Objectives:
• Planetary motion
Reading: Schutz, 11; Hobson 9; Rindler 11
Figure: Newtonian effective potential: centrifugal barrier always wins
• Centrifugal barrier always dominates as r → 0
• 2 types of orbits: unbound, hyperbolic E > 0; bound, elliptical E < 0.
64

LECTURE 16.
SCHWARZSCHILD ORBITS 65
• Circular: ˙ = 0, r = r
C such that ¨ = 0 = ⇒ dV /dr = V ′ ( r ) = 0.
• Newtonian elliptical orbits do not precess.
To see last point, expand potential around r = r
C
:
V ( r ) ≈ V ( r c
) +
1
2
V ′′ ( r
C
)( r − r
C
) 2 .
cf potential/unit mass of a spring kx 2 / 2 m , then r must oscillate with angular frequency (“epicyclic frequency”)
ω r
2 = V ′′ ( r
C
) .
Given the Newtonian effective potential h 2
V ( r ) =
2 r 2 so
V ′ ( r
C
) = 0 = ⇒ h 2
− h 2
V ′ ( r ) = r 3
= GM r
C
, therefore
−
GM r
,
+
GM
.
r 2
V ′′ ( r
C
) =
3 h r 4
C
2
−
2 GM r 3
C
=
GM
.
r 3
C
However, ω 2
φ
= GM/r 3
C
, thus same φ , so no precession.
ω r
= ω
φ
.
= ⇒ always reach minimum r at
Case 1. Large angular momentum h Reminder:
V ( r ) = h 2
2 r 2
¡ 1 −
2 µ r
¢
−
µc 2 r
.
Units of h on plots are µc .
Figure: Schwarzschild effective potential for a large values of h

LECTURE 16.
SCHWARZSCHILD ORBITS
• Essentially Newtonian behaviour as small r is inaccessible.
• This case applies to the planets. e.g. for Earth h ≈ 10 4 µc .
Case 2. Intermediate angular momentum h
66
Figure: Schwarzschild effective potential for an intermediate value of h
• Bound near-elliptical and circular orbits still exist
• Qualitatively different capture orbits possible.
Case 3. Low angular momentum h
Figure: Schwarzschild effective potential for a low value of h

LECTURE 16.
SCHWARZSCHILD ORBITS
• No bound orbits.
67
The Schwarzschild effective potential is
V ( r ) = h 2
2 r 2
µ
1 −
2 µ ¶
− r
µc 2
.
r
At the radius of circular orbits, dV ( r ) /dr = V ′ ( r ) = 0 = ⇒
V ′ ( r ) = − h 2 r 3
+
3 h 2 µ
+ r 4
µc 2 r 2
= 0 , or
µc 2 r 2
− h 2 r + 3 h 2 µ = 0 , so r
C
= h 2
± p h 4 − 12 h 2 µ 2 c 2
.
2 µc 2
The smaller root is a maximum of V and unstable. The larger root is stable while h 2 > 12 µ
At this point
2 c 2 , but once h 2 ≤ 12 µ 2 c 2 there are no more stable circular orbits.
r
C
=
2 h 2
µc 2
= 6 µ =
6 GM c 2
= 3 R
S
.
In accretion discs around non-rotating black-holes no more energy is available from within this radius. Calculate energy lost using E = kmc 2 .
Since ˙ = 0, r = 6 µ and h 2 = 12 µ 2 c 2 : c 2 2 ( k
C
− 1) = h r
2
2
µ
1 −
2 µ ¶
− r
2 µc 2
, r
12 µ 2 c 2
=
36 µ 2
= −
1
9 c 2 .
µ
1 −
2 ¶
−
6
2 c 2
,
6
Thus k 2
C
= 8 / 9. A mass dropped from rest at r = ∞ starts with k = 1, and thus 1 − k
C
= 5 .
7 % of the rest mass must be lost to radiation. Compare cf “Newtonian” with ∼ 0 .
7 % H → He fusion.
value of
Accretion power from black-holes is thus a conservative hypothesis in many cases as it requires much less fuel than fusion, e.g. 1 star per week rather than
7 or 8. Rotating black-holes can be more efficient still, with a maximum of
GM/ 6 R
S
=
1 / 12 = 8 .
3%.
42% (Kerr metrics). In realistic cases it is thought that about 30% efficiency is possible.