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Object leaving the atmosphere

  1. May 2, 2013 #1
    hi all, me and my friend were arguing today. We read that a cricket ball requires 11 km/s to leave the atmosphere. Now not taking friction/drag into account would this velocity be sufficent for any mass to leave the earth. Thanks.
     
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  3. May 2, 2013 #2

    A.T.

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  4. May 2, 2013 #3
    so it is mass dependant. Thanks.
     
  5. May 2, 2013 #4
    using equatiion of acceleration
    [itex]a_{s}=\frac {v^{2}{r}[/itex] and [itex]g=\frac {GM}{r^{2}}[/itex]

    the velocity is enought to escape a mass from gravity, we must ensure that no resultant of acceleration there
    [itex]a_{s}=g[/itex]
    [itex]\frac {v^{2}}{r}=\frac {GM} {r^{2}}[/itex]
    [itex]v^{2}=\frac {GM}{r}[/itex]
    [itex]v=\sqrt {\frac {GM}{r}}[/itex]

    as [itex]G\ =\ 6.673(10)\ \times\ 10^{-11}\ m^{3} kg^{-1} s^{-2}[/itex]
    and mass of earth [itex]M=5.97219 × 10^{24} kg [/itex]
    radius of earth [itex]6371 m [/itex]
    so the result
    [itex]v=\sqrt {\frac {GM}{r}}=7909 m/s[/itex]
     
  6. May 2, 2013 #5

    Bandersnatch

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    No, no. You got it wrong. It's value depends on the central mass, i.e., the Earth, or any other source of gravity you might have in mind.
    All objects need the same escape velocity to leave the central mass' gravitational influence.

    Also, the 11 km/s in not about leaving atmosphere, but about going so fast that you never get pulled back. You need less to e.g., put an object in orbit.
     
  7. May 2, 2013 #6
    no.. it's not about mass dependant..
    you know the acceleration and free fall mass is not depend on mass..
    It's depend gravitational acceleration that must be canceled.
    For ilustration, on moon, you need smaller acceleration, because acceleration of gravitation there is smaller that on the earth.
    for [itex]\frac {GM}{r} [/itex] is properties of planets.. It's variabel that cause how big gravitational acceleration
    thanks
     
  8. May 2, 2013 #7
    ya.. I mean like that.
     
  9. May 2, 2013 #8

    Bandersnatch

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    arifz2303, you've provided calculation for the First Cosmic Velocity, i.e., the speed needed to put an object in a circular orbit, not the escape velocity(aka the Second Cosmic Velocity), which needs to be calculated from energy considerations, as shown in A.T.'s wikipedia link.

    Granted, the OP did not specify what he meant by "leaving the atmosphere", but let's stick to the velocity value he provided so as not to breed any more confusion.
     
  10. May 2, 2013 #9
    thank you so my friend was wrong! Sorry for being so vauge.
     
  11. May 2, 2013 #10

    Danger

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    I'm just going to repeat something here that Russ pointed out a couple of years ago. The term should be "escape speed"; velocity is a vector measurement, but you can escape in any direction if you have enough speed.
     
  12. May 3, 2013 #11

    Bandersnatch

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    Sure it should, but it's been traditionally called "velocity", so using "speed" instead might confuse people just as much.
    Although, perhaps pushing the nomenclature in that direction is worth the effort of having to occasionally explain what you mean, hmmm.
     
  13. May 3, 2013 #12

    Danger

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    That's the way I see it. Ever since he mentioned that, I've never gone back to the old way.
     
  14. May 3, 2013 #13

    DEvens

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    Folks in the thread talked about the speed you need to go just to overcome gravity. Nobody talked about the speed you need to "leave the atmosphere." That requires going fast enough to overcome air resistance in addition to gravity. It's a way harder calculation not handily available from folks like Newton.

    Notice it will depend on lots of stuff. Like the altitude you start from. If you start from the top of a tall mountain where you are above a non-trivial part of the atmosphere, the required speed is less. If you fire at a very shallow angle, very close to horizontal, the required speed is more.

    A friend once did several such calculations for a hobby interest. He was connected with HARP at the time (but since has left all consideration of it) and wanted to send masses to low Earth orbit (LEO) using a really BIG gun.

    http://en.wikipedia.org/wiki/Project_HARP

    His calculations seemed to indicate that from about 15,000 feet altitude and at a 45 degree angle you'd need something over 15 km/s to get a mass to LEO, depending on how dense it was. He expected you could make steel ingots survive the trip. Then you'd need some kind of vehicle already in orbit to catch them and adjust their orbit. You could thus put up quite a lot of building material for very cheap.

    The project was cancelled because people complained about a cannon that could target any place on Earth.
    Dan
     
  15. May 4, 2013 #14
    intresting 0_o. Me gusta.
     
  16. May 4, 2013 #15
    yes but that speed would be direction dependant. The escape speed straight up is surely less than if you went down through the earth. We ARE talking about minimum escape speed right?
     
  17. May 4, 2013 #16

    Bandersnatch

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    Assuming you can pass through Earth without interacting with it, the speed is the same.
     
  18. May 4, 2013 #17
    fair play, i did say "not including friction". Right you are bandersnatch
     
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