Planet Mass and angular radius

[Maradoc]

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.

Homework Equations

Keppler's Third Law:
r³/T² = GM/4(pi)²

The Attempt at a Solution

I think that I need to determine the linear radius of the moons orbit and use that combined with it's period to solve for M in the equation r³/T² = GM/4(pi)².

I am not sure exactly how to do so, first i converted arcminutes to radians:
r_moon= 1.2/60*pi/180= 3.5*10^-4rad.

Then using Keppler's Third Law and the information given in the question I determined that the distance from the Earth (where the orbital radius is viewed from) to the planet is the semi-major axis of the planet - the semi major axis of the earth:
A_planet= (18^2)^(1/3) = 6.87 AU A_earth=1 AU

So the distance from Earth to the planet would be 5.87 AU.

This is where I'm not really sure what to do. To get the radius of the moon's orbit do I just convert 5.87AU to meters and then multiply by the angular radius of
3.5*10^-4rad? (Or perhaps my method is completely wrong in which case any advice on where I am going wrong would be helpful)

Thanks.

 

[anathema]

Thank you for your question. Your approach is on the right track, but there are a few things that need to be corrected in order to solve for the mass of the planet.

Firstly, your conversion from arcminutes to radians is correct, but it should be noted that the radius of the moon's orbit is actually 1.2 arcseconds, not arcminutes. This means that the radius in radians would be 1.2/3600*pi/180 = 2.1*10^-6 rad.

Next, your calculation for the distance from Earth to the planet is slightly off. Kepler's Third Law states that the ratio of the cube of the semi-major axis to the square of the orbital period is constant for all planets in the solar system. Therefore, we can write:

(Aplanet/Aearth)^3 = (18/365)^2

Solving for Aplanet, we get Aplanet = 1.7 AU.

Now, to calculate the radius of the moon's orbit, we can use the following formula:

rmoon = Aplanet*sin(2.1*10^-6 rad)

This gives us a radius of approximately 5.1*10^8 meters.

Finally, we can use Kepler's Third Law to solve for the mass of the planet:

(5.1*10^8)^3/(10 days)^2 = GM/4(pi)^2

Solving for M, we get M = 1.2*10^26 kg.

I hope this helps clarify the steps needed to solve for the mass of the planet. Let me know if you have any further questions. Good luck with your studies!