The Sun is about 300,000 times as massive as the Earth. Calculate the radius of the Sun's orbit about its common center of mass with Earth. How big would the radius of the Sun's orbit appear if the...
- The Sun is about 300,000 times as massive as the Earth. Calculate the radius of the Sun's orbit about its common center of mass with Earth.
- How big would the radius of the Sun's orbit appear if the solar system were viewed from a distance of 1 pc?
Question one regards barycentric coordinates. The center of mass in the Sun-Earth relationship is not at the center of either body; it is a sort of average between the two. I can tell you without even doing the math that, because the Sun is so large and so much more massive, that their barycentric point will probably be inside the Sun, but definitely not at the center of the Sun. This is an important concept in astronomy because the "wobble" produced by this displacement can tell us that an object is being influenced by another; this is how some of the first extrasolar planets were discovered.
The equation we can use to find the barycentric coordinates is r = a/(1 + m1/m2).
R represents the distance from the center of the primary (the sun)
A is the distance between the centers of the two bodies (approx. 150,000,000km)
M1 and M2 are the two masses relative to each other (i.e. let the Earth equal a mass of 1, the Sun will be 300,000)
So, r = 1.5 x 10^8/(1 + 3 x 10^5/1) = 150,000,000 / 300,001 = 450km
The sun has a radius of almost 700,000km, so this is still relatively close to the Sun's center.
This is phrased a little strangely to me: I'm not sure we're asking about the size of the orbital path, or the total area which the Sun occupies during this orbital path, which would be much larger. I'm going to assume that we're talking only about the orbital path, which would be a circle of radius 450km and diameter 900km.
Because stars are extremely far away, their sizes seem extremely small. We use a small unit, the arcsecond, to describe how large things appear to be in the sky. The arcsecond is 1/3600 of one degree of a circle, so there are 1,296,000 arcseconds in the sky.
The formula for the apparent size of an object, the angular diamater, is: Ad = d/D
Ad is the angular diameter, expressed in radians. There are 2pi radians in a circle, and 1296000 arcseconds in this circle, so the answer expressed in arcseconds will require a coefficient of 206,265.
d is the actual size of the object, 900km
D is the distance to the object, 1 parsec or 3x10^13 km
206,265 x (900 / 3 x 10^13) = 6.2 x 10^-6 or .0000062"