Detecting Invisible Objects:
a guide to the discovery of Extrasolar Planets and Black Holes

    general theory: how to determine the mass of an invisible object in a "binary" system
   
    a worked example:
    determination of the mass, orbital radius, temperature, and composition of the planet accompanying the star  HD 209458

    black hole data for determining the mass of an invisible black hole in a binary system
   

general theory: how to determine the mass of an invisible object in a "binary" system


    Single-line spectroscopic binaries are special cases of binary stars  in which
    only one star is visible, yet the binary nature of the system is apparent because
    the single star that is visible shows periodic doppler variation in its spectral lines.
    For simplicity, let's
assume that

    1) the orbits are circular, in which case the radial velocity curves (determined from the
        Doppler effect) have a sinusoidal shape.  The spectroscopic binary applet will help
        give you a visual picture of how the shape of the radial velocity variations is
        determined by  the shape of the orbits, the size of the orbits, the masses of the
        orbiting objects in the system, and the orientation of the orbits relative to Earth.
        (Once the applet opens, set the parameters to the following values:
             e = 0      which means the orbits are circular
            i  =  90    which means that the orbits are in the plane of Earth's line of sight
                            (see below),
             and then click on Enter button to update the changes that you have made.

      

    2) that the plane of the two orbits is inclined to the plane of the sky by an inclination
        angle of 90 degrees. 
[This means that the plane of the orbits is
        inclined by (90 - i) to the earth's line of sight to the system.]
       
In general, the actual value of this inclination angle  unobservable and unknown.  

{diagram showing the angle i goes here}

    The two equations that describe the physics of binary systems are

        1) the center of mass condition:

(1)
    Mi ai  =  Mv a

            where
                a = radius of the circular orbit
                M  =  mass of the orbiting object
                and the subscripts  i  and  v  stand for invisible and visible object, respectively.

        2) Kepler's 3rd law defines the relation between the common orbital period P of the
             two objects around the center of mass, the orbit sizes (
ai  and av ),
            and the masses of the two objects ( Mi  and Mv )

            (From here on, special units will be used:  solar masses for M; years for P;
            astronomical units for a .  In these units, 4p2/G = 1):

(2)
    (Mi +  Mv) P2  =  (a + av)3

    For circular orbits, the orbital speed of each object is constant throughout
    the orbit and given
by

(3)
    v  =  2 p a /P

    The only things that can be determined from observations of a single-line spectroscopic
    binary are

        a) the orbital period  P  of the radial velocity of the visible object
            (this is actually the orbit period of both the invisible and visible object)

        b) the orbital speed   vv sin i   of the visible object
            (the amplitude of the radial velocity variation, determined from the Doppler
            effect formula)
        c) the mass Mv of the visible object (inferred from its spectral type)

    Therefore, any information derived about the invisible object must use only these
    3 determinable
quantities.  Combining equations (1) - (3) above gives:

(4)
    Mi3 sin3i / (Mi + Mv)=   P (vv sin i)3/(2 p)3

    [hint: first combine equations (1) and (2) to eliminate  a :  i.e., solve equation (1) for ai and
    use the result to replace 
ai  in equation (2); then use equation (3) with your new equation
    to eliminate
av :  i.e., solve for av  in equation (3) and use the result to to replace  av  with 
    the variable
P and vv  ]

    equation (4) is useful because only quantities that are able to be determined from observations
   
are present on the right-hand side of this equation.

Extrasolar Planets

determining the planet's mass

    Let's use equation (4) along with the observed properties of a planet's parent star
    to determine the mass and the orbit size of the invisible planet.

    The number of detected extrasolar planets now exceeds 130;
   a complete list can be found at either

 Extrasolar Planets Catalog or Known Planetary systems   

    the quantity on the left-hand side of equation (4) is called the mass function; it doesn't seem,
    at first glance, to be particularly useful, since it is a function of 3 things (1 known: Mv and
    2 unknowns: Mi  and  i ) and involves an inequality rather than an equality. 
    It turns out, however, that a useful lower limit to Mi can be found.

    (remember that the period should be in years; the orbit speed, in au/yr; and the masses,
    in solar units; a table of stellar
masses according to spectral type will be needed)

    The star HD 209458 was the first to have its planet detected both by spectroscopic
    and photometric methods.  The radial velocity of the star varies with time over a regular
    period of 3.52 days.



The observed quantities for this system are

star with 
extrasolar planet



 star's radial velocity amplitude



period of radial velocity variation

star's absolute magnitude 
(to get the star's luminosity 
relative to the sun, use
Lstar/Lsun   2.512(4.7 - M) 
don't forget to convert to actual luminosity of the sun 
in watts by multiplying by Lsun
M
( Lstar/Lsun)
star's spectral class

and 
mass 
(solar units)
 

SC
 
( M/Msun)

HD209458
86.5 m/s
=
.0182 au/yr
3.52 days
=
.00965 yr
4.6
G0 V

(1.05)

    Entering the observed quantities for the symbols on the right side of equation (4)
    results in a value of the mass function MF of

MF  =  2.4 x 10-10  (solar masses is the unit, assuming you used the units above)

 
    Therefore,
(5)

MF  =     Mi3 sin3i / (Mi + Mv)=   2.4 x 10-10  Msun

    Because   sin i  <  1,

(6)

Mi3 / (Mi + Mv)>    2.4 x 10-10   Msun

    Furthermore, we know the mass Mv of the visible star (from the observed spectral class
    of the star; it is 1.05 solar masses.   The
previous equation then becomes

(7)

 Mi3 / (Mi + 1.05 Msun)>   2.4  x 10-10 Msun

    We now have an equation in a single unknown; although it cannot be solved analytically,
    it can be easily solved by trial and error (guessing values) or by using a graphing calculator.
    Can you find the solution to this inequality?

(answer:  approximately  M > 0.00064 Msun  or   0.67 MJupiter)


determining the planet's orbital radius

    Once an upper limit to the planet's mass is known, its orbit radius (or distance from
    its parent star) can be found from equation (2) above.  The planet's mass is very much
    smaller than its parent star's mass; therefore, the 
Mi  term on the left-hand side can be
    ignored.  Similarly [because of the center of mass condition, equation (1)], the star's
    orbit size around the system center of mass is much smaller than the planet's orbit size. 
    Therefore, the
av term on the right-hand side can be ignored, and

(8)
    Mv P2  =  (ai)3

     Using the values of  Mv  and  P above, we find  ai   =  0.046 au.  This is about 9x smaller
      than Mercury's orbit about the sun.


determining the planet's temperature

    Next, let's calculate the equilibrium blackbody temperature of a planet.   We assume that
    thermal equilibirium (i.e., constant temperature) applies, and consequently that the power
    ( = energy/time)
emitted by the planet is the power absorbed from its parent star:

(9)

    Pabsorbed  =   Pemitted

    The left hand side is found from geometry, corrected by a coefficient that takes into account
    reflected light; the right hand side is given by the Stefan-Boltzmann law:

(10)

    Lstar  (1  -  A)  (p Rp/4 p dp)2  =   4 p Rp2s Tp4

     Lstar  =  luminosity (power) of the parent star
    A  =  planet's albedo  =   (light reflected)/(light incidnet)
    Rp  =  planet's radius
    Tp =  planet's temperature
    dp =  distance of planet from parent star
    =   Stefan-Boltzmann constant

    Solving for T gives

(11)

Tp4   =   Lstar(1 - A)/(16 p s dp2)

    Notice that the equilibrium temperature depends on the "guessed"  albedo of the planet; the
    ratio of the temperature derived with albedo = 0.95 to the temperature derived with an
    albedo of 0.05 is approximately 2.  Albedos of planets in our solar system are listed in the
    table below.  The lowest albedo is around 0.05 (Earth's moon); the highest, around 0.7 (Venus).

    This calculation doesn't take into account thermal energy released from the planet's interior,
    tidal energy released via a star-planet interaction, the greenhouse effect in the atmosphere,
    etc.  It would be a good idea to have students calculate the equilibrium temperature of the
    solar system's planets first to have students have some sense of how close this calculation
    comes to actual temperatures.

    data for solar system objects:


planet
orbit size d
(a.u.)
A
(albedo)
calculated  Tp
(Kelvin)
actual  Tp
(Kelvin)
Mercury
 0.387
0.1
440 
100-620 
Venus
0.723 
0.7 
250 
750 
Earth
1.0
0.4 
250 
290 (equatorial) 
Moon
1.0
0.05 
270 
90-400 
Mars
1,52 
0,2 
220 
130-290 
Asteroid Ceres
2.77 
0.1 
160 
 
Jupiter
5.22 
0.5 
100 
160 (cloudtops) 
Saturn
9.55 
0.5 
75 
95 (cloudtops) 
Uranus
19.9 
0.5 
53 
55 (cloudtops) 
Neptune
30.1 
0.5 
43 
55 (cloudtops) 
Pluto
39.5 
0.5 
37 
50 (cloudtops) 

 
   determining the planet's composition

    with the termperature, the physical status (solid, liquid, or gas) of various chemicals in the
    planet's make-up can be determined... the types of materials that make up (typical solar
    system) planets include
 
 
material
condensation
(gas, liquid  --> solid)
temperature
(at ~ one atmosphere)

uncompressed
density
(g/cm3)
relative abundance
in solar nebula
Ca, Ti, Al silicates 1400 - 1800 K 3
.00001
Fe/Ni
1300 - 1500 K 7
.00001
Mg silicates
1300 K
3

other metalic silicates & sulfides
900 - 1200 K 3

hydrous silicates
500-600 K
2

ices
(water, ammonia, methane)
150 - 300 K
1
.0001
hydrogen, helium < 20 K
1

    Warning:  the make-up of a planet may more likely represent the composition (and physical
    state) at birth rather than at present.  In general, stellar luminosities were higher during their
    pre-main sequence phase (when the planets formed) than at present (when the star is in the
    main sequence phase).

    Unfortunately, the size of the planet can be determined (unless the planet can actually be
    imaged or until it eclipses or is eclipsed by another object), no further information
    (gravitational field, escape velocity, atmospheric retention,  density) can be determined. 

    However,  the planet around HD209458 happens to be one of those that transits its star.
    The transit data is here.  From the transit data, it can be determined that the radius of this
    planet is 1.27 RJupiter.  Can you figure out 2 different methods of determining the planet's
    radius from the light-curve data given in the link?  Because both the mass and radius of
    the planet, the average density of this planet can be determined:  0.41
g/cm3 .  What type of
    substances in the table above match a density such as this?
 

Black Holes

The number of strong black hole candidates of stellar-mass size is still small;
the observed properties of some of the black hole's companion are listed below


black hole candidate
(click on link for
radial velocity curve)

CM radial velocity

velocity amplitude
period of
radial velocity curve
visible star information
orbit inclination
(degrees)
derived minimum mass of the
invisible black hole
Cygnus X-1,
the first candidate




72 km/s
5.6 days
O9/B0 supergiant 
(M/Msun  =   33)
(see  Universe, section 24-5)



7.0  Msun
V404 Cygni


-0.4 km/s

211 km/s

6.473 days
late G or early K
(K0IV?)

55 + 4
6.3  Msun
A0620-00
(V616 Mon)

10 km/s


433 km/s
7.75 hours K3.5 V
37 + 5
3.2  Msun

QZ Vul
GS2000+25


19 km/s

518 km/s
8.3 hours
K3 - K5 V
(M/Msun  =   0.55)

10 + 4
5.9 Msun
Nova Scorpii 1994
(GRO J1655-40)

-142 km/s

216 km/s

2.92 days
(or 2.4 or 2.62 days)
G0 V or F6 IV
(M/Msun  =   1.1)

67 + 3
5.5 Msun

GU Muscae
(Nova Mus 1991)
GS1124-683


16 km/s


409 km/s

10.4 hours
K0 - K4 V
(M/Msun  =   0.70)

54 + 20
4.2 Msun
4U1543-47
-87 km/s
124 km/s
1.12 days
A2V
21 + 2

Nova Ophiuchi 1977
H1705-250

-54 km/s

420 km/s

12.51 hours

K5V

60 - 80

Nova Velorum 1993

475 km/s
0.285 days
K6 V - M0 V

3.6 - 4.7 Msun
GRS1915+105

140 km/s
33.5 days
K - M III
(M/Msun  =   1.2 )


14 Msun

XTE  1650-500





3.8 Msun
lowest mass;
found by quasi periodic oscillations

for another list of black hole candidates, go here


    How do we know that these objects are black holes and not something else like planets, or
    neutron stars, or white dwarfs, or brown dwarfs, or just some type of star?  Notice that all the
    black holes candidates are at least 3 solar masses... the upper limit to neutron star masses is
    approximately 3 solar masses; the upper limit to white dwarf masses is 1.4 solar masses; the
    upper limit to brown dwarf masses is 0.8 solar masses; the upper limit to planetary masses is
    even smaller.... therefore each of these candidates is ruled out as a possible companion to
    these stars...

    A different argument is required to rule out another star as the possible "invisible"
    companion.  Here is a table of stellar mass objects arranged by temperature and
    brightness.
    Notice that all of the objects in the stellar table that have mass more than this minimum
    3 solar masses are bright (in most cases, in fact, brighter than the visible star in the
    system) and therefore should be visible
-- unless it is a black hole .