Solutions to Intermediate Astronomy (7)

The problems again:

1) Find, approximately, the periods of revolution of the following binary star systems in which each star has the same mass as the Sun, and in which the semi-major axis of the relative orbits has the value:(a) 1 AU(b) 6 AU(c) 100 AU.

Solution:

For each system we have: m1 = m2 so total mass = 2M_s (solar masses)

Then:

m1+ m2 = (a)^3/ P^2 and

P = [(a^3)/ 2]^½

(a) a = 1 AU

P = [1/2]^½ = 0.707 yr.

(b) a = 6AU

P = [(6)^3/2]^½ = 10.3 yr.

(c) a = 100 AU

P = [(100)^3/2]^½ = 707.1 yrs.


(2) For each of the systems in (1), at what distance would the two stars appear to have an angular separation of 1"?

We require:

m1 + m2 = (d a")^3/ P^2

and need to find d for different a" = 1"

Simplifying:

(a) P = 0.707 yr.

(d a") = [(m1 + m2) P^2]^1/3

Then: d = [(2) (0.707)^2)]^1/3 = 1 pc

(b) P = 10.39 yr.

Then: d = [(2) (10.3)^2)]^1/3 = 6 pc

c) P = 707.1 yr

Then: d = [(2) (707.1)^2)]^1/3 = `00 pc


(3) The true relative orbit of Epsilon Ursae Majoris has a semi-major axis of 2½" and the parallax of the system is 0."127. If its period is 60 years, find the sum of the components in solar mass units.

First, find distance: d = 1/p"

d = 1/0."127 = 7.87 pc

Apply:

m(A) + m(B) = (d a")^3/ P^2

where a" = 2½" and d = 7.87 pc with P = 60 yrs.

Then:

m(A) + m(B) = ((7.87) (2½" )]^3/ 60^2 = 2.1 solar masses


(4) A hypothetical spectroscopic-eclipsing binary system is observed and its period is 3 years. The maximum radial velocities with respect to the center of mass of the system are:

Star A: 4π/3 AU/yr

Star B: 2π/3 AU/yr

A graph of the radial velocity curves for each star matching its physical situation is given in the diagram. The important thing is to have the maxima and minima in the correct directions at the key points in their respective orbits.

Positions (1) and (2) in the graphs allow us to obtain the max. relative velocity for the system, for which:

V = (4π/3 + 2π/3)AU/yr = 6π/3 AU/yr = 2π AU/yr

P = 3 yrs.


(a) Find the ratio of the masses of the components.

The ratio of the masses will be in the ratio of the radial velocities or:

4π/3 : 2π/3 = 2: 1

(b) Find the mass of each star in solar units.(Assume the eclipses are central)

Recall to get distance:

a = (V x P)/ 2π = (2π AU/yr x 3 yr)/ 2π = 3 AU

and: m1 + m2 = (3AU)^3/(3 yr) ^2 = 3 solar masses

The masses are inversely proportional to the radial velocities in their orbits. Since:

Star A has v(A) = 4π/3 AU/yr = 2 v(B) where v(B) = 2π/3 AU/yr

then:

m(A)/ m(B) = v(B)/v(A) = 1/2

or m(A) = m(B)/2 or m(B) = 2m(A)

But: m(A) + m(B) = 3

so: m(A) + 2m(A) = 3

3 m(A) = 3

and m(A) = 1 solar mass, m(B) = 2m(A) or, 2 solar masses