Selected Questions - Answers From All Experts Astronomy Forum (Binary Stars)

Question: Could you explain the concept of a binary star system and how we get its orbit parameters correctly oriented to our view?  (Since the system is not likely to be in the plane of our viewing axis initially)  Also,  how  might one find the mass of each, say for a specified binary system?

Answer:

Below I sketch a basic binary star system for reference:
















The diagram above shows the simplest type of system, with the plane of the binary components A (of mass m(A)) and B (of mass m(B)) perpendicular to the viewing plane. This visual binary system, it's important to note, indicates the apparent relative orbit of the 2 stars - since ordinarily (as you pointed out) -  the plane of the real orbit will not lie in the plane of the sky, i.e. perpendicular to the line of sight as portrayed. Usually, what we do is observe the motions of the fainter member about the brighter over a period of time sufficient to determine the orbit period, and then obtain the apparent relative orbit.

    Geometrically,  it is straightforward to show that an ellipse in one plane when projected onto another plane (say oblique to it) will yield another ellipse but of different eccentricity, e = c/a. Most importantly, the foci of the original ellipse do not project onto the foci of the projected one. This means the primary (brightest star) though it is located at one focus of the true relative orbit, is not at the focus of the apparent relative orbit.

It is this circumstance which makes it possible to determine the inclination of the true orbit to the plane of the sky. Basically, the problem reduces to finding the angle at which the true relative orbit must be projected in order to account for the amount of displacement of the primary from the focus of the apparent relative orbit.

  If the semi-major axis of the true relative orbit (e.g. the one it would have if displayed face-on) has an angular distance of a" (seconds of arc) and if the system is at a distance d parsecs, then the semi-major axis in astronomical units is:

a = (a" x d)

Then the sum of the masses of the two stars is given by Kepler's  3rd law:

m(A) + m(B) = (d a")³/ P²

where P is their period.

Of course, obtaining the total mass is only the first step. One then wishes to obtain the individual masses for each star. This is done by analyzing the motions of each member with respect to the center of mass of the system which ordinarily will be apart from either member. For example, if m(A) = m(B) then the center of mass (cm) will be exactly midway between them.

Consider the visual binary system for Sirius composed of Sirius A and B. The semi-major axis of the true relative orbit is observed to be  7½" and the distance from the Sun to Sirius is 2.67 pc. If the period of the Sirius binary system is 50 years and the component B is found to be twice as far from the center of mass as component A, then we can compute the mass of each component.

We first obtain the total mass from Kepler's 3rd law:

m(A) + m(B) = (d a")³/ P² =

(2.67 pc x 7.5") ³/(50 yr)² = 3.2 solar masses, i.e.  3.2 M_s

So:

m(A) + m(B) = 3.2 M_s

But in terms of the center of mass:


A O------------x cm-------------------------o B



where: xB = 2 (xA)

Then: xB/xA = 2, and: m(A)/m(B) = xB/xA = 2/1

so: m(B) = ½ m(A)

(since the more massive star is always closer to the center of mass)

Thus, m(A)  » 2.13 M_s, and m(B) » 1.06 M_s

Spectroscopic binaries are also of much interest and derive their name because spectroscopic analysis is needed to obtain the radial velocity curve for the system and hence the relative orbital velocity, V, for the pair. The distance around the relative orbit,  i.e. its circumference, is just the relative orbital velocity V (deduced from the radial velocity profile) multiplied by the period P. Then the distance between the stars a, is just:

a = (V x P)/ 2π 

 If, for instance, the relative velocity is a lower limit, then the separation we obtain is a lower limit for the system. If this is then applied to Kepler's 3rd law one can obtain a lower limit to the sum of the masses of the components:

m1 + m2 = a³/P²

To fix ideas, say a spectroscopic binary system is found to have a relative velocity of 100 km/sec and a period of 17.5 days.   Then let component (1) be 3 times the mass of component (2) and find: a) a lower limit for the separation of the components, a, and thence a lower limit to the sum of the masses,  and b) the lower limits on the masses of the components.

First convert 100 km/sec to AU/yr. (Astronomical units per year)

Over one year: t = 3.156 x 10⁷ s

total distance covered:

d = v x t = 100 km/s (3.156 x 10 ⁷ s) = 3.156 x 10⁹ km

But 1 AU = 1.495 x 10⁸ km

Then, the AU in this total distance:

d/AU = (3.156 x 10⁹ km)/ (1.495 x 10⁸ km) = 21.1 AU

The period in yrs. for 17.5 days:

P = 17.5/ (365.25) = 0.048 yr.

Then:

a = (21.1 AU x 0.048 yr/AU)/ 2π = 0.161 AU

The lower limit to the masses is therefore:

m1 + m2 = (0.161)³/ (0.048)² = 1.8 solar masses

Since star m1 has 3x the mass of star m2, then:  m1 = 3m2

And: m2 + 3m2 = 1.8 or 4m2 = 1.8

so:

m2 = 1.8/4 = 0.45 solar masses, and

m1 = 3(0.45) = 1.35 solar masses

Basically, in working any binary star problem, say to get the relative masses, one needs to take care the units used are consistent, and also that Kepler's 3rd law is applied in the correct format. 

Applied Algebra Solutions (2)

We pick up now with the solutions to applications of  algebra to simple machines, applications of the lever principle:

1) a) If F will move 6 times as far as the load w, then the M.A. = 6. So: F = w/6. And w = 40 kg x (9.8 ms^2) = 39.2 N. Then: F = 39.2N/6 = 6.5 N.

The work done is: W = F s = (6.5N) (10m) = 65 N-m = 65J.
(b) The mechanical advantage in this case is increased such that: Fs = 2 wd = w

2) By the principle of the lever, the combined vertical forces must balance a total clockwise torque or moment of 40 lbs. x 6 ft. = 240 lbs.-ft.  The right arm exerts a counterclockwise torque at its position of 3' x F1 = 240 ft-lb. So that: F1 = (240 ft.-lbs.)/ 3 ft. = 80 lbs..

The left arm, meanwhile,  exerts a second counterclockwise moment at its position (6' from the load, at the center of gravity of the plank- since it's uniform) of 6' x F2. Therefore: F2 = (240 ft.-lbs.)/(6 ft.) = 40 lbs. is the upward force exerted by his left hand.

3) Sketch is below:


-------------(4 m)-------------------^----(1m)-------x (mg)
!
\!/
F

The new set up to lift the block with less work done is shown in the diagram. This is done by lengthening the effort distance to 4 m (from 3.5 m) and shortening the load distance to 1 m (from 1.5m). Then the new force exerted needed to lift the block is found from the law of levers: F x 4m = 500N x (1m) = 500 N-m. (The 'notch' represents the pivot point, and the weigh mg is directed downwards at the end point x)
Then: F = 500 N-m/ 4m = 125 N. (Compared to the original force exerted of 214 N). The new work done is: Fs = (1/4) 500N x 1.0 m = 125 J.

4)  Apply the lever principle:

m(A) + m(B) = 3.2 M_s 

And :  m(B) = ½ m(A) 

Since m(B) is one half the mass of m(A) it must be situated at twice the distance from the center of mass, so:  Bx = 2 Ax


Use the above algebraic relations, substituting m(A) = 2 m(B):

m(B) +  2 m(B) =    3.2 M_s  


3m(B) = 3.2  M_s   


Then: m(B) =   3.2 M_s    / 3 m(B)   =  1.06 M_s 

Then:  m(A) =  3.2 M_s     - 1.06 M_s      = 2.14 M_s 



5) Given:   m1 + m2 = a³/P²


We need to find the period if a = 0.16 AU, and m1 + m2 =  1.8 solar masses

Write:

 P²    =    a³/ (m1 + m2)  =   (0.16 )³ /   (m1 + m2) =  (0.00409)/ 1.8

P²    =   0.00227

Taking the square root of both sides:


P =  0.048 yrs.
      

Teaching Algebra II The Right Way: By Applications! (2)

3. Simple Machines

We continue examining more applications to elicit student interest in Algebra II. The great thing about these applications, to do with simple machines, is that most can easily be constructed by the teachers to achieve a hands on effect.  We look first at a simple pulley system:

The pulley shown in 3(a) is a single movable pulley, in contrast to the Atwood machine which is a single fixed pulley. In operating such a pulley, say to lift a weight w, the force applied (F) must move twice as far as the weight w = mg. The mechanical advantage (assuming no friction) is s/d -  which is the displacement (s) of the applied force, how much it moves, divided by the distance (d)  the weight is moved. Since for Fig. 3(a) if the weight w is moved 1 m then the force F is moved 2 m. Thus, s = 2' and d = 1' so: w/F = s/d or F = wd/s = ½ w.

Example: A student sets up a pulley system using a mass of 0.5 kg which moves 0.5m. What is the displacement for an applied force F = 4N? (Take g = 10 ms^-2  )

In this case, the weight w = mg = (0.5 kg) (10 ms^-2  ) = 2 N
One needs to make the displacement s, the subject, from the equation F = wd/s.

Then: s = wd/ F  =   (2N) (0.5m)/ 4N  =   1.0 N-m/ 4N  =  0.25m

In Fig. 4A below, a variant of the earlier pulley system (a bit more complex) is depicted, called the "wheel and axle" (A) and we see also grouped pulleys (C) and multiplied strings(B). The wheel and axle is of particular interest in that it makes use of two different radii, an inner small one, r and a larger outer one R. If the depicted wheel (Fig. 4(A)) moves through one complete revolution, the distance the force will move is just d = 2πr. Meanwhile, the distance the force moves will be s = 2πR. If we take the mechanical advantage: M.A. = s/d = (2πR)/ (2πr) = R/r, then:

mg/ F = w/F = R/r and so: F = (r/R)w which is the law of the wheel and axle.
Example problem: In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.
Solution:

In any verbal algebra problem (I or II) it is essential to identify the unknowns. In this case, we seek the applied force, F and the mechanical advantage, M.A. We also need to identify what we already know: thus load = weight = mg - 80N. R = 23 cm, and r = 1 cm.

Since: M.A. = s/d = (2πR)/ (2πr) = R/r, then:
M.A. = R/ r =  23 cm / 1 cm = 23

The applied force F can then be obtained via the ratio relationship:

w/F = R/r  =  23

Or:  F = w/23   = 80 N/ 23 = 3.47 N


Lastly, we come to perhaps the most famous machine of all, the lever. Archimedes, the ancient Greek physicist and mathematician, is quoted as saying: "Give me a lever long enough and I will move the Earth!". A basic depiction of a workable lever is shown in Figure 5.

The lever principle is applicable to everything from figuring out where two people ought to sit on a teeter totter to achieve balance, to the respective distances of two stars in a double star system, from their mutual center of gravity.

Basically a load L is placed at one end which we wish to lift by applying a force F. Let the load be a distance a from the pivot, and the applied force acts at a distance, b. Then:

Force x distance from axis = load (mg) x distance from axis or:

F x b = L x a or F = (a/b) L = (a/b) mg.

This is called the "law of the lever". It helps to illustrate using a simple problem how it works:

Example problem:

A 50 kg concrete block has to be moved from the ground to a wheelbarrow and a workman is provided with a board 5 m in length. If the workman pivots the block at 3.5 m from one end and lifts from the other (assume g = 10  ms^-2 ) What applied force is needed to lift the block? What is the work done?

We have the effort distance, a = 5.0 m - 3.5 m = 1.5 m, and the load is:

w = mg = 50 Kg (10 ms^-2 ) =  500 N, with load distance a = 3.5m.

Then, since:

F x b = L x a, we have:

F = (a/b) w and F = (1.5 m/ 3.5m) 500 N = 214 N.

The work done is

Fs = (mg)d but (d/s) = (a/b)


 so Fs = (a/b) mg x 1.5 m = 321 J.


Additional problems:

1. a) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 ms^-2, find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done? (Work is defined as the force times the distance moved, i.e. against gravity, or W = F x)

b) Examine the pulley system shown in Fig. 3(b). How would the basic applied equation for force and mechanical advantage be changed compared to the pulley shown in Fig. 3(a)? (Given that:  F = wd/s = ½ w  for the single moveable pulley in 3(a))


2. A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.

3. In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.

4. The "line of centers" for a binary star orbit, in reference to the center of mass x, is shown below:

A O------------x cm-------------------------o B

The sum of the two stellar masses is given by:

m(A) + m(B) = 3.2 M_s

where M_s   denotes solar mass units.  If mass  m(B) = ½ m(A)  then find the distances Ax and Bx. Thence also find the masses of m(A) and m(B) in terms of solar masses (M_s ).


5. The sum of masses for a binary star system can also be obtained using Kepler's 3rd law:

m1 + m2 = a³/P²


Using this, where a = the separation of the stars in astronomical units (AU) , and P is the period for their revolution about the center of mass in years, find P if:

m1 + m2 =  1.8 solar masses and a = 0.16 AU.