Physics 320: Orbital Mechanics
(Lecture 7)
Dale Gary
NJIT Physics Department
Reminder of Kepler’s Laws
Kepler's Three Laws (quantitative version)
First Law:
Planets travel in elliptical orbits with the Sun at one focus, and obey
the equation 𝑟 = 𝑐/(1 + 𝑒 cos 𝜃), where 𝑐 = 𝑎(1 − 𝑒 2 ) for 0 < e < 1.
(Comets and other bodies can have hyperbolic orbits, where 𝑐 =
𝑎(𝑒 2 − 1), for e > 1.)
Second Law:
The radius vector of a planet sweeps out equal areas in equal times
(planet travels fastest when near perihelion).
Third Law:
The square of the orbital period of a planet is proportional to the cube
of its semi-major axis:
𝑃2 = 𝑘𝑎3, where k is a constant
We also showed how Newton found the value of the constant 𝑘 =
So
𝑃2
=
4𝜋2 3
𝑎 .
𝐺𝑀
Today, we will give a refinement of this result.
September 25, 2018
4𝜋2
.
𝐺𝑀
Center of Mass Reference Frame
We are now going to discuss the notion of center of mass, with which you
are certainly already familiar. Think of a system of N particles a = 1, …, N,
with masses ma and positions ra. The center of mass (or CM) is defined to
𝑁
N
be the position
m1r1    mN rN
1
where 𝑀 =
𝑚𝛼
R
m
r


a a
M a 1
M
𝛼=1
 Like any vector equation, this represents separate equations for each of the
components (X, Y, Z):
1 N
1 N
1 N
X
ma ya , Z 
ma za .
 ma xa , Y  M 

M a 1
M
a 1
a 1

You can think of the center of mass as a weighted average of the positions
of each mass element, i.e. weighted by the mass of that element, or
equivalently it is the vector sum of the ra, each multiplied by the fraction of
mass at that location.
 To get a feeling for CM, let’s look at the center of mass for a two particle
system, which might, for example, represent the Sun and Earth, or two
stars in orbit around each other.

September 25, 2018
Center of Mass and Equation of Motion


1 N
m1r1  m2r2
In this case, R 
m
r

 a a m  m , which can be seen in theCMfigure.
M a 1
1
2
m1
It is easy to show that the distance of the CM from m1
and m2 is in the ratio m2/m1. The figure shows the case
r1 R
where m1  4m2. In particular, if m1 >> m2, then the CM
r2
will be very close to m1.
m2
O

Note that the time derivative of the center of mass for N particles is just the
N
N
CM velocity
1
1
 
R
ma ra 
pa Here I am using “dot-notation”


𝑑𝑟
M a 1
M a 1
for time derivative. = 𝑟
𝑑𝑡
.
so the momentum of an N-particle system is related to its CM by P  MR
 Differentiating this expression, we get the very useful relation for the

Newton’s second law, the equation of motion of a system: Fext  MR
 This says that the CM of a collection of particles moves as if the external
forces on all of the individual particles were concentrated at the CM. This is
why we can treat extended objects (e.g. a planet) as a point mass.
September 25, 2018
Reduced Mass
If we now move our coordinate system to the center of mass (i.e. set R = 0),
𝑚1 𝐫1 + 𝑚2 𝐫2
𝐑=
=0
𝑚1 + 𝑚2
 Then we can write r1 and r2 in terms of the vector r = r2 – r1 between the two
masses:
𝑚2
𝐫1 = −
𝐫
𝑚1 + 𝑚2

𝐫2 =
𝑚1
𝐫
𝑚1 + 𝑚2
𝑚1 𝑚2
 We now introduce the concept of reduced mass: 𝜇 =
𝑚1 + 𝑚2 so
𝐫1 = −
𝜇
𝐫
𝑚1
𝜇
𝐫2 =
𝐫
𝑚2
September 25, 2018
Reduced Mass and the CM
Reference Frame


When two objects of similar mass orbit each other, they both move around
the common center of mass as in the figure below, left.
We can reduce this complicated looking problem to an equivalent problem,
where there is a single body of reduced mass m, orbiting a central body of
mass M = m1 + m2, with a separation r = |r1- r2| .

m1 r O
1
r2
CM
r1
r2
m2
r = |r1- r2|
m
M

O
Arbitrary
Origin atorigin
CM
Path relative to CM
Equivalent one-dimensional problem
Note, as 𝑚1 ∞, 𝜇 𝑚2
and the problem
corresponds to a relatively
small mass planet orbiting a
massive star, as we have
been implicitly assuming up
to now.
In many of our expressions
from now on, we will be
using reduced mass instead
of the mass of one body,
since it is more general.
September 25, 2018
Kepler’s First Law Derived

To derive the fact that a planet has to orbit in an ellipse, we have to begin
with the orbital angular momentum (in the center of mass and using the
reduced mass):
𝐋 = 𝐫 × 𝐩 = 𝐫 × 𝜇𝐯

An important fact about the gravitational force is that it is a central force,
meaning it only acts radially, and cannot produce any change in angular
momentum, i.e. 𝑑𝐋/𝑑𝑡 = 0. You can show this directly:
vr and F
p =are
mvparallel,
are parallel,
so
𝑑𝐋 𝑑𝐫
𝑑𝐩
=
×𝐩+𝐫×
=𝐯×𝐩+𝐫×𝐅
so
cross
product
is
thethe
cross
product
is zero.
𝑑𝑡 𝑑𝑡
𝑑𝑡
zero.
The text starts with the expression for angular momentum, and goes through
a purely mathematical derivation of Kepler’s first law, which is not at all
obvious and uses several obscure mathematical tricks. I encourage you to go
through the steps, but we will not do so here.
 We just write the final result: The equation for an ellipse in terms of L,

Kepler’s

1st
Law
𝐿2 /𝜇2
𝑟=
𝐺𝑀(1 + 𝑒 cos 𝜃)
𝑎(1 − 𝑒 2 )
cf. our earlier polar
equation for an ellipse 𝑟 = 1 + 𝑒 cos 𝜃
Note that this means 𝐿 = 𝜇 𝐺𝑀𝑎(1 − 𝑒 2 ) , which we will make use of later.
September 25, 2018
Kepler’s 2nd Law Derived

Kepler’s 2nd law, about the radius sweeping out equal areas in
equal times, can be derived by considering the area swept.
The radius sweeps out an area given by half of this rectangle,
1
or 𝑑𝐴 = 2 𝑟 2 𝑑𝜃.
𝑑𝐴
𝑑𝑡
1 2 𝑑𝜃
𝑟 𝑑𝑡
2
1
𝑟𝑣𝜃 .
2

Taking the derivative wrt time,

But angular momentum is 𝐿 = 𝜇 𝐫 × 𝐯 = 𝜇𝑟𝑣𝜃 , so the rate of
change of area can be expressed in terms of L
=
=
dq
r
Sun
𝑑𝐴
𝐿
nd Law
=
Kepler’s
2
𝑑𝑡 2𝜇
 So Kepler’s finding, that this rate of change of area is a
constant, is merely an expression of the fact that the angular
momentum is a constant.
September 25, 2018
orbit
Velocities in the Orbit
At two particular locations in the orbit, namely perihelion (when the planet
is closest to the Sun) and aphelion (when farthest from the Sun), there is
no radial part of the velocity—the planet is moving only in the theta
direction.
 At perihelion, 𝜃 = 0, so our earlier equation for the ellipse becomes,
𝐿2 /𝜇2
𝑟𝑝 =
𝐺𝑀(1 + 𝑒)


But also here, the angular momentum is just 𝐿 = 𝜇𝑟𝑝 𝑣𝑝 , so
(𝜇𝑟𝑝 𝑣𝑝 )2 /𝜇2
(𝑟𝑝 𝑣𝑝 )2
𝑟𝑝 =
=
𝐺𝑀(1 + 𝑒)
𝐺𝑀(1 + 𝑒)

A couple of lectures ago, we showed that 𝑟𝑝 = 𝑎(1 − 𝑒), so solving for 𝑣𝑝 ,
we have
and likewise
𝑣𝑝 =
𝐺𝑀 1 + 𝑒
𝑎 1−𝑒
1/2
Perihelion speed
𝑣𝑎 =
𝐺𝑀 1 − 𝑒
𝑎 1+𝑒
1/2
Aphelion speed
September 25, 2018
Kepler’s Third Law Derived
Kepler’s third law can easily be obtained from his second law, by noting that
the area swept out in a given time is constant, and the area swept out over
one period is just the entire ellipse, which has an area 𝐴 = 𝜋𝑎𝑏.
 Thus, integrating the second law equation over an entire period,
𝐿
𝐿
𝐴=
𝑃=
= 𝜋𝑎𝑏
2𝜇
2𝜇


So Kepler’s third law becomes
𝑃2

4𝜋 2 𝑎2 𝑏2 𝜇2
=
𝐿2
But we have an expression for b in terms of a, i.e., 𝑏2 = 𝑎2 (1 + 𝑒 2 ), and we
also have our expression for L, which when squared is
𝐿2 = 𝜇2 𝐺𝑀𝑎(1 − 𝑒 2 ).
 Putting these both into the above, we have our final form of Kepler’s third
law:
4𝜋 2
2
𝑃 =
𝑎3 Kepler’s 3rd Law
𝐺(𝑚2 + 𝑚1 )
September 25, 2018
Potential Energy
We learned earlier that Newton’s law of universal gravitation provided the
gravitational force equation:
𝑀𝑚
𝐅 = −𝐺 2 𝐫
𝑟
 You may also recall that energy is force through a distance, and that
potential energy is the negative of work done, i.e.

𝑟𝑓
𝑈𝑓 − 𝑈𝑖 = ∆𝑈 = −
𝐅 ∙ 𝑑𝐫
𝑟𝑖
Inserting the gravitational force into this equation (and being careful with
signs!):
𝑟𝑓
𝑀𝑚
∆𝑈 =
𝐺 2 𝑑𝑟
𝑟
𝑟𝑖
 Evaluating the integral:

1 1
−
𝑟𝑓 𝑟𝑖
 I stated earlier that we take the zero of potential energy at infinity, so we
have the final result for any r:
𝑀𝑚
A massive object M creates a
𝑈 = −𝐺
gravity well around it
𝑟
𝑈𝑓 − 𝑈𝑖 = −𝐺𝑀𝑚
September 25, 2018
Escape Speed
As a side note, since the potential energy at infinity is zero, an object falling
from infinity will lose potential energy, which is everywhere negative inside
the gravity well, and consequently the object will gain a kinetic energy
given by the well-known K = ½ mv2. But it does this in such a way that its
total energy remains zero! It simply converts its lost potential energy into
kinetic energy, in exactly the same way that a ball dropped from a height
on Earth does.
 Of course, it works the other way for an object moving outward. It will lose
kinetic energy as it moves toward infinity (climbs out of the gravity well),
and provided it has just enough energy to get to infinity, it will arrive there
with 0 total energy.
 The speed that an object needs to go from some point in a gravity well
(say, the surface of a planet) to infinity with zero energy is called the
1
𝑀𝑚
escape speed, derived from:𝐸 = 𝐾 + 𝑈 = 2 𝑚𝑣 2 − 𝐺 𝑟 = 0, so

𝑣esc =
2𝐺𝑀/𝑟
September 25, 2018
Orbital Energy
It was mentioned a few lectures ago that any bound orbit has a total
energy of less than 0. Let’s write the energy of a planet when it is at
perihelion:
1
𝑀𝜇
𝐸 = 𝜇𝑣𝑝2 − 𝐺
2
𝑟𝑝
 Using our values for perihelion distance and velocity:
𝐺𝑀 1 + 𝑒
𝑣𝑝 =
𝑟𝑝 = 𝑎(1 − 𝑒)
𝑎 1−𝑒
 We have

1/2
1
𝑀1+𝑒
𝑀𝜇
1
𝑀
𝑀𝜇
𝜇𝐺
−𝐺
= 𝜇𝐺
𝑒 − 1 = −𝐺
2
𝑎 1−𝑒
𝑎 1−𝑒
2
𝑎 1−𝑒
2𝑎
 If you calculate the same thing at aphelion, you will find the same
expression. In fact, the energy is everywhere constant (and negative) over
the orbit, and has this value:
𝑀𝜇
𝐸 = −𝐺
2𝑎
𝐸=
September 25, 2018
Velocities in the Orbit Again

Since the energy is everywhere constant, we can equate it to the kinetic plus
potential energy at any point:
𝑀𝜇 1 2
𝑀𝜇
𝐸 = −𝐺
= 𝜇𝑣 − 𝐺
2𝑎 2
𝑟

We can then get a general expression for velocity in an orbit
2 1
= 𝐺(𝑚1 + 𝑚2 )
−
𝑟 𝑎
 In the above, we have used the fact that M is the total mass of the system,
M = m1 + m2.
𝑣2

This equation has a special name, the vis-viva equation.
September 25, 2018

What
We’ve
Learned
We learned that Kepler’s laws are just a consequence of the properties of a
central force law, and that angular momentum is conserved in such a system.
 You should know that a star+planet system has a rather complex orbital
relationship when the two masses are similar, but that this two-body system is
𝑚 𝑚
equivalent to a simpler system with a planet of reduced mass 𝜇 = 𝑚 1+𝑚2
1
2
orbiting a central object of mass 𝑀 = 𝑚1 + 𝑚2 ., at a distance 𝑟 = 𝑟1 + 𝑟2 , where
r1 is the distance of m1 from the CM, and r2 is the distance of m2 from the CM.
 The equations governing Kepler’s three laws are:
𝐿2 /𝜇2
Kepler’s 1st Law (equation for an ellipse)
𝑟=
𝐺𝑀(1 + 𝑒 cos 𝜃)
𝑑𝐴
𝐿
Kepler’s 2nd Law (conservation of energy)
=
𝑑𝑡 2𝜇
2
4𝜋
rd
𝑃2 =
𝑎3 Kepler’s 3 Law (period-distance law)
𝐺(𝑚2 + 𝑚1 )
𝑀𝜇
 We also have the orbital energy equation, 𝐸 = −𝐺
constant (and negative)
2𝑎
for any bound orbit.
2 1
 We derived the useful equation for speed in an orbit 𝑣 2 = 𝐺(𝑚 + 𝑚 )
−
1
2
𝑟 𝑎
September 25, 2018