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Escaping earth's atmosphere with a balloon

  1. Dec 4, 2014 #1
    Hi all,

    So my question is, could you use a weather balloon to assist with escaping the atmosphere and attaching a rocket to take it the rest of the way, and we're talking a really small payload obviously.

    Would the weight of the fuel/rocket weigh down the balloon enough to significantly reduce its impact?

    Or simply am I (literally) miles off and the altitude a weather balloon reaches will help in no way?
    On average a weather balloon can surpass the boundary of space (100km), so it's done most of the work?

    Thanks,
    Nick
     
  2. jcsd
  3. Dec 4, 2014 #2
    The bulk of the work is not gaining the altitude, but getting enough speed to get in an orbit around the earth, and not fall back immediately.
    If we can believe wikipedia, the altitude record for unmanned balloons is 53 km. Getting 1 kg to 53 km takes mgh = 1 * 9.8 * 53000 = 5.19 * 10^5 J.

    Minimum Orbital speed is 7.5 km/s = 75000 m/s. The kinetic energy needed is (1/2)mv^2 = (1/2) * 1 * (7500)^2 = 2.81 * 10^7 J. If you start on the equator and use the 450 m/s rotational speed of the earth, you still need (1/2) * (7050)^2 = 2.48 * 10^7. Still about 50 times more energy needed than the boost you can get from getting to 53 km.
     
  4. Dec 5, 2014 #3
    I'm not sure I understand what you're comparing here?

    It would make sense that the problem you have now is you have to reach the speeds required to go from the 53 km to outer space?
    And you're suggesting that this is still a significant amount of the work left to do?

    Wouldn't the thinner atmosphere and weaker gravity make it far easier to reach those speeds?
     
  5. Dec 5, 2014 #4

    Bandersnatch

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    Hi Dead Fish,

    willem2 is refering to the speed required to get in orbit, which is usually what you want to achieve with spacecraft.
    While this speed is dependent on the strength of gravity and gravity falls with the square of distance, the difference between 6378km(Earth radius) and 6431km (radius + 53km) is minuscle, and the speed required doesn't differ much.

    However, the atmospheric drag is significant.
    If not for the atmosphere, rockets could be in principle launched into orbits barely high enough to clear any mountaintops in their way. With atmosphere, it is more fuel-efficient to go nearly straight up until the air gets thinner, before starting to accelerate sideways.

    Have a look at this infographic for the Zenit3SL rocket launch profile:
    http://www.spaceflight101.com/uploads/6/4/0/6/6406961/2349991_orig.gif?537 [Broken]
    The record height a ballon could get you (53km as of today), could potentially allow you to reduce the first stage by a good bit, as most of it is used just to propel the rest of the rocket above the atmosphere.

    However, what's left is still a humongous weight in fuel and metal, and lifting it up with a balloon to any height, let alone the record 53km is beyond daunting.
    Buoyancy equation states that ##F_B=(\rho _{air} - \rho _{gas})*g*V##, so at sea level you get enough lift from 1 cubic metre of helium to move 1 kg of payload. This goes down quickly as air density decreases. At 6km you'd need twice the amount to stay afloat.
    The size of a balloon needed just to lift a couple hundred tonnes of payload off the ground is about the same as Hindenburg, and this is not taking into account the weight of the balloon itself.
     
    Last edited by a moderator: May 7, 2017
  6. Dec 5, 2014 #5

    russ_watters

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    You don't gain speed because you need speed to get into outer space, you gain speed because you want to be in orbit once you reach space. So gaining altitude has no impact on that requirement; they are completely separate things.
     
  7. Dec 5, 2014 #6
    What is Q [kgf/m2] here ?
     
    Last edited by a moderator: May 7, 2017
  8. Dec 5, 2014 #7

    Bandersnatch

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  9. Dec 5, 2014 #8
    I was wondering what kind of unit is "kgf". Now I see it means "kilogram force", lol.
     
  10. Dec 6, 2014 #9
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