate, and gives a number, which mine does not. $\endgroup$ – HDE 226868 Dec 17 '14 at 1:18
  • $\begingroup$ Yes, it suits the needs of the story. I am grateful for your help. -A $\endgroup$ – adamholtwrites Dec 17 '14 at 1:23
  • $\begingroup$ @adamholtwrites No problem. Gave me a chance to practice my skills with air resistance. $\endgroup$ – HDE 226868 Dec 17 '14 at 1:24
  • $\begingroup$ Win/win then...have good one. $\endgroup$ – adamholtwrites Dec 17 '14 at 1:27

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