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Posers and Puzzles

Posers and Puzzles

  1. 13 Apr '09 01:01
    Minas, a moon of Saturn, has an orbital radius of 1.87x 10^8m and an orbital period of about 23h. Use Newton's version of Kepler's third law and these data to find the mass of Saturn.
  2. 13 Apr '09 07:45
    Originally posted by wormer
    Minas, a moon of Saturn, has an orbital radius of 1.87x 10^8m and an orbital period of about 23h. Use Newton's version of Kepler's third law and these data to find the mass of Saturn.
    If we don't know anything more than this, and the Newton's version of Kepler's third law, then we cannot find a reliable answer.

    Using the same formulae to the Moon of Earth, or the moon of Pluto, I don't think we will get a very good result.

    ...unless we assume that the mass of Mimas is near to nothing compared with the mass of Saturn. And that information is not given.
  3. Standard member sven1000
    Astrophysicist
    13 Apr '09 20:08
    Originally posted by FabianFnas
    If we don't know anything more than this, and the Newton's version of Kepler's third law, then we cannot find a reliable answer.

    Using the same formulae to the Moon of Earth, or the moon of Pluto, I don't think we will get a very good result.

    ...unless we assume that the mass of Mimas is near to nothing compared with the mass of Saturn. And that information is not given.
    It is certainly true that the mass of Mimas is much less than Saturn. All of Saturn's moons together are much less.
  4. Standard member sven1000
    Astrophysicist
    13 Apr '09 20:19
    And the answer...

    P^2 = C*a^3 is Kepler's third law, with P = period, a = radius of orbit

    C = 4*pi^2 / G*M, with G as Newton's constant and M as Saturn's mass, in the Newtonian version.

    Using the given numbers for Mimas, I get a mass for Saturn of 5.72x10^26 kg.
  5. 14 Apr '09 00:52
    I got 4.214x10^36 as my anwser but I know its wrong and I am trying to figure out my mistake.
  6. 14 Apr '09 05:42
    Originally posted by sven1000
    It is certainly true that the mass of Mimas is much less than Saturn. All of Saturn's moons together are much less.
    True. I know it, you know it, but I'm sure not all of us knows it without looking it up. Mimas is certainly a small moon.

    I ask you: Does it matter if the orbit of Mimas is circular, or does it gives the same result if the orbit is highly elliptical? Do we have to know its eccentricity before we can rely on the answer?
  7. Standard member sven1000
    Astrophysicist
    14 Apr '09 21:26 / 1 edit
    Originally posted by FabianFnas
    True. I know it, you know it, but I'm sure not all of us knows it without looking it up. Mimas is certainly a small moon.

    I ask you: Does it matter if the orbit of Mimas is circular, or does it gives the same result if the orbit is highly elliptical? Do we have to know its eccentricity before we can rely on the answer?
    We should technically use the semi-major axis of the elliptical orbit, in place of the radius. But the eccentricity itself doesn't matter (so long as the moon doesn't collide with the planet ).

    Edit: It occurs to me to wonder if the combination of a known period and known semi-major axis around a fixed mass would determine a unique eccentricity. I don't recall off-hand the derivation for eccentricity in terms of these variables.
  8. 18 Apr '09 09:53
    5..68xtentothe26th
  9. Standard member bill718
    Enigma
    26 Apr '09 03:36
    Originally posted by sven1000
    And the answer...

    P^2 = C*a^3 is Kepler's third law, with P = period, a = radius of orbit

    C = 4*pi^2 / G*M, with G as Newton's constant and M as Saturn's mass, in the Newtonian version.

    Using the given numbers for Mimas, I get a mass for Saturn of 5.72x10^26 kg.
    You do this...and you'll clean it up!!!