A simple description of the mass distribution and gravitational potential of a galaxy is to represent the stars as point masses. This is a good approximation for elliptical galaxies, since these usually contain very little gas. It is an adequate approximation for spiral/disk galaxies, because they usually contain significantly more stars than gas.
We can represent the potential Φ at a distance R from a point mass by
Φ(R) = −GM/R.
and the force acting F(r) around the point mass as
F(r) = −GMr/R3
where r is the vector from the point mass to the point at which we evaluate the force. This may be more familiar in the form
F(r) = −GMer/R2
where er is the unit vector directed along r.
In a galaxy, we have a large number of point masses, N, which we denote by the subscript i, where i =1,2,...,N.
The gravitational potential is additive, so that the total potential at some position in space denoted by the vector r, is given by
Φ(r) = − ∑Gmi / |r-r'|
where we denote the particle masses by mi
In galaxies, there are typically 109 to 1011 stars (or point masses) so that the sum above contains 109 to 1011 terms, clearly an impractical number to do by hand, and even today still impractical to do directly with computers (the largest simulations of "N-body" systems of this type on computers are presently for about 107 particles, and we clearly have a few years to wait before simulations will be able to represent each and every star in a galaxy.
There are two ways around this impracticallity. Firstly, the stars can be represented by point masses in a computer simulation. Secondly, the stars can be considered to have a smooth distribution in space, so that we may apply Poisson's Equation in order to recover the potential from the matter distribution. Poisson's equation is derived in Binney and Tremaine, section 2.1
The following results are from Binney and Tremaine's textbook "Galactic Dynamics" (Princeton University Press 1987, section 2.1 --- hereafter BT).
Poisson's Equation : ∇2Φ = 4πGρ.
The force F(r) generated at a position r is given by
F(r) = −∇Φ(r)
These two equations in words are as follows : from the distribution of
the stars in space ρ we can determine the total potential Φ, and from
the gradient of the potential we can determine the force on any given star.
Newton derived two useful theorems concerning spherical shells of matter.
(1) A body inside a spherical shell of matter experiences no net gravitational force from the shell.
(2) A body lying outside a closed spherical shell experiences a gravitational force which is the same as if all the matter in the shell were concentrated at a point at its center.
These two theorems allow us to deduce that the gravitational attraction of a spherically symmetric density distribution ρ(r') on a test mass at radius r is determined only by the mass interior to r. The mass exterior to r has no effect.
A spherical mass distribution can therefore be characterised in terms of a circular speed vcirc(r), which is the speed a test particle would have in a circular orbit at radius r (note the particle mass has been normalised out)
F(r) = v2circ(r) / r
F(r) = − ∇ Φ(r) = −dΦ/dr
v2circ(r) = r dΦ/dr = GM(r)/r
where M(r) is the mass interior to r.
The kinetic energy of a particle with mass m and velocity v is defined as KE = mv2/2. Gravitational fields are conservative, which means that the energy used to move between two points within the field is independent of the path taken. Thus, the total energy of the system is the sum of the kinetic energy and gravitational potential energy:
E = KE + Φ.
In a conservative field the potential Φ can be defined to within a constant with no effect on the dynamics of the system. Consequently, it is quite common to define the total energy of the system to be identically zero
KE + Φ = 0.
If a test particle of unit mass at infinity falls into this gravitational potential field, then its KE at radius r is
KE = −Φ(r)
i.e. v2 = −2Φ(r).
Note that with this definition, Φ is always negative, so that v2 is positive.
The escape speed vesc is defined by
v2esc = 2 |Φ(r)|
This is just the kinetic energy a particle needs at radius r to move infinitely far from the center of the potential field.
Some examples with easy potentials will help clarify matters! We'll derive the circular speed in each case.
(1) Consider a point mass of mass M.
The potential is given by
Φ(r) = −GM/r.
This is called a Keplerian potential, since it is that obeyed by the planets in the solar system and yields Kepler's three laws.
The velocity of a circular orbit at radius r is
vcirc(r) = √GM/r.
(2) Now consider a uniform density sphere of mass M and radius a:
|⌈||−2πGρ(a2−r2/3)||for r < a|
|⌊||−GM/r||for r > a|
note that the density of the sphere is given by ρ = M/ (4 π a3 / 3).
Inside the sphere the potential grows parabolically to the surface, while outside the sphere it falls off like 1/r, as in the Keplarian case (and as we should expect).
(3) An isothermal sphere has a density profile given by
ρ(r) = ρ0(r/r0)−2
and the potential is
Φ(r) = 4πGρ0r20 ln(r/r0).
The circular velocity is
vcirc = (4πGρ0r20)1/2.
The rotation speed is constant (unchanging with radius), as the RHS contains only constants of the potential! This makes it a useful potential when trying to understand flat rotation curves (see end of lecture 2).
Verify that the rotation velocity for each potential above is correct.
For the point mass potential, figure out what the rotation velocity of the Earth is around the Sun. What period do you derive for this velocity?
Make a plot of the escape speed as a function of radius from the Sun for the Solar System.