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Thread: calculating moon size via distance

  1. #1
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    calculating moon size via distance

    So I was wondering if this was a thing that is possible. A moon in this series has a phase cycle of roughly 36 days and it hangs in the sky appearing larger than Earth's Moon. From my understanding objects farther away rotate at a slower rate than objects closer, so it stands to reason that a moon (lets call her Cathy) that appears larger in the sky than Earth's but has a greater phase cycle, and therefore is farther away than the Moon, must be massive to still appear so large in the sky. I was wondering if its possible to determine how big it is or what distance it is? We know how big the Moon appears in the sky, its distance, and phase cycle. What can we learn about Cathy knowing its phase cycle and relative size in the sky?

  2. #2
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    I think that, assuming you mean revolution rather than rotation, you can indeed make that assumption, with the caveat that you donít know the mass from the size, unless you know the density of the object.


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  3. #3
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    Quote Originally Posted by winkingowl View Post
    So I was wondering if this was a thing that is possible. ...
    You would also need to know the mass of the planet it is orbiting. If the planet itself was half the mass of the Earth, the "Cathy" could be the same size as our Moon, but just a little closer.
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  4. #4
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    The distance to the moon in question can be measured by triangulation from opposite sides of the planet, as we have been doing with our own Moon for centuries. That, along with the observed orbital period, will yield the total mass of the system. Both bodies orbit their common barycenter (center of mass), each at a radius inversely proportional to its mass. If we have a means of observing and measuring the planet's barycentric motion, we can find the mass ratio and thus the moon's mass in proportion to the planet's mass. If I am not mistaken we determined Earth's barycentric motion by observing its effect on the apparent motions of Mars and Venus. That was technically difficult to do with great accuracy by visual means, and may or may not be possible with our hypothetical planet.

  5. #5
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    The nice thing about a calculation like this is it can all be done with ratios to the current situation, where Pm is the current 29.5 day period of the Moon's orbit, Me is the 6 x 1024 kg mass of the Earth, and am is the current 384,000 km average distance to the Moon. Then Newton's modification to Kepler's law gives:
    a / am = (P / Pm)2/3 (M / Me)1/3
    where a is the distance you want, P is 36 days, and M is the mass of the hypothetical planet. If we take the latter to be the same as Earth, then a / am is 1.14, which is farther away just as you expected. The angular diameter theta, compared to that of the current Moon (call it thetam), then obeys
    theta / thetam = (r / rm) / (a / am)
    where rm is the current Moon radius, so that's the other effect you mentioned. To convert that into a mass of the moon M, given mass of our Moon Mm, we must compare the average densities, call that ratio rho / rhom. Then we have
    r / rm = (M / Mm)1/3 (rho / rhom)-1/3
    so
    theta / thetam = (M / Mm)1/3 (rho / rhom)-1/3 (a / am)-1
    and we have all the effects mentioned above. Solving for the mass, since that's what you want to know, we have
    M / Mm = (rho / rhom) (theta / thetam)3 (a / am)3
    Substituting the term with a, we get finally
    M / Mm = (rho / rhom) (theta / thetam)3 (P / Pm)2 (M / Me)
    and you see that the mass of the moon will be proportional to the assumed mass of the planet, given what it looks like from the ground.

    Thus if you assume the same moon density and the same Earth mass, you get
    M / Mm = (theta / thetam)3 (P / Pm)2 = 1.5 (theta / thetam)3
    so it's quite sensitive to how big it looks and you are probably talking about a moon that is at least 3 times the mass of our Moon, if not more.

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