ALL 13 CHAPTERS COVERED
SOLUTIONS MANUAL
,Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
Problem solutions – Chapter 1
Kepler, Newton, & the mass function
Problem 1.21. External force on binary; collisions in globular cluster
(a) Consider external forces on a binary
Assume wide binary, rs = 10 AU = 1.5 ⫻ 101 2 m. Find ratio of external force difference on the
two partners to the force exerted by the binary partners on each other. Each star has mass Ms = 1
M.
(i) Effect of galactic center. Mg ⬇ 106 M, point source at rg = 25 000 LY = 2.4 ⫻ 102 0 m. The
disruptive effect is due to the difference of the external force on the two stars in the binary. It will be
maximum when the two stars are aligned with the galactic center.
First consider the ratio of the two forces on one of the two stars,
Fg Mg Ms –1 Mg rs2 (1.21.s1)
= =
Fs rg2 rs2 Ms rg2
The radial component of Fg is krg – 2 and its gradient is dFg /dr = –2kr– 3. The ratio of the latter to the
former is
dFg /dr
= – 2 rg– 1
Fg
or
dFg (1.21.s2)
= – 2 dr
r = –2 rs
Fg g rg
where dr = rs when all three objects are coaligned.
Multiply (s1) and (s2) to find the desired ratio,
dFg Mg rs3 12 3 (1.21.s3)
= –2 = 106 1.5 ⫻10 = 4.9 ⫻ 10–19
Fs Ms r g
3 2.4 ⫻ 10 2 0
The difference force is negligible compared to that holding the binary together.
(ii) Effect of nearby star in spherical Globular cluster of N = 106 stars and radius Rglob = 15 LY.
= 1.42 ⫻ 101 7 m.
Average distance d between adjacent stars is (volume per star)1/3,
4 1/3 (1.21.s4)
d = Vs1/3 = Rglob = 2.29 ⫻ 101 5 m = 15 000 AU
3N
1–1
,Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
The force ratio is thus, from (s3),
Fn-s 12 3
= M 1.5 ⫻ 10 = 5.6 ⫻ 10–10
Fs M 2.3 ⫻101 5
Again, the external difference force is negligible for our assumption of uniform star density. But, in
the more dense center of a globular cluster, occasional close encounters can occur that could disrupt
the binary.
(b) Time for collision, within 10 AU, to occur in globular cluster, on average.
Find typical speed of star in globular cluster of radius Rglob = 1.4 ⫻ 1017 m and mass Mglob= 2
⫻ 103 6 kg. From virial theorem (2.14),
GMglob m
⬇ mv2
Rglob
GMglob
v⬇ = 3.0 ⫻ 104 m/s
Rglob
Cross section for collision, = (10 AU)2 = 7.0 ⫻ 1024 m2 .
The spatial density of stars, from (s4), is n = d–3 = 8.3 ⫻ 10–47 m– 3.
Now the flux of stars is nv, and the number of collisions per second with a single binary is nv.
Hence the time between collisions is the inverse of this, t = (nv )– 1 = 5.7 ⫻ 1016 s = 1.8 ⫻ 109 yr,
which is ~1/10 the age of the globular cluster.
This is the typical time for a given star (or binary) to suffer such a collision. Since there are 106
stars, one expects an encounter every 2 ⫻ 103 years, for our assumptions. The increased density at
the center would markedly decrease the collision time.
– – –
Problem 1.22 – Finding and observing binary star systems (no solution)
– – –
Problem 1.23 – Find distance range where binary system is detectable as
both a visual and a spectroscopic binary.
Two 1-M stars in circular binary orbit with i = 90° at distance D. Star separation s. Each star
orbits the barycenter with radius s/2 with period P. Let M be the system mass, M = 2M.
Spectroscopy
Require line shift during the course of an entire orbit be at least (⌬/)min = 3 ⫻ 10– 5. From
classical Doppler shift (2),
⌬ = 2 v = 2 2(s/2) (1.23.s1)
c c P
1–2
, Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
From Kepler III (76), GMP2 = 42 s3 . Solve for the period, P = 2 (GM)–1/2 s3/2. Substitute into
(s1), to find
GM 1/2 –1/2
3 ⫻ 10– 5 < s
c
Solve for s,
5 2 (1.23.s2)
s < 10 GM = 3.3 ⫻ 101 2 m = 22 AU
3 c2
Separation s must be small so velocities are high enough to get the required ⌬/.
Imaging
Require the angular separation of the images at greatest separation to be ⌬ 1.45 ⫻ 10– 5 rad
(=3⬙). Since ⌬ = s/D,
s > 1.45 ⫻ 10– 5 D (1.23.s3)
The separation s must be large so the images are resolved. The requirement is most severe at
large D. At the largest separation allowed by spectroscopy (s2), the maximum distance for imaging
is, equating (s1) and (s2),
3.3 ⫻ 101 2 = 1.45 ⫻ 10– 5 D
D = 2.2 ⫻ 101 7 m = 23 LY
Since the nearest stars are about 4 LY distant, the range where it might be observed as both a visual
and spectroscopic binary is only 4 to 23 LY. The latter distance is attained only if the separation is
at the maximum value allowed by spectroscopy.
Optical interferometry with resolution 10– 3 arc second (1 mas), increases this distance to 23 000
LY, but since interferometry requires bright stars, the distance is much more limited. See Armstrong
et al. AJ 104, 2217 (1992) for an example ( Cygni).
– – –
Problem 1.24– Doppler shifts – 1st and 2nd order.
(a) What is range of fractional frequency shift around orbit, for Fig. 6?
Data: Circular orbit, observer in plane of orbit.
Star 1: Max. frequency is when m1 approaches observer, at time t1 . Minimum frequency is at
t3 . From (2)
max – 0 v t – v t
= – r c1 ; min 0 = – r c3
0 0
Thus,
max – min v t –v t
0 = – r 1 c r 3
1–3
SOLUTIONS MANUAL
,Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
Problem solutions – Chapter 1
Kepler, Newton, & the mass function
Problem 1.21. External force on binary; collisions in globular cluster
(a) Consider external forces on a binary
Assume wide binary, rs = 10 AU = 1.5 ⫻ 101 2 m. Find ratio of external force difference on the
two partners to the force exerted by the binary partners on each other. Each star has mass Ms = 1
M.
(i) Effect of galactic center. Mg ⬇ 106 M, point source at rg = 25 000 LY = 2.4 ⫻ 102 0 m. The
disruptive effect is due to the difference of the external force on the two stars in the binary. It will be
maximum when the two stars are aligned with the galactic center.
First consider the ratio of the two forces on one of the two stars,
Fg Mg Ms –1 Mg rs2 (1.21.s1)
= =
Fs rg2 rs2 Ms rg2
The radial component of Fg is krg – 2 and its gradient is dFg /dr = –2kr– 3. The ratio of the latter to the
former is
dFg /dr
= – 2 rg– 1
Fg
or
dFg (1.21.s2)
= – 2 dr
r = –2 rs
Fg g rg
where dr = rs when all three objects are coaligned.
Multiply (s1) and (s2) to find the desired ratio,
dFg Mg rs3 12 3 (1.21.s3)
= –2 = 106 1.5 ⫻10 = 4.9 ⫻ 10–19
Fs Ms r g
3 2.4 ⫻ 10 2 0
The difference force is negligible compared to that holding the binary together.
(ii) Effect of nearby star in spherical Globular cluster of N = 106 stars and radius Rglob = 15 LY.
= 1.42 ⫻ 101 7 m.
Average distance d between adjacent stars is (volume per star)1/3,
4 1/3 (1.21.s4)
d = Vs1/3 = Rglob = 2.29 ⫻ 101 5 m = 15 000 AU
3N
1–1
,Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
The force ratio is thus, from (s3),
Fn-s 12 3
= M 1.5 ⫻ 10 = 5.6 ⫻ 10–10
Fs M 2.3 ⫻101 5
Again, the external difference force is negligible for our assumption of uniform star density. But, in
the more dense center of a globular cluster, occasional close encounters can occur that could disrupt
the binary.
(b) Time for collision, within 10 AU, to occur in globular cluster, on average.
Find typical speed of star in globular cluster of radius Rglob = 1.4 ⫻ 1017 m and mass Mglob= 2
⫻ 103 6 kg. From virial theorem (2.14),
GMglob m
⬇ mv2
Rglob
GMglob
v⬇ = 3.0 ⫻ 104 m/s
Rglob
Cross section for collision, = (10 AU)2 = 7.0 ⫻ 1024 m2 .
The spatial density of stars, from (s4), is n = d–3 = 8.3 ⫻ 10–47 m– 3.
Now the flux of stars is nv, and the number of collisions per second with a single binary is nv.
Hence the time between collisions is the inverse of this, t = (nv )– 1 = 5.7 ⫻ 1016 s = 1.8 ⫻ 109 yr,
which is ~1/10 the age of the globular cluster.
This is the typical time for a given star (or binary) to suffer such a collision. Since there are 106
stars, one expects an encounter every 2 ⫻ 103 years, for our assumptions. The increased density at
the center would markedly decrease the collision time.
– – –
Problem 1.22 – Finding and observing binary star systems (no solution)
– – –
Problem 1.23 – Find distance range where binary system is detectable as
both a visual and a spectroscopic binary.
Two 1-M stars in circular binary orbit with i = 90° at distance D. Star separation s. Each star
orbits the barycenter with radius s/2 with period P. Let M be the system mass, M = 2M.
Spectroscopy
Require line shift during the course of an entire orbit be at least (⌬/)min = 3 ⫻ 10– 5. From
classical Doppler shift (2),
⌬ = 2 v = 2 2(s/2) (1.23.s1)
c c P
1–2
, Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
From Kepler III (76), GMP2 = 42 s3 . Solve for the period, P = 2 (GM)–1/2 s3/2. Substitute into
(s1), to find
GM 1/2 –1/2
3 ⫻ 10– 5 < s
c
Solve for s,
5 2 (1.23.s2)
s < 10 GM = 3.3 ⫻ 101 2 m = 22 AU
3 c2
Separation s must be small so velocities are high enough to get the required ⌬/.
Imaging
Require the angular separation of the images at greatest separation to be ⌬ 1.45 ⫻ 10– 5 rad
(=3⬙). Since ⌬ = s/D,
s > 1.45 ⫻ 10– 5 D (1.23.s3)
The separation s must be large so the images are resolved. The requirement is most severe at
large D. At the largest separation allowed by spectroscopy (s2), the maximum distance for imaging
is, equating (s1) and (s2),
3.3 ⫻ 101 2 = 1.45 ⫻ 10– 5 D
D = 2.2 ⫻ 101 7 m = 23 LY
Since the nearest stars are about 4 LY distant, the range where it might be observed as both a visual
and spectroscopic binary is only 4 to 23 LY. The latter distance is attained only if the separation is
at the maximum value allowed by spectroscopy.
Optical interferometry with resolution 10– 3 arc second (1 mas), increases this distance to 23 000
LY, but since interferometry requires bright stars, the distance is much more limited. See Armstrong
et al. AJ 104, 2217 (1992) for an example ( Cygni).
– – –
Problem 1.24– Doppler shifts – 1st and 2nd order.
(a) What is range of fractional frequency shift around orbit, for Fig. 6?
Data: Circular orbit, observer in plane of orbit.
Star 1: Max. frequency is when m1 approaches observer, at time t1 . Minimum frequency is at
t3 . From (2)
max – 0 v t – v t
= – r c1 ; min 0 = – r c3
0 0
Thus,
max – min v t –v t
0 = – r 1 c r 3
1–3
