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Homework Help: Comet around sun

  1. Mar 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.5×104m/s when at a distance of 2.1×1011m from the center of the sun, what is its speed when at a distance of 5.1×1010

    2. Relevant equations
    Conservation of energy

    3. The attempt at a solution
    I was able to solve this using conservation of energy, but I was wondering why I don't the get same answer from conservation of angular momentum. There is no external force because the only force is the force of gravity, which is part of the system.

    Here is the attempt with conservation of angular momentum:
    m(r x v) = m(rfinal x vfinal)
    (r x v) = (rfinal x vfinal)
    The velocities are tangential so sin(90) = 1
    (r)(v) = (rfinal) (vfinal)
    (1.5x104)(2.1x1011) = (5.1x1010)(vfinal)
    (vfinal) = 6.18 x 104 m/s

    The correct answer, obtained using conservation of energy is (vfinal) = 6.5 x 104 m/s

    The answers should be the same though. There is no external force or non-conservative forces.
  2. jcsd
  3. Mar 16, 2015 #2
    Even earlier in my assignment, it mentions that the angular momentum is conserved about the center if the distance between them is changed.
  4. Mar 16, 2015 #3


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    Staff: Mentor

    You've assumed that the velocity vectors for the two given distances are perpendicular to the radius vectors, i.e., you've assumed perihelion and aphelion for the given points. But this is NOT specified in the problem statement! You cannot take the angle in the cross product to be 90°.
  5. Mar 16, 2015 #4

    rude man

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    Homework Helper
    Gold Member

    Kepler's 2nd law (constant rate of area being swept out, or 1/2 ωr2), is equivalent to the conservation of momentum, which is mωr2. The key is a central force but not a velocity perpendicular to that force.

    ω is continually changing and = dθ/dt where dθ is the differential angle formed by the comet's radius vector at two differentially short times dt apart. In other words, at any instant, ω = v/r.

    So, differentially, v IS perpendicular to r. Your answer of 6.18e4 is correct. Your answer based on energy conservation had to use the mass of the sun which is not necessarily the mass of the sun as the problem implies!

    EDIT: after looking at this some more I've concluded that there is no way to solve this problem other than by energy conservation. The reason is that
    v2 = (r2)(dθ/dt)2 + (dr/dt)2 .... (1)
    & I left out the second term in computing v2. Unfortunately I see no way to determine the second term above with the given data. The solution would not be unique.

    Unless points 1 and 2 correspond to the following points on the elliptical orbit:
    (a) aphelion and one of the two points where the semi-minor axis intersects the orbit;
    (b) perihelion & one of the above-mentioned two points; or
    (c) aphelion and perihelion.
    In all three cases the second term in my equation (1) vanishes and your answer would be correct.

    Sorry for the confusion.
    Last edited: Mar 17, 2015
  6. Mar 16, 2015 #5


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    Staff: Mentor

    v is perpendicular to r only for circular orbits or at aphelion and perihelion for elliptical orbits. This is why angular momentum involves a cross product rather than it being a simple scalar product of |r| and |v|.
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