Transcript Gravity and Orbits

The gravitational force between two objects:
•The “r-hat” is a unit vector pointing directly
away from the source of gravity
Gravitational Potential Energy
•The potential energy is the (negative of the)
integral of the force
EP    Fdr  
GMmdr
r2
GMm
EP  
r
M
F
r
m
GMmrˆ
F
r2
F
GMm  r  rM 
r  rM
3
Simple Orbits
Circular orbits:
•If the velocity is exactly right,
you get a circular orbit
•Gravitational force must match
GM
centripetal force
2
v 
R
GMm mv 2
F

2
R
R
Other orbits:
•If the velocity is smaller, or a bit
bigger, you get an elliptical orbit
•If the velocity is a lot greater, the object leaves
•Depends only on speed, not direction
•Kinetic energy must be sufficient to overcome potential energy
•Minimum speed is escape velocity
M
R
m
GMm
EK  mv 
R
2GM
2
vesc

R
1
2
2
Gravitational Field
•In PHY 114, you learned a lot about electric forces and fields
•We introduced the Electric Field as the force per unit charge
keQrˆ
E r   2
r
keQqrˆ
EF q
F
2
r
•Compare this with gravity:
•By analogy, introduce the gravitational field
GMmrˆ
F
r2
gF m
GMrˆ
GMr
g r    2   3
r
r
F  gm  ma
•If there are many sources of gravity, their effects must be added up
g  r   
i
Gmi  r  ri 
r  ri
3
ag
Gauss’s Law for Gravity:
•In 114, you learned the total electric flux out of a
region was related to the total charge in that region:
 E   E  nˆ dA
 E  nˆ dA  4 ke q
 g   g  nˆ dA  4 GM
What’s the gravitational flux
from the region in this case?
E
b
z
r
a
r
m4
m1
m2
E
b
q
b
a
m3
 g  4 G  m1  m2  m3 
s
n̂
•There is an exactly analogous formula
for the gravitational field:
n̂

Applying Gauss’s Law: Spherical Symmetry
•Gauss’s Law can be used to find the gravitational field
when there is a lot of symmetry
 r     r 
•Example: Spherical symmetry
•Mass density depends only on distance from center
•Draw a spherical Gaussian surface
•Logically, the gravitational field will be radial everywhere
•Gauss’s Law tells you the flux is proportional to the contained mass
 g  R   4 GM  R 
 g  R   g  R  4 R2
GM  R  rˆ
•The mass contained inside the sphere is just the sum of the g  R   
2
R
masses on each spherical shell inside it
•The volume of a thin spherical shell is the area of a sphere of radius r
times the thickness dr.
R
R
R
R
ˆ
G
r
2
2
g
R


4

r
  r  dr


M  R    dM     r  dV   4 r   r  dr
2 
R 0
0
0
0
Sample Problem:
A gravitational source takes the form of a
uniform sphere of density0 and radius a
(a) What is the gravitational field everywhere?
(b) What is the corresponding orbital velocity
for circular orbits?
 0
 r   
0
if r  a,
if r  a.
Grˆ
g  R    2  4 r 2   r  dr
R 0
R
Grˆ
g  R   2
R
min  a , R 

0
4
Grˆ 0 4 3 min  a , R 
ˆ


3  G 0rR


r
4 r 0 dr
g 4
0
3 2
R2 3
ˆ


G

r
a
R
0
 3
2
rˆv 2
2
v
 Rg
ag
R
4

if R  a,

3  G 0 R
v
3 1/ 2
4

G

a
R
if R  a.

0
 3
if R  a,
if R  a.
Applying Gauss’s Law: A Slab
•Consider a slab source, spread out uniformly in two
 r     z 
dimensions
•Density depends only on z.
•Assume the top half has same mass
distribution as the bottom half
•No gravity at z = 0.
•Draw Gaussian surface
•Box from z = 0 to z = Z, of area A.
•Use Gauss’s Law:
4 GM  Z  zˆ
gZ   
A
 g  Z   4 GM  Z 
g  Z   g  Z  A
Z
Z
Z
0
0
0
z
M  Z    dM     z  dV   A  z  dz
•For uniform density, we find:
g  Z   4 Gzˆ    z  dz
g  Z   4 G0 Zzˆ
0
Gravitational Potential
•For electric fields, it is often more useful to work with the electrostatic potential
ke qi
VE
VE
VE
VE  r   
E  VE  r   
xˆ 
yˆ 
zˆ
VE  r     E  ds
x
y
z
i r  ri
•Exactly the same thing can be done for the gravitational field
  r     g  ds
g    r   



xˆ 
yˆ 
zˆ
x
y
z
  r   
i
•The potential energy for one particle, then, is
EPi  mi   ri 
Gmi
r  ri
Sample Problem (1)
What is the gravitational potential everywhere for a uniform sphere of density
0 and radius a? What is escape velocity from the center of the sphere?
•Because the gravitational field is only in the
if r  a,
  43  G 0rˆr
r-direction, the potential should depend only on r. g  r    4
3 2
ˆ


G

r
a
r
if r  a.
0
 3
•Therefore the relationship between  and g is:
d
g r   
rˆ
  r     g  r  dr
dr
Must do inside and outside separately!   r   4  G  rdr
in
0
3

 in  23  G 0 r 2  Cin
•Don’t forget the constant of integration!
 out  r    43  G 0 a 3r 2 dr
 out   43  G 0 a 3 r 1  Cout
Cout  0
How can we find the constants of integration? 0  out     0  Cout
•Potential at infinity is zero
in  a   out  a 
•Potential at the boundary must be continuous
2
3 1
2
2
4

G

a

C



G

a
a
C


2

G

a
0
in
0
3
3
in
0
Sample Problem (2)
What is the gravitational potential everywhere for a uniform sphere of density
0 and radius a? What is escape velocity from the center of the sphere?
2
2
2


G

r

3
a


0
 r   3
3 1
4


G

a
r

0
3

if r  a,
if r  a.
•To find escape velocity, match potential
and kinetic energy at the origin
EP  mi   0  2 G0a 2m  EK  12 mv 2
2
vesc
 4 G 0 a 2
 3GM a
Global Conservation Laws:
•Consider a galaxy/structure which is not interacting with other galaxies/structures:
•Often a good approximation
•The following must be conserved:
•Total energy, Total momentum, Total angular momentum
P   mi v i
Total momentum describes overall motion
i
•We can eliminate it by working in the “center of mass frame”
Energy is more complicated
E  EP  EK  Eother
•Potential and kinetic
•But there can be other contributions E  Gmi m j
2
1
E

m
v

K
i i
2
•If gravity is the only force involved, P 
r

r
i j
i
i
j
then global Ep + EK energy is conserved
•Stars, for example, rarely collide
•If there are other effects, like collisions, energy can be transferred and lost
•Gas clouds, in contrast, commonly collide
L   mi ri  vi
•Finally, the total angular momentum of the galaxy is conserved
i
Potential from a distribution
Gmi
•Suppose mass is distributed in a continuous manner
  r   
•How do we calculate potential and potential energy?
i r  ri
•Divide into many small regions of size dV
G i  ri  dVi
  r   
•These will each have mass dV
r  ri
i
•Convert to an integral:
 r 
  r  3
 r   
d r
r  r
•Potential energy for the whole system is:
EP   
i j
Gmi m j
ri  r j
1 G   ri  dVi   r j  dV j
1 Gmi m j
 
 
2 i j
2 i  j ri  r j
ri  r j
•Can be rewritten in terms of potential:
EP 
1
2
3
d
 r  r    r 
G   r    r 
1 3
3
EP    d r  d r 
2
r  r
Sample Problem
EP 
What is the total gravitational binding energy
for a uniform sphere of mass M and radius a?
2
2
2

 3  G 0  r  3a 
 r   
3 1
4


G

a
r

0
3

EP    G  d r0 0  r  3a
1
2
3
2
3
2
1
2
if r  a,
if r  a.
2

  G 4
1
3
a
2
0 0

r 2 dr  r 2  3a 2 
a
2
2 5
  G   r  a r    16

G

0a
15
0
4
3
2
2
0
1
5
5
2 3
M  43  a 3 0
0 
16 Ga  3M 
EP  


15  4 a 3 
2
5
3M
4 a 3
2
3GM 2
EP  
5a
3
d
 r  r    r 
Virial Theorem (1)
•For circular orbits, there is a simple relationship between
the potential energy and the kinetic energy:
•For non-circular orbits, this is not true, because energy
keeps changing between the two components.
•However, if you average over time, this will still be true
GMm
r
GM
2
1
1
EK  2 mv  2 m
r
2 EK  EP  0
EP  
•If you have many objects, some of them will be at their
2 EK  EP  0
maximum, and others at their minimum
•Could this expression be true if you add everything up?
•Consider a complicated combination of many masses acting gravitationally
•Galaxy or Globular cluster, for example, consists of 104 to 1014 stars
•First, find the total kinetic and potential energy
•And the force on any one object
Gmi m j  ri  r j 
Gm
m
i
j
Fi  
3
EK   12 mvi2 EP  
j

i
ri  r j
i  j ri  r j
i
EK   12 mvi2
Virial Theorem (2)
i
EP   
i j
Gmi m j
ri  r j
•We will assume that the system isn’t changing much; i.e., though the individual stars
are moving, there will be as many moving one way as another
•Galaxy has no net motion
Gmi m j  ri  r j 
Fi  
•Any quantity that “adds up” effects of all components
3
j i
ri  r j
will be constant
i miri  vi  constant
•Consider the following quantity:
•Time derivative should vanish:
d
dr
dv
0    mi ri  vi     mi i  v i  mi ri  i     mi v i2  ri  Fi 
dt
dt 
i dt
i
i 
 r  r  r  r  r  r  
ri   ri  r j 
i
i
j
j
j
i


 2 EK   Gmi m j
 2 EK   Gmi m j

3
3
3
 r r

i j i
i j
ri  r j
r

r
i
j
j
i


 2 EK   Gmi m j
i j
r  r 
i
2
j
ri  r j
3
 2 EK  EP
2 EK  EP  0