Solutions To Binary Star Problems

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

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

(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.

(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   and  Star B: 2π/3 AU/yr

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

(b) Find the mass of each star in solar units. 

(Assume the eclipses are central)

Solutions:

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

Then:  m1+ m2 = (a)³/ P² and

P = [(a³)/ (2)]^½

(a) IF  a = 1 AU

P = [1/2]^½ = 0.707 yr.

(b)IF  a = 6AU

P = [(6)³ / 2]^½ = 10.3 yr.

(c) IF a = 100 AU

P = [(100)³/2]^½ = 707.1 yrs.

(2)We need to determine the distances for the two stars in each system of (1) to appear to have an angular separation of 1". 

We require:

m1 + m2 = (d a")³/ P²

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

Simplifying:

(a) P = 0.707 yr.

(d a") = [(m1 + m2) P²]^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 = 100 pc

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

First, find the distance d from the parallax method:

d = 1/p"  =  1/ 0."127 = 7.87 pc

Apply Kepler’s 3^rd law for binaries separated by a”.

m(A) + m(B) = (d a")³/ P² 

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

Then:

m(A) + m(B) = ((7.87) (2½" )]³/ 60² = 2.1 solar masses 

(4)The graphs of the radial velocity curves for the stars, matching the physical situation of each, are given in the diagram below. 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 maximum 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 =  4π/ 2π  =  2: 1

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

Recall that 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

Use Of Basic Celestial Mechanics To Obtain Some Earth Orbital Parameters

In this post, basic celestial mechanics will be used to obtain some basic information on Earth's orbital dynamics, including: velocity at aphelion and perihelion points, magnitude of energy given by energy constant a,  eccentricity e, specific relative angular momentum h, and derivation of the "vis viva" equation which enables one to obtain the velocity at any point in the orbit. (The latter is also known as the "orbital energy invariance law" because the total energy is the same despite how the velocity v varies over the orbit. )

To fix ideas we focus on the diagram shown above. Let the points A and P denote the aphelion (farthest point) and perihelion (closest point) to the Sun (S), respectively. We let   V_A, V_P  be the respective velocities at those orbital extrema. As may be deduced here, points A and P are the only ones in the whole orbit for which the velocities are truly tangential or at right angles to the radius vectors for those positions. Consequently, we can write:

V = (2π/T) r

where r is the radius vector at the point, and T is the period.

 If Kepler's 2nd (equal areas)  law holds at every point (i.e. equal areas swept out in equal intervals of time) we also have:

r² (2π/T) = h 

where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when A1 = π(r1)². Hence, at aphelion and perihelion only we have: 

V = h/r    for which r = a(1   +  e)  OR:   r =  a (1  -  e)

For the perihelion velocity we have:

V_P = h/ a(1 - e)

where a is the semi-major axis, and e is the eccentricity.

For the velocity at aphelion:

V_A = h/ a (1 + e)

Then the ratio of velocities is:

(V_P/V_A) = (1 + e)/ (1 - e)

The correct energy ("vis viva")  equation can be written:

½V² - m /r = a

where a is an energy integration constant.   More conventionally it is written
 without  the energy constant (see derivation at end of post):

V² = m   (2/r - 1/a )


Energy constants in celestial mechanics are very useful for quickly coming to terms with specific properties of an orbit such as shown in the  more detailed accompanying sketch- designating a generic orbit in x-y-z space, e.g.. 


In the diagram, w   is the argument of the perihelion, W is the longitude of the ascending node , f is the true anomaly and i is the inclination of the orbit. The critical or key parameter here is h, the angular momentum vector for the orbiting system.

Getting specific, assuming r and r' are r (radius vector) and d r/dt, respectively, the magnitude h, of the angular momentum vector is:

h = r x r’ =

(y z’ - z y’)

(z x’ - x z’) = (c1 c2 c3)

(x y’ - y z’)

so:   (r x r’) = (c1/ h, c2/ h, c3/h)

Where c1, c2 and c3  are integration constants that determine the orientation of the orbital plane.

Inserting angular orbital elements (i, W) one finds:

c1/ h = sin W sin (i)

c2/ h = - cos W sin (i)

c3/h = cos(i)

Now since the inclination of Earth's orbit to the ecliptic  (i) is known (23.5 deg) and therefore cos(i) can be determined, then sin(i) can be as well.  Also, h can be determined, since: h = c3 / cos(i) .  (Also h =  [c1 ² +  c2 ²   +  c3 ²]  ^½)   We also know  W =  11.26 deg.

Since for any bound system of masses m1 and m2, m = G (m1 + m2), where G is the Newtonian gravitational constant (G = 6.7 x 10^-11 Nm²/kg²) then if we know V_P and V_A, along with a and e, we can compute a, viz.

a = ½V_P² - m /a(1 - e)

at perihelion, and

a = ½V_A² - m/a(1 - e)

at aphelion. 

For the Earth-Sun system :

m= 1.33 x 10²⁰ Nm²/kg

(Note: for m, we already know G and m1= 1.99 x 10³⁰ kg (Sun's mass) and m2 = 6.4 x 10²⁴ kg, (Earth's mass)

Also: a (semi-major axis)  = 1.496 x 10¹¹ m

Then h = + [m a(1 - e²)]^½ =    4.46 x 10¹⁵ N-m/kg = 4.46 x 10¹⁵ J/kg 

The energy constant a =   - m/ 2a   for an elliptical orbit

So:   

a =    -(1.33 x 10²⁰ Nm²/kg) / 2 (1.496 x 10¹¹ m)  =  = -4.45 x 10⁸ m²/s² 

The eccentricity of the orbit e, can now be obtained from:

e =   [1   +   (2 h ²  a )/ m ² ] ^½ =


[1   +    2(4.46 x 10¹⁵ J/kg ) ² (-4.45 x 10⁸ m²/s²)/ (1.33 x 10²⁰ Nm²/kg)² ] ^½ =

0.016

What about the velocities at perihelion and aphelion?

Since we have obtained h and e,    the velocity at perihelion is easy to calculate

from:

V_P = h/ a(1- e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 - 0.016)]

V_P = 3.03 x 10⁴ m/s  = 30, 300 m/s

and the velocity at aphelion can be obtained using:

V_A = h/ a(1 + e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 + 0.016)]

V_A = 2.93 x 10⁴ m/s    = 29, 300 m/s

 Now, how would the vis viva equation (given earlier) be derived? 

From the earlier energy constant equations (at aphelion, perihelion):

a = ½V_A² - m /a(1 +  e)

at aphelion. 

a = ½V_P² - m /a(1 - e)

at perihelion.

Then,   we may write without  loss of generality:

 ½V² - m/r   = a  =   - m/ 2a

Or:

½V² =   m/r    - m/ 2a

And:   

V² =   2 [  m/r    - m/ 2a ]    

Whence:

V² = m   (2/r - 1/a)

Comprehension Problems:

1) Show that the energy constant  a   is the same at aphelion and perihelion.

2)  The Earth's aphelion distance is 1.01671 AU and its perihelion distance is 0.98329 AU. Use the vis viva equation to obtain the difference in velocity between the two points.

3) Calculate the three integration constants applicable to the orientation of the Earth's orbital plane.: c1, c2 and c3,    In standard practice these are already computed, then used to obtain the longitude of the ascending node, W   and the inclination, i.  Show how this could be done.

4) Derive the independent expression for h of the form:

h = + [m a(1 - e²)]^½

And show it is equal to:  h =  [c1 ² +  c2 ²   +  c3 ²]  ^½

4) The orbital period of Jupiter's 5th satellite is 0.4982 days about the planet. Its orbital semi-major axis is 0.001207 AU. The orbital period and semi-major axis of Jupiter are 11.86 yrs. and 5.203 AU. Estimate the ratio of the mass of Jupiter to that of the Sun.

5)  For the Pluto-Charon system,  the orbit of Pluto's moon Charon has an eccentricity e = 0.0020. The semi-major axis of the orbit is 19, 450 km. The mass of Pluto = 1.27 x 10²² kg and the mass ratio (Charon to Pluto) is found to be m(c)/m(P) = 0.12. From this information, find:

a) The mass of Charon

b) The ratio of the velocity of Charon at perihelion to aphelion

c) The period of Charon, and its velocity

Selected Questions - Answers From All Experts Astronomy Forum (Ratio of Planets' Orbital velocities)

Question: How would a person calculate the ratio of the orbital velocities of two planets, say Earth and Venus?

Answer:    This is a relatively straightforward astronomical computation, using the ratio relation:

V2/V1 = (a2/a1) (T1/T2)

Where a2, and a1 are the respective semi-major axes of the orbits (i.e. the mean distances from the Sun) and T2, T1 are the respective periods.

By convention we assign '1' to the inner planet (e.g. Venus) and '2' to the outer (Earth).

Then we have a2 = 1 AU,  and for Venus  we need to obtain T1 from Kepler's third law:

(T1/ T2) 2      = k(a1/ a2) ³ 

Now, obtaining Venus' period in days (224.69) from planetary data, we can convert it to years, e.g.

T1 = (224.69/365.25) yr. = 0.615 yr.

And Venus' semi-major axis is:
 
a1 = {[T1] ² }1/3   = [(0.615)²] 1/3 


a1 = 0.723 AU

Therefore:

V2/V1 = (1/ 0.723)(0.615/ 1)


V2/V1 = 0.615/ 0.723   = 0.851

Or, in terms of the ratio of Earth's orbital speed to Venus':   1/ 0.851  =1.175

This can be checked  using a Table of Orbital Velocities found in Astrometric & Geodetic Data, from which one finds:

V(Venus) = 35.02 km/s

V(Earth) = 29.78 km/s

Take the ratio:

V(Venus)/V(earth) = (35.02 km/s)/ (29.78 km/s) = 1.175

So, Venus' orbital velocity is 1.175 times Earth's

Looking At Analytic Geometry (2): The Ellipse

The ellipse is a geometric figure - curve we examined before in the context of real planet orbits. For example, in the diagram shown above, a planet at point P(x,y) in its orbit is referenced to the two foci of the ellipse at f1 and f2. As the point P moves along the total connecting distance f1-P-f2 remains constant. By inspection what we call the semi-major axis is a = 4 and the semi-minor axis is approximately 2.8.

If we know the general form of the equation for an ellipse, e.g.

x² / a ² +    y²  / b ² = 1

We could make a fair guess at what the actual equation of the orbit shown would be, say taking

b =   Ö 8  =   2.82

The ellipse equation would then be:

x² /16 +    y²   / 8 = 1

And the graph would be plotted as shown below:

Which is a solid match for the earlier sketch orbit.

What is c, mathematically?

It is the square root of the difference between the semi-major axis squared, and the semi-minor axis squared:

c =   Ö (a ²  -   b ²)

In this case, we'd get:    Ö (16  -  8) =   Ö 8  =   2.82

That means the coordinates of the foci are f1 (- Ö 8, 0)  and f2 (0,  - Ö 8)

Note that when c = 0 we have a = b. (One constant fixed radius: e.g. r = a = b)

This means the figure is a perfect circle. For a circle the eccentricity e = 0. For our own example here: e =  Ö 8  / 4   =   2.82/ 4  = 0.705

When, however, a > b then the shape alters to elliptical, but also when b > a.

If we juxtapose the values for a, b in the general equation we get:

x² / 8  +    y²   / 16 = 1

And the ellipse orientation changes to one with major axis now in the vertical direction:

From the preceding it is possible to sketch the graph of a given ellipse, as well as give its eccentricity e (= c/a) from just the coordinates of the eclipse center,. the focus F (at least one) and the semi-major axis a.

Example problem: Sketch the graph and find e for the ellipse with center coordinates C(0,0), focus coordinate F(0, 2) and a = 4

Solution: Given the coordinate F(0,2) we must have a > b   and hence, c = 2. Then the semi-minor axis is found from:

 b =   Ö (a ²  -   c ²)    =   Ö (16  -   4 )    =  Ö 12

The eccentricity e = c / a    = 2/ 4   = 0.5

The equation for this ellipse will be:

x² /16 +    y²   / 12 = 1

As in the case of the circle, ellipses aren't always centered at (0,0) or the origin of an x-y grid. Consider the equation below:

9 x²  +   4  y²   + 36 x   - 8 y + 4 = 0

To get this into a form that can easily be inspected and graphed, first re-arrange the terms to get:

9(x²  +   4 x)  +  4 (y²   - 2 y)  =  - 4

Complete the square for each term in parenthesis:

9 (x²  +   4 x  +  4)  +  4 (y²   - 2 y   + 1)  =  - 4   + 4   + 36

Or:

9 (x²  +   4 x  +  4)  +  4 (y²   - 2 y   + 1)  =  36

Divide through by 36:

(x + 2) ² / 4 +    (y - 1 ) ²   / 9 = 1

To simplify further, let x' = (x + 2) and y' = ( y - 1)

Then write:

x' ² / 4 +    y ' ²   / 9 = 1

Which yields the ellipse shown below once the x, y- coordinate offsets are taken into account:



We also find: 

c =   Ö (a ²  -   b ²)     where in this case,  a ²  =  9   and   b ²  =   4

So:  c =   Ö (9   -   4 )    =   Ö 5

So the foci are situated at the points (0,  + Ö 5 ) in the new coordinate system or at:

(-2,  1 +   Ö 5) in the original. Which values can be checked from the graph.

Problems:

1) In the last example problem worked above, we saw that the ellipse equation of form:

(x + 2) ² / 4 +    (y - 1 ) ²   / 9 = 1

Could be simplified to: x' ² / 4 +    y ' ²   / 9 = 1

Using:   x' = (x + 2) and y' = ( y - 1)

Show how the offset affects the location of the simplified ellipse form by graphing both ellipses on the same axes. Thus, satisfy yourself that the graph shown is justified once the x, y- coordinate offsets are taken into account.

2) An ellipse has center coordinates at C(-3,0) and focus at F(-3, -2) with a = 4. Find the eccentricity e and sketch the graph.

3) An ellipse has center coordinates at C(2, 2) and focus at F(-1, 2) with a = Ö 10. Find the eccentricity e and sketch the graph.

4) Sketch the following ellipses from their equations below, and give the semi-major axes, semi-minor axis and eccentricity e.

a) 4 x²  +   9  y²   = 144

b) 4 x²  +    y²   =    1

5) Look up the definition of the "directrix" of an ellipse. Then find the eccentricity and directrices of the ellipse:

x² /7 +    y²   / 16 = 1

And sketch the graph showing the locations of the foci.