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Huygens & non-orbital re-entry speed

  1. Jan 19, 2005 #1

    DaveC426913

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    So, I notice in the animations of Huygens, they seem to suggest the probe enters the atmosphere more or less vertically, as opposed to almost horizontally, as in the case of re-entry of Earth-based vehicles. While I grant the the animations are idealized, I thought that perhaps it might be possibly be doable.

    Earth re-entry vehicles must decelerate from Mach25+, a velocity they needed to reach to reach in orbit in the first place. But Huygens was not in orbit, and theoretically could have made a relatively slow planetfall, thus it might not have need a huge aerobraking phase.

    And then while watching another venerable science-fact show, Futurama, I saw Bender fall straight to Earth, and I thought to myself:

    (self, I thought...)

    How high above Earth would one have to start, to reach Mach 25 by the time one reached the outer atmosphere, if one were plunging straight down under gravity?

    Or, alternately,

    How high could one start from and still survive a direct, falling entry into the atmosphere? For this, we must assume some point in the descent when it can supposedly switch to conventional touchdown procedures, such as gliding or parachuting. I understand that Huygens was going about Mach 1.5 when it deployed its shute for a soft landing. We must aslo guess at the aerobraking effct of the atmosphere over only a hundred kilometres or so.

    Any guesses?
     
  2. jcsd
  3. Jan 20, 2005 #2
    The escape velocity is [tex]\sqrt {2} [/tex] times the orbital velocity, so that is even more. I get V_escape for the Earth:

    [tex] V_e = \sqrt {2 \mu /r} = \sqrt{\frac {2 \times 4E14} {6.3E6 }} = [/tex]

    11000 m/sec.

    For Saturn [tex] \mu [/tex] is much greater and there is an additional velocity coming from the trip from Earth so the approach velocity to the top of the atmosphere will be far higher.
    There are some pretty smart guys in the astrodynamics threads who you can get a real education from. See https://www.physicsforums.com/forumdisplay.php?f=69
     
    Last edited: Jan 20, 2005
  4. Jan 21, 2005 #3

    DaveC426913

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    Um. Appreciate all the effort, but. Meaningful answers to my questions would be distance values, not velocity values.
     
  5. Jan 25, 2005 #4
    I have a formula:
    (V^2)/2=GM (1/r1 - 1/r2)
    Let GM be the constant for the Earth.
    r1 is the height of the top of the atmosphere from the center of the Earth.
    r2 is the drop height.
    If R2 was infinite then V would be about 25000 mph. (the excape velocity)
    Mach 25 I guess is 800 mph X 25 = 20000 mph.
    So your drop height would be a pretty good distance out.
     
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