Announcements
•Exam 1 is in two weeks. Will cover
material in textbook up through
Chapter 3 plus additional material on
sidereal time and Julian date
•Homework Set 3: Supplemental
Problems
Kepler’s Laws of
Planetary Motion
Empirical laws developed by
Johannes Kepler based on the
observational data of Tycho Brahe
Kepler’s Three Laws of
Planetary (and stellar) Motion
1st Law: The
planets move
in elliptical
orbits with the
Sun at one
focus.
2nd Law: A line
drawn from a
planet to the
Sun sweeps
out equal
areas in equal
times
3rd Law: The ratio of the
square of the orbital
period to the cube of
the semimajor axis is
the same for all planets
Using Kepler’s
rd
3
Law
2
P
k
3
A
So what is k? Newton eventually showed that k is related
to the mass of everything inside the orbit of the planet.
P
4

3
r
GM *
2
2
M* is the mass of everything inside the
orbit of the planet. Since planetary
masses are so small, this is effectively
the mass of the Sun
For objects orbiting other
stars?
P
4

3
r
GM *
2
2
This can be solved for the
mass
4 r
M* 
2
GP
2 3
For a planet orbiting another star, if r is in meters and P is
in seconds, this gives the mass of the star in kilograms. If
you want the mass in solar masses, use r in AU and P in
years then
3
r
M*  2  in solar masses 
P
Example Problem
The first extra-solar planet discovered orbits the
star 51 Pegasi. If the semimajor axis is 0.052 AU
and the orbital period is 4.23 days, what is the
mass of 51 Pegasi (in solar masses and in kg)?
Example Solution 1
Since G has units of (Nm2/kg2), distances must be in
meters and periods in seconds so do unit conversions
first 0.052 AU 1.49598 1011 m  7.779 109 m
AU
4.23days  86400 s day  365472 s
Next, chose the equation to use.
4 2 r 3
M
GP 2
3
2 numbers
9 to solve
Finally,
plug
in
2 3
4

7.779

10
m


4 r
M
GP 2


6.673 1011 Nm
2.09 1030 kg
 1.05M Son
30 kg
1.99 10
M Sun
2
kg
2
 365472s 
2
 2.09  1030 kg
Example Solution 2
If we want the mass in solar masses, use AU and years.
Orbital radius is already give in AU so just convert the
period into years. 4.23days  365.25 days  0.011581years
year
Now choose the appropriate equation and plug in numbers.
M* 
3
 0.052 AU 
3
r
in solar masses  
 1.058M Sun  1.06M Sun
2
2 
P
 0.11581years 
Suppose we have two stars
orbiting a common center?
Kepler’s 3rd Law will give the combined mass of the system.
To get the individual masses, we need more information
We could use their speeds as
the second piece of info
In this case it is an
inverse relationship
vA : vB  mB : mA
Example
A binary system is observed for a number of years and it
is found that one star appears to orbit the other at a
distance of 10.0 AU every 5.00 years. From spectroscopic
data it is found that one star moves at 25.0 km/s while to
other star moves at 100.0 km/s. What are the masses of
the two stars?
This is the position
measurements for the star
70 Ophiuchi showing how
one star appears to move
around the other
Example Solution 1
First find the combined mass of the system using
Kepler’s 3rd Law
10.0 AU   40M  M  M
r3
M*  2  in solar masses  
Sun
A
B
2
P
 5.00 years 
3
Now use the ratio of their velocities to find the
individual masses
M A vB 100.0 km s


 4.00M B  M A
km
M B vA 25.0 s
M A  M B  4.00M B  M B  5.00M B  40M Sun
M B  8.00M Sun
M A  32.0M Sun