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Finding the minimum speed for putting an object in orbit

  1. Mar 15, 2009 #1
    yet another doubt about gravitation. we must affirm that it's also not a homework question, so, we won't need any number, only the formulas.

    1. The problem statement, all variables and given/known data
    1) suppose that an object is located in a distance r from the center of the Earth (not necessary to say that r is larger than the Earth's radius). we want to know the formula of the speed necessary (launching it horizontally) for that object to describe (A) a circular orbit of radius r.
    we additionally want to know how to make this object describe (B) an elliptical, (C) a parabolic and a (D) hyperbolic orbit.
    2) we would also like to know how to calculate the speed necessary to launch an object from the surface of the Earth, so it can enter in a circular orbit of radius r (bigger than the radius of the Earth).

    2. Relevant equations
    gravitational force:
    Fg = GmM/r², where G is the gravitational constant, m is the mass of the object, M is the mass of the Earth and r is the distance between the center of the Earth and the center of the object (in this case, the distance r mentioned above).
    centripetal force:
    Fcp = mv²/r (where v is the speed of the object).
    conservation of energy
    [tex]\frac{mv^{2}_{0}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1}}{2}-\frac{GmM}{r}[/tex], where R is the radius of the Earth and r is the distance we mentioned above.

    3. The attempt at a solution
    1) (A) for the object to describe a circular orbit, its centripetal force has to balance with the gravitational force:
    Fcp = Fg
    mv²/r = GmM/r²
    v²/r = GM/r²
    v² = GM/r
    [tex]v = \sqrt{\frac{GM}{r}}[/tex]
    for B, C and D, we have no clue, because we don't know the mathematical condition to obtain those orbits.
    NOTE: does v have to be horizontal? why wouldn't an object thrown with vertical v = [tex]v = \sqrt{\frac{GM}{r}}[/tex] also respect that Fcp = Fg?
    2) we don't have much idea where to start, but we think that this velocity will have a vertical component and a horizontal component which is [tex]v_x = \sqrt{\frac{GM}{r}}[/tex].
    the vertical velocity would be v0y: [tex]\frac{mv^{2}_{0y}}{2}-\frac{GmM}{R} = \frac{mv^{2}_{1y}}{2}-\frac{GmM}{r}[/tex].
    in this case, we think that v1y would have to be zero.

    thank you advance.
  2. jcsd
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