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Homework Help: Conservation of energy and momentum in planetary motion

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data

    The period of a comet is 75.8 years. The perihelion distance is 0.596 AU (1 AU = 1.5 ⋅ 1011 m).
    The velocity at perihelion is vp = 5.45 ⋅104 m/s.

    a) Find the length of the major semi-axis of the elliptical orbit.
    b) Find the aphelion distance and the velocity at aphelion for the comet.

    2. Relevant equations

    T2 = k a3 (Kepler's third law)
    m v r = const. (conservation of angular momentum at aphelion and perihelion)

    [tex]1/2 m v^2 - \frac{G m M}{r} = const[/tex] (conservation of energy)

    3. The attempt at a solution

    I used the 3rd Kepler's law to find the semi-major axis a. For the constant I used
    [tex]k=\frac{4 \pi^2}{G M}[/tex]
    where M = 1.99 ⋅1030 kg as provided by the book.

    I find
    a = 17.9 AU = 2.68⋅1012 m
    ra = 35.2 AU = 5.28 ⋅1012 m

    Then, from [tex]m v_a r_a = m v_p r_p[/tex] I find va=923 m/s.

    So far, I'm in complete agreement with the solution provided by the book.

    Now, I think that the same result could be derived from the conservation of energy.

    [tex]v_a = \sqrt{v_p^2+2GM \left(\frac{1}{r_a}-\frac{1}{r_p}\right)}[/tex]

    However, this formula gives a completely different result: 7.15 ⋅103 m/s.
    Much larger than the previous one.

    For the life of me, I cannot see the mistake. Here we are talking about a small comet orbiting the Sun, so the latter should be basically at rest. Center of mass should not be relevant.
    The other thing I can think of is the constant in Kepler's law. But that should be ok in this case (again, the comet has a very small mass)

    Can anyone help? Could it perhaps be a matter of roundoff? Difference between very small numbers?

    EDIT: corrected typo in initial formula for energy
    Last edited: May 16, 2016
  2. jcsd
  3. May 16, 2016 #2


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    There is a typo (?) in your initial equation for the energy, it should be r, not r2.

    The approach later looks fine. The calculation via energy is sensitive to rounding errors (you subtract two large quantities to get a small one), so check that you keep enough significant figures. 7 km/s is way too fast.
  4. May 16, 2016 #3
    Yes, sure, thanks, I've corrected the typo.

    So you think it could be a rounding error... It occurred to me at the end of the post. With those big and small numbers multiplied by each other.
    I tried collecting all of the powers of ten and managing them algebraically. Not much luck with that.

    I'll give it another try. Thanks a lot for your reply. Rounding error is less scary than not understanding why it won't work.
    Interesting problem...
  5. May 16, 2016 #4


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    The orbit is completely determined from the perihelion distance and speed. So, the period is determined from ##v_p## and ##r_p##. It could be that the period specified in the problem is inconsistent with the data given for ##v_p## and ##r_p##. I haven't checked this, though.
  6. May 16, 2016 #5


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    Going from 54.5 km/s to 54.4 km/s or 54.6 km/s changes the result significantly. Yes, rounding issue, not necessarily in the actual calculations.
  7. May 17, 2016 #6
    Thanks a lot to both of you for your replies.
    I'm not sure I know how to check whether the ##v_p## and ##r_p## are consistent with the period.
    I can't simply equate the centripetal force and the gravitational force at perihelion, as if the orbit was circular, can I?

    I'd like to go to the bottom of this. I'm not sure whether
    1) using the conservation of energy is correct in principle but impractical because it would need data with a lot of significant digits.
    2) the conservation of energy doesd not work in this specific case due to inconsistent data, but it would for consistent data (I mean, even with a reasonable number of significant digits).

    Thanks again for your insight.
  8. May 17, 2016 #7


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    You can use the Vis-viva equation
    $$v =\sqrt{GM\left(\frac 2r - \frac 1a\right)}$$ to solve for ##v_p##, ##r_p##, or ##T## after choosing the other two parameters.

    I found the parameters you were given were inconsistent, but even with more accurate parameters, the energy calculation is very sensitive to the number of digits you use because the orbit is very eccentric.
    Last edited: May 17, 2016
  9. May 17, 2016 #8


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    Just to add to what vela and mfb have said regarding the sensitivity of the data in this problem,

    If I take the given value for the period and perihelion distance, I find that the perihelion speed should be 54.1 km/s to 3 significant figures. Although this is fairly close to the given value of 54.5 km/s, this makes a significant difference in the aphelion distance and speed. I find the escape speed at the given perihelion distance to be about 54.6 km/s. So changing vp from 54.1 to 54.6 km/s makes an "infinite" change in the aphelion distance!
  10. May 17, 2016 #9
    Following vela's suggestion, I used the vis-viva equation by choosing ##T=75.8\, \rm yrs ## and ## r_p = 8.94\cdot 10^{10} \, \rm m ##.
    I found ## a = 2.68 \cdot 10^{12} \,\rm m ## and ## v_p = 54.0 \,{\rm km/s} ##.

    Repeating the original exercise with the original set of (corrected) data, I find ## v_a = 917 \,{\rm m/s} ##.
    The last value is the same irrespective of the equation I use (conservation of a.m., vis-viva equation again or conservation of energy, which is basically the same as v.-v.).
    Also, the energy at aphelion and perihelion are virtually the same.

    Thus, as far as the exercise is concerned, if the data are compatible the result won't depend on the equation used to derive ## v_a ##.
    I guess that if the data are compatible, the above discussed sensitivity to the velocity should not matter, right?
    All in all it seems to me that it was a matter of sloppiness in the preparation of the exercise.

    Thanks to all of you!
  11. May 17, 2016 #10


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    You saw how much the difference between 54.1 and 54.4 km/s made. Between 54.050 and 54.149, which both get rounded to 51.4, the difference is still notable.
  12. May 17, 2016 #11
    Yes, I was too hurried.
    The problem is that everything "worked" because I used all of the decimal digits given by the calculation with the vis-viva equation.

    As soon as I truncate ##v_p## to fewer significant digits, the result from the conservation of energy starts to drift away from that of the conservation of angular momentum.
    So I guess that in this situation it is preferable to use the conservation of angular momentum, which does not involve those products of huge and small numbers.
    I mean, if ##v_p## is truncated, the conservation of a.m. still gives a reliable result for ##v_a##, right?
    It compares reasonably with the result of the vis-viva equation, which does not contain the truncated datum.

    Thanks for your patience
  13. May 17, 2016 #12


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