Determining the Mass of a Planet

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To determine the mass of a planet with an 18.0-year orbital period and a moon with a 10-day period, Kepler's Third Law can be applied. The assumption of circular orbits simplifies calculations, allowing for the use of the formula P² = a³, where P is the period in years and a is the semi-major axis in astronomical units (AU). The angular size of the moon's orbit, measured at 1.2 arcminutes, can be converted to physical dimensions to find the moon's orbital radius. By calculating the total mass of the planet-moon system and assuming the moon's mass is negligible, the planet's mass can be approximated. The discussion emphasizes the importance of correctly interpreting the problem and applying the relevant equations.
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Homework Statement


Suppose that we see a planet in our Solar System that we measure to have
an orbital period (around the Sun) of 18.0 years. We look at it with a
telescope and see that it has a moon. From repeated observations, when
the planet is near or at opposition, we note that the orbit of the moon is
approximately circular, with an observed radius of about 1.2 arcminutes
and a period of 10 days. What is the mass of this planet?

Pplanet = 18.0 years = 567024668 seconds
Pmoon = 10 days = 864000 seconds
emoon = 0
θmoon = 1.2 arcminutes



Homework Equations


P2G(M1 + M2) = (4∏2)a3

L = mvr

M = rv2 / G


The Attempt at a Solution



I simply have no idea where to begin. I want to say that,

mplanetvplanet = mstarvstar

But that ignores the moon and the Earth's positions and what not. Also, I'd have to assume that its orbit is circular, when it doesn't say so in the question.
 
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I think that it would be safe to assume that the mass of the planet is insignificant compared to that of the Sun. Similarly, the mass of the moon will be insignificant compared to that of the planet.

Can you find the orbital radius for the planet? What then will be the Earth-planet distance when the planet is at opposition?
 
Here's a tip:

when the period is in years, and the semi-major axis in AU, then in those units, the constant of proportionality is just 1. Therefore, Kepler's 3rd Law becomes:

P2 = a3

I think you have to assume that the planet's orbit around the sun is roughly circular. Solar system planets' orbits all have fairly low eccentricity, since the orbits tend to circularize over time. I mean, if you didn't assume circularity, then I don't know how you'd solve it, since saying that "the planet is at opposition" doesn't tell you anything about where it is in its orbit.

If you do assume its circular, then at opposition, the distance between Earth and the planet is approximately equal to the difference between their orbital radii (draw a diagram).

Given this distance, and the angular size of the orbit of the fictitious planet's moon, you can figure out the physical size of this moon's orbit. Given that, and the moon's orbital period, you can use Kepler's Third law (applied to the planet-moon system, rather than to the sun-planet system) to figure out the total mass of the system. If the moon is much less massive than the planet, than this is pretty much equal to the planet's mass.

It could be there's a way to solve for the moon's mass and the planet's mass separately after this. I have to admit I'm not totally sure.
 
How can we determine the size of the moon's orbit? If we use the arc length formula a = θr, wouldn't that give us the moon's radius and not it's orbit?
 
Dakkers said:
How can we determine the size of the moon's orbit? If we use the arc length formula a = θr, wouldn't that give us the moon's radius and not it's orbit?

No, it would give you the radius of the moon's orbit. The 1.2 arcmin is the angular size of the orbit, not of the moon itself.
 
Well. Next time, I think I should read the question properly.

Thanks much :)
 
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