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Third cosmic velocity

  1. Sep 23, 2005 #1
    i have a question-
    Find approximately the 3rd cosmic velocity v3, i.e. the minimum velocity that has to be imparted to a body relative to the Earth's surface to drive it out of the solar system.The rotation of the Earth about its own axis is to be neglected.

    just as we deduce escape velocity,i went for the following equation-


    -(GMm)/R -(GM'm)/r +mv3^2/2 =0

    or,v3={2GM/R +2GM'/r}^1/2

    However,the answer is v3~{2GM/R +(2^0.5-1)^2.GM'/r}^1/2
    ~17 Km/s [~means approximately;can't get the real sign]

    Here,M=earth's mass
    M'=Sun's mass
    R=earth's radius
    r=radius of earth's orbit.
    -----------------------------------------------------------------------
    this is not a homework.thanking you all in advance :smile:
     
    Last edited: Sep 23, 2005
  2. jcsd
  3. Sep 23, 2005 #2

    Doc Al

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    Staff: Mentor

    Don't forget that the body is already moving with the earth, so its initial speed is that of the earth around the sun.
     
  4. Sep 23, 2005 #3

    HallsofIvy

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    Is the formula for a body, with mass m, in orbit of radius R, about a body of mass M. In order to "deduce Earth's velocity", in orbit around the sun, R would be the radius of earth's orbit, not "earth's radius"

    In any case, this problem has nothing to do with orbiting. An object going "out of the solar system" is not in orbit around either the earth or the sun.

    You can calculate the potential energy difference between an object at earth's distance from the sun and an object "at infinity". To do that, integrate the gravitational force function [itex]F= \frac{GMm}{r^2}[/itex] from earth's orbit to infinity. Finally, set the kinetic energy of the object (relative to the sun) equal to that and solve for the speed (relative to the sun).

    Doc Al's point is that since you want the speed relative to the earth, you will have to subtract earths speed relative to the sun from that.
     
  5. Sep 23, 2005 #4
    My mistake :blushing: i meant the escape velocity for a body on Earth.i am going to edit that part :biggrin: Thanks for the help.I will see if I can solve the problem now.
     
  6. Sep 23, 2005 #5

    O.K. as you said,I deduced the potential energy difference which is GM'm/r.
    So,mv^2/2=GM'm/r -->v=(2GM'/r)^1/2
    Therefore,v3=v-(GM'/r)^1/2.
    But this is nowhere near the answer.I have not considered any influence of the Earth which is most probably wrong.So,where & how should I take the Earth in account?
     
    Last edited: Sep 23, 2005
  7. Sep 24, 2005 #6
    I am still waiting for an answer
     
  8. Sep 24, 2005 #7

    Janus

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    Okay, so far you have determined the velocity with respect to the sun needed to escape from the sun for an object r distant from the sun (v').
    You also need to determine how much addtional velocity would be needed to escape from the Earth starting at the Earth's surface(v).

    Then you must consider that that Earth is already moving with respect to the Sun, and thus any object that just escapes the Earth will share that velocity and will already be moving at some fraction of v' with respect to the sun.
     
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