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AST 121S: The origin and evolution of the Universe
Introduction to Mathematical Handout 1
Please read this first....
This is an unusually long hand-out and one which uses in places mathematics that you may not be
familiar with. Section D is not part of the course and will not be examined - it is provided only for
your interest. I have included Section D in the handout because it represents one of the most important
breakthroughs in science. Newton's demonstration that all of Kepler's three Laws of Planetary Motion
could be explained by a simple force f = GMm/r2 and that this force was the same phenomenon as
familiar terrestrial gravity was important because:
(a)
It was the first time that the underlying cause of diverse complex natural phenomena were
explained in terms of a single very simple process, i.e. inverse square law gravity.
(b)
Gravity was the first of the fundamental forces of Nature to be identified and described (even if
we now know that Newtonian Gravity is only a very good approximation to gravity as described
by Einstein's General Relativity).
(c)
It was the first time that processes in the remote Universe were understood in terms of the same
phenomena seen on Earth. Newton's Gravity was a Universal Force. Thus the science of
astrophysics was born. The fact that the Universe follows the same physical laws as are seen in
terrestrial laboratories makes cosmology possible.
In this handout, Sections A and B will be familiar from the lectures. Everyone should be able to cope
with the mathematics in Section C in which we derive the force required for a body to move in a
circular path. The proof, in Section D, that Newtonian Gravity produces orbits which satisfy all of
Kepler's Three Laws of Planetary motion, is enormously easier if we use vectors and vector calculus.
Many of you will not be familiar with these tools. As stated above, I do not expect students to be able
to reproduce Section D in an exam or elsewhere and you should read it for interest only. I have
provided a short introduction to vectors and vector calculus for those who will want to work through
this derivation. I hope many of you will try this, and "experience the breakthrough" first-hand. In
Section E, I state without proof a couple of simple results that Newton proved that enabled him to
calculate the gravitational field near to extended objects like the Earth.
What you need to know from this handout:
(a)
You should know Sections A and B, be able to apply the results of Section E and be able to
reproduce Section C.
(b)
In Section D, you need to know only
(i)
that the angular momentum, J = mr2 d/dt, is conserved around an orbit.
(ii)
that the total energy, E, is also conserved around an orbit
2
2
GMm 
 1  dr  1 J

 1 2 GMm  
E   mv 

   m  

2
r  
r 
2
 2  dt  2 mr

In the first assignment, we will use these two results, (i) and (ii), and we will obtain the same energy
equation (ii) when we come, later in the course, to construct the equation to describe the evolution of
an expanding Universe.
AST 121S: The origin and evolution of the Universe
Mathematical Handout 1: Orbits in a simple gravitational field.
A. Kepler's Three Laws of planetary motion
From careful analysis of Tycho Brahe's voluminous observations of the apparent positions of the
planets as seen from Earth, Kepler formulated his three "laws" of planetary motion that described how
the planets, including the Earth, moved in space.
Kepler's 1st Law: Each planets moves on an orbit around the Sun that is an ellipse with the Sun located
at one focus of the ellipse
Kepler's 2nd Law: A line drawn from the Sun to a given planet sweeps out equal areas in equal time
intervals as the planet moves around the orbit.
Kepler's 3rd Law: For the various planets in the Solar System, there is a relation between the period of
the orbit T (the time taken to complete one orbit) and the semi-major axis of the ellipse,
A: T 2  A3
It should be noted that Kepler did not understand why the planets moved in this way though he
suspected that it was related to some terrestrial phenomenon and in fact was the first to coin the term
astrophysics.
B. Newton's three laws of motion
Newton formulated three laws that described the motions of objects, based in part on many of Galileo's
ideas:
Newton's 1st Law: Bodies move in a straight line at constant speed (i.e. at constant velocity) unless
acted upon by a force.
Newton's 2nd Law: The magnitude of a force is given by the mass and the acceleration the force
produces.
dv
F  ma
where a 
dt
Newton's 3rd Law: For every force there is an equal and opposite reaction.
C. Acceleration required for circular motion
The velocity of a particle is continually changing as it moves around a circular path (remember that
velocity is a vector quantity incorporating directional information, as distinct from speed, which does
not) and it must therefore, according to Newton's First and Second Laws, be accelerating and
experiencing a force.
Consider a particle moving with speed v around a circular orbit of radius r. Let us look at the change in
velocity while the particle moves through a very small angle  around the orbit. Suppose that the
particle is initially moving in the y direction. In time t it moves a distance y corresponding to the
angle .
motion in absence of force
circular orbit
x


y
r
Now, in this same time interval, the particle must stop moving purely in y and must acquire a small
velocity in the x direction, i.e. towards the centre of the orbit (remember we can make  arbitrarily
small). On the figure above, it moves x
x     y   r2
with a velocity
vx 
x
2
r
t
t
Therefore the acceleration required is
a
v x
  
 r  
t
 t 
2
Now we can write (t) as
 d
v


t dt
r
where w is the angular velocity of the orbit (measured in radians per second). Thus the acceleration
required to produce a circular orbit is
(1.1)
v2
a
  r 2
r
Thus in order to produce a circular orbit we must continuously apply a force, directed towards the
centre of the orbit, of mr2 or mv2/r.
A number of people, including Newton's great rival Hooke as well as Newton himself, realized that a
force field in the Solar System that fell off with the square of the distance r (and was proportional to
the mass of the planet) would reproduce Kepler's 3rd Law. This is because the period, T, of a planet in
a circular orbit is related to  by
(1.2)
T
2

Now, Kepler's 3rd Law states that
T 2  r3
since A  r for a circular orbit
Thus, from (1.1) and (1.2)
(1.3)
a  r 2 
1
r2
Thus in the case of idealized circular orbits, the force on a given planet must be proportional to the
mass of the planet and inversely proportional to the square of the distance between the Sun and the
planet.
But Kepler's 1st Law states that planetary orbits are not circles, but rather are ellipses. Newton's great
achievement was to demonstrate that such an inverse square force field would also produce elliptical
orbits, with the Sun at one focus, and with T2  A3, thereby satisfying all of Kepler's three Laws
describing planetary motions. The circular orbits considered above are just a special case of more
general circular orbits. Furthermore, he then showed that the force that causes the Moon to orbit the
Earth was the same as causes apples to drop from apple-trees. It is these two to which we now turn.
D: Elliptical orbits in a simple gravitational field
(Section D is for interest only and is not a formal part of the course)
The derivation of the general orbit in a inverse-square law gravitational field is most easily seen using
vectors and vector calculus:
Box 1: Vectors and vector calculus
For those students who (a) are not familiar with vector calculus and (b) wish to go
though this next section, the following is a brief overview of vectors and the calculus of
them.
Recall that whereas many quantities, such as mass, speed, temperature, are scalars (i.e.
simply have a magnitude), others are vectors and have both magnitude and direction.
Examples are the displacement from the origin of some coordinate system, velocity,
force etc. We will write vectors in bold print and scalars (and the magnitudes of vectors)
in regular print.
The direction of a vector can be represented by dividing it by its magnitude, i.e. a/a. The
magnitude of a vector can be represented as a = a
Vectors can be combined in various ways. Addition (and subtraction) is straightforward:
b
a+b
a
Box 1 (continued)
Vectors can also be multiplied in two ways: The dot-product is a scalar given by
a. b  ab cos 
where  is the angle between the vectors. The dot-product is the magnitude of one
vector times the component of the other vector that is parallel to the first (and vice
versa). It is easy to see that a.b = b.a.
The cross-product is a vector that has direction perpendicular to both a and b and has
magnitude
a  b  ab sin 
The direction of a  b is set by the right-hand rule such that a, b and a  b are like the
thumb, first and middle fingers of your right hand. This means that (a  b)  (b  a) .
Clearly parallel vectors (where  = 0) have zero cross-product (as they must since the
direction perpendicular to both is then not defined!).
Vector differentiation occurs in the normal way. If we have a time dependent vector
a(t). Then da/dt. is given by
da a(t  t )  a(t )

dt
t
a(t+dt)-a(t)
a(t)
a(t+dt)
da/dt is thus itself a vector.
Finally, differentiation of cross-products proceeds in the usual way by the chain rule
d
a  b    a  db    da  b 
dt
dt   dt


Consider the simple case where the gravitational field is dominated by a single point-like object of
mass M (i.e. in the case of the Solar System, the Sun). For an orbiting object of mass m, whose
instantaneous position, relative to the central object is given by r and whose instantaneous velocity is v
= dr/dt, Newtonian gravity describes the force and resulting acceleration of the orbiting object as:
(1.4)
m
d 2r
dv
GMm
m
 3 r
2
dt
dt
r
If we take the cross product of this equation (1.4) with r, and cancel out m, we get:
(1.5)
r
d 2r
GM
  3 rr  0
2
dt
r
since r  r  0 by definition
If we now look at the quantity r  v (i.e. r  dr dt  ) we get
(1.6)
d  dr   d 2r   dr dr 
 r     r  2       0 from (1.5) and a  a  0
dt 
dt  
dt   dt dt 
Thus the quantity r  v  is constant for motion in a gravitational field. The quantity r  v  is the
specific angular momentum (i.e. angular momentum per unit mass), and the angular momentum, J, is
simply m r  v  .
(1.7)
J  mr  v  constant
Remembering that the direction of r  v  is perpendicular to r and v, the constancy of r  v  means
that the orbit lies in a plane.
Equation (1.6) now explains Kepler's Second Law since the rate at which area swept out by the vector
r is just
(1.8)
dA 1
1
 rvsin   r  v 
dt 2
2
If we now take the dot-product of equation (1.4) with v ( = dr/dt) , we get:
(1.9)
m v.
dv
GMm
  3 v. r
dt
r
With a few moments thought you should be able to convince yourself that
(1.10)
d 1 2
dv
d  GMm 
GMm dr
 mv   m v. and 
   3 r.
dt  2
dt
dt  r 
dt
r

Therefore, we can rewrite (1.9) as
(1.11)
d  1 2 GMm 
 mv 
0
dt  2
r 
You may recognize that the quantity in brackets is the total energy, E, of the orbiting object (i.e.
kinetic energy plus negative potential energy).
GMm 
1
E   mv 2 

r 
2
Equation (1.11) therefore simply proves the conservation of energy.
Because the orbit is in a plane, from (1.6), we can conveniently adopt (r,) coordinates and drop the
vector notation. We have defined the angular momentum, J, which we have seen is constant, to be
(1.12)
J  m vr

J  mr 2
d
dt
We can decompose v into a component parallel and perpendicular to r using Pythagoras.
J2
 dr   d 
 dr 
v2     r



 
m2r 2
 dt   dt 
 dt 
2
2
2
Thus, we can rewrite our energy equation as
2
(1.13)
E
1  dr 
1 J2
GMm
m  

 constant
2
2  dt 
2 mr
r
Rearranging (1.13) we get:
1/ 2
(1.14)
dr  2 E 2GM
J2 


 2 2
dt  m
r
m r 
Note in passing that the left hand side of (1.14) is just the radial velocity of the particle towards or
away from the central object. If E is positive then this is still finite as the distance tends to infinity and
the particle can completely escape from the influence of the central object. On the other hand, if E is
negative, then the radial velocity must become zero at some finite distance from the central object and
in this case the particle never gets further away, and is bound to the central object. The energy E is
thus often referred to as the binding energy of the particle.
Dividing (1.14) by J ( = mr2d/dt) we get:
1/ 2
(1.15)
1 dr dt  2 E
2GM
1 
 2  2  2 2
2
mr dt d  mJ
J r
m r 
If we now make the following substitutions:
1
1
 du   2 dr
mr
mr
GMm
wu
 dw  du
J2
u
2
2 E  GMm 
w 

  constant
mJ 2  J 2 
2
0
then (1.15) becomes
(1.16)

dw
  w02  w2
d

1/ 2
Equation (1.16) has solution
w  w0 cos 
i.e. substituting back into the above,
(1.17)
1 1
 1   cos  
r r0
with
1/ 2
(1.18)

J2
2 EJ 2 

r0 
and


1

 G 2 M 2 m3 
GMm2


The quantity  is known as the eccentricity of the orbit. A circle is an ellipse with  = 0. For E < 0, i.e.
for an object which has negative net energy (i.e. which is bound to the central object), equation (1.17)
is in fact the equation of an ellipse with the origin at one focus. Thus is Kepler's First Law explained.
You may be more familiar with the equation of an ellipse in Cartesian coordinates:
(1.19)
x 2 y2

1
a2 b2
In this case, the origin of the Cartesian system (x,y) is offset from the origin of the polar coordinate
system (r,) by (a,0). The semi-major and semi-minor axes of the ellipse are given by:
a
(1.19)
b
r0
1 2
r0
1   
2 1/ 2
The E = 0 case corresponds to a parabola and the unbound E > 0 case to a hyperbola. The circular  =
0 case has E = -G2M2m3/2J2
We can easily now recover Kepler's Third Law by noting that the area of an ellipse is
(1.20)
A  ab
Now, the time taken to go around the orbit is, from (1.8)
(1.21)
T   dt 
2m
2mA
m
dA 
 2ab

J
J
J
Now, from (1.18), we know r0 = J2/GMm2, and from (1.19) we have that b2 = ar0, it follows that:
(1.22)
42 3
T 
a
GM
2
This is exactly Kepler's Third Law with a = A. Note that the mass of the planet does not enter in to
equation (1.22) so it applies for any planetary system in which the gravitational field is dominated by a
single massive object.
E: Gravitational fields from symmetrically distributed mass systems.
Newton's analysis in Section D beautifully explained Kepler's Three Laws of planetary motion in terms
of a simple 1/r2 force field emanating from the Sun. Newton next wanted to demonstrate that this was
the same force as familiar gravity on Earth. Galileo's observations that the motions of objects on Earth
was independent of their masses, could be explained by Newton's new force since the size of the force
was proportional to the mass of the body it was acting on, so the acceleration would be independent of
mass. However, for several years he was stymied by the fact that the Earth is obviously not a point
mass and he did not know how to calculate the gravitational field of an extended distribution of mass
such as the Earth.
After several years he was able to prove the following two theorems which will here be stated without
proof: For any spherically symmetric shell of matter, the gravitational field arising from the mass on
the shell is (a) zero within the shell and (b) the same, outside the shell, as if all the mass of the shell
was a point mass at the centre of the shell.
Thus, the gravitational field at the surface of the Earth is the same as if the Earth's mass was
concentrated into a point at the centre of the Earth, so that one can in fact use the simple equation for
the point mass (1.4). Furthermore, the gravitational field at some radius r within an extended
spherically symmetric object (such as an idealized galaxy) is given by (1.4) with M the mass enclosed
within the radius r. The effect of mass further out in the galaxy is zero.