Planet Mass and angular radius

  • Thread starter Maradoc
  • Start date
  • #1
1
0

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:
rmoon= 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:
Aplanet= (18^2)^(1/3) = 6.87 AU Aearth=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.
 

Answers and Replies

Related Threads on Planet Mass and angular radius

  • Last Post
Replies
6
Views
29K
  • Last Post
Replies
1
Views
6K
Replies
3
Views
5K
Replies
1
Views
1K
Replies
10
Views
4K
  • Last Post
Replies
5
Views
1K
Replies
8
Views
3K
Replies
4
Views
23K
Replies
1
Views
3K
Top