Problem solutions, Ch. 1 Astrophysics Processes
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
1–1
, 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 1012 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 1020 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 2
r g 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 = – 2 dr = –2 rs (1.21.s2)
Fg rg rg
where dr = rs when all three objects are coaligned.
Multiply (s1) and (s2) to find the desired ratio,
dFg Mg r 3 (1.21.s3)
s = 106 1.5 1012 3 = 4.9 10 –19
= –2
Fs Ms r g3 2.4 1020
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 1017 m.
Average distance d between adjacent stars is (volume per star)1/3,
1/3 1/3 (1.21.s4)
4 Rglob = 2.29
15
d = Vs = 10 m = 15 000 AU
3N
1–1
“Kepler, Newton, and the mass function” ©8/8/08 Hale Bradt
1–1
, 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 1012 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 1020 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 2
r g 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 = – 2 dr = –2 rs (1.21.s2)
Fg rg rg
where dr = rs when all three objects are coaligned.
Multiply (s1) and (s2) to find the desired ratio,
dFg Mg r 3 (1.21.s3)
s = 106 1.5 1012 3 = 4.9 10 –19
= –2
Fs Ms r g3 2.4 1020
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 1017 m.
Average distance d between adjacent stars is (volume per star)1/3,
1/3 1/3 (1.21.s4)
4 Rglob = 2.29
15
d = Vs = 10 m = 15 000 AU
3N
1–1
