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Thread: Falling on Venus or Titan

  1. #1
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    Falling on Venus or Titan

    If you ignore temperature, etc. and just focus on terminal velocity, could I survive a fall from a great height on Titan or Venus?
    SHARKS (crossed out) MONGEESE (sic) WITH FRICKIN' LASER BEAMS ATTACHED TO THEIR HEADS

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    Wow, that's a fun question! Thanks Tom.
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    Quote Originally Posted by Tom Mazanec View Post
    If you ignore temperature, etc. and just focus on terminal velocity, could I survive a fall from a great height on Titan or Venus?
    If I am not mistaken, Titan's ground level atmospheric density is close to ours, so I would say no. Venus has several hundred times ours, so it might be a different story. I hereby defer to aerodynamics experts to estimate the terminal velocity.

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    Hornblower:
    Titan's density may be close to ours (I think it is slightly higher?), but its gravity is a lot lower.
    SHARKS (crossed out) MONGEESE (sic) WITH FRICKIN' LASER BEAMS ATTACHED TO THEIR HEADS

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    Terminal velocity for something the size of a person is proportional to the square root of [gravitational acceleration divided by atmospheric density]. I leave the calculation for the interested student.

    (Venus's atmosphere is so dense there will be slight buoyancy factor, too, knocking a few kilograms off the apparent weight of a person - but given the uncertainty in the quantity "survive", I doubt if it's worth bothering about.)

    Grant Hutchison

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    Define ďa great height. ď

    10 feet? 20 feet? 50 feet?

    Titanís pressure is about 2 bars at the surface. Venus is about 90 bars. But in both cases the fall is through a gas, not a liquid. So does pressure or density even mean anything specific to those two bodies?

    This fellow has built a table of terminal velocity for various planetary bodies.

    https://www.google.com/amp/s/amp.red...falling_human/

    The table says Earth is 54 m/s and Venus is 46 m/s whereas Titan is 9.6. So...thereís the basis for your answer.

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    Quote Originally Posted by grant hutchison View Post
    (Venus's atmosphere is so dense there will be slight buoyancy factor, too, knocking a few kilograms off the apparent weight of a person - but given the uncertainty in the quantity "survive", I doubt if it's worth bothering about.)
    Since were playing with temperature, couldn't we use a surface density to match the pressure. At 93 bar for Venus, this is a significant air density but not quite most people's weight, Earth of Venus. Admittedly, it's been said that some people shrink under pressure, but those that don't would have significant buoyancy, right?

    Added: Ok, the air density is only about 10% that of the human body, so that won't help that much after all. It wouldn't take much of an added balloon to float, however. A human + balloon volume of just under 1 cu. meter is all this is required.
    Last edited by George; 2019-Jul-17 at 03:11 PM.
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    Quote Originally Posted by grant hutchison View Post
    Terminal velocity for something the size of a person is proportional to the square root of [gravitational acceleration divided by atmospheric density]. ...
    Thanks Grant! As a very quick estimate that puts terminal velocity on Venus as roughly 15-20 MPH, which if you landed right might merely break your legs, or maybe not even that.
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    It depends on if you lie flat or dive.

    Quote Originally Posted by grant hutchison View Post
    Terminal velocity for something the size of a person is proportional to the square root of [gravitational acceleration divided by atmospheric density].
    This might be correct for a spherically-shaped body.

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    Quote Originally Posted by CaptainToonces View Post
    It depends on if you lie flat or dive.
    Sure does. The proportional dependence on g and rho remains the same however. If sqrt(g/rho) comes out to be 0.5, then a spreadeagled human will fall half as quickly as they would on Earth, and a head-down human will fall half as quickly as they would on Earth.

    Quote Originally Posted by CaptainToonces View Post
    This might be correct for a spherically-shaped body.
    It certainly will be, but it's actually applicable across a range of masses, cross-sectional areas and coefficients of drag - falling humans and cows and cars and anvils, just so long as the flow is turbulent and there are negligible buoyancy or lift forces.
    The full formula is:

    VT = sqrt[(2.m.g)/(rho.A.CD)]

    A given body, falling in a given orientation, is going to have constant m, A and CD, so (as I said) all we need to do is consider the gravity and the atmospheric density.

    Grant Hutchison

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    Good stuff, Grant. In the case of falling through a planet's atmosphere, there's also the complication that g and rho change with altitude. Since they both increase as altitude decreases, they might cancel each other out, but I would hesitate to assume an equal balance there.

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    From an ideal gas law calculator on-line, and using R for CO2, the density at the surface would be about 167 kg/m^3, if we allow a temp. of 70 deg F. A 70kg body holding an empty high pressure tank that is 6 ft long and 3 ft. in diameter (ie 1 cu meter) could weigh as much as 100 kg and still allow the person holding it to float along the surface. [Though this is another napkin calculation, so it could be off.]
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    Quote Originally Posted by schlaugh View Post
    Define “a great height. “
    It is defined in the OP: high enough to reach terminal velocity.

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    Quote Originally Posted by antoniseb View Post
    Thanks Grant! As a very quick estimate that puts terminal velocity on Venus as roughly 15-20 MPH, which if you landed right might merely break your legs, or maybe not even that.
    A quick Google suggests standard parachute landing velocity is about 12mph. And that's enough to ensure a high probability of walking away unharmed.

    So a landing of 15-20mph would be - what? - about double the energy? (It increases as the square of the velocity, does it not?)

    On the other hand, a parachute guarantees that you will land in-control and feet first - not so on Titan or Venus.

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    Quote Originally Posted by DaveC426913 View Post
    It is defined in the OP: high enough to reach terminal velocity.
    On Venus, any velocity achieved while falling is considered "terminal" velocity.

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    The terminal velocity on Earth, 1 g, 1 bar, is about 50 m/s. Kinetic energy equivalent to falling without friction from 125 m height.
    If parachutists think it acceptable to land at 6 m/s, that has about 1,4 % the kinetic energy of free fall, equivalent to falling from about 1,8 m height.

    On Titan, the air has about 1,5 times the pressire of Earth, but it is colder, so about 4,4 times density. The local gravity is 0,138 g, so the kinetic energy is about 32 times less than on Earth - about 3,1 %, equivalent to 4 m fall on Earth.
    On Venus, the air has about 90 times the pressure of Earth, but is hotter, so about 50 times the density. The local gravity is 0,95 g, so the kinetic energy is about 52 times less than on Earth - about 1,9 %, equivalent to 2,4 m fall on Earth.

    You can probably survive, but there are reasons why parachutists prefer slower landing speeds, rather than save by having smaller parachutes. There are some nonfatal injuries, like damage to feet on landing.

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    We know time flies, we just can't see its wings.

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