Some info and problems

1
Some info and problems to go with last nights
presentation on Extrasolar planets.
I worked on this most of the day today. I plan to
use some of this in my class presentation . Jim
Honeycutt
2
Properties of Ellipses
semi-major axis a 2 times a is the length of
major axis
semi-major axis
a
2a
2a is major axis
3
Most planetary orbits are fairly close to being a
circle. Comet orbits are the most elliptical.
Perihelion is when the object is closest to the
Sun, and Aphelion is when the object is farthest
from the sun. Add these two numbers together ,
and divide by 2 gives you a, which is the average
distance. If a comet came as close to the Sun as
.5 au but travels as far away as 7.8 au, then a
(.5 7.8)/2 a 4.15 au
aphelion
perihelion
Sun
a aphelion perihelion
_______________________
2
4
Using Keplers Third Law
P2 a3
The square of the period of a revolving objects
equals the distance of one of the objects from
the other cubed.
This is a simple problem that only works if you
do the following P must be in earth years, and
must be in au. An au is the distance from the
earth to the Sun (93 million miles). Mars is 1.5
au , and Jupiter is 5 au. So lets use the
formula on the comet that was 4.15 au from the
Sun. a 4.15, find the period.
a
The period is 8.45 years
5
Lets work in the opposite direction The period
of Halleys Comet is 75 years. What is the
comets average distance from the Sun?
P2 a3
P 75, find a
Since in my text I find that Uranus is 19.19 au
from the Sun, Halleys Comet goes almost all the
way out to Uranus orbit. This formula works not
only in our solar system, but for binary systems
of stars, or star, and planet.
6
Orbits and Masses of Binaries
The primary importance of binaries is that they
allow us to measure stellar parameters
(especially mass).
Both stars in a binary system revolve around
their center of mass. The center of mass is the
place where a fulcrum would be placed to balance
the two stars. This location depends upon how
much more massive one object is over the other.
For our solar system the center of mass is inside
the Sun.
7
This formula is Newtons modification of Keplers
Law.
A must be in au, P in years, and Mass (M) must be
in solar masses. So, when we calculate mass we
will be comparing it to the mass of the Sun. This
simplifies things.
Epsilon Eridani has a planet orbiting around
it.The planet is 3.4 au from the star, and has a
period of 7.1 years. Find the mass of the
star.First, since the mass of the planet,
compared to the star is extremely small we can
ignore the planet.
8
Sirius is orbited by a White Dwarf star.The star
is 20 au from Sirius , and has a period of 50
years. Find the mass. Since the mass of the WD
can not be neglected, we will find the combined
mass.
We could now look up information about bright
stars.A spectrum of Sirius reveals it is a A0,
and plotted on an HR diagram its mass corresponds
to 2.6 . So, the White Dwarf has a mass of
3.2-2.6 0.6 solar masses.
9
The Masses of Stars
Spectral Type Mass in Solar Masses
O5 40
B5 7.1
A5 2.2
F5 1.4
G5 0.9
K5 0.7
M5 0.2
10

Clevis in the program , Marcy said that the short
curve after Jupiter was due to Saturn, this
should be it.
12 years
. . . Jupiter
(Meters/sec)
Need Telescope . . .
11
Here is one of the graphs from last night 16
Cygni B Note p 2.2 years


Mass 1.7 MJUP (Min)
Not Sinusoidal
Orbit Period
Velocity Wobble
2.2 yr
16 Cygni B
12
Lets do our calculations , and see what happens.
P 2.2 years
P2 a3
Other information stated that the closet the
planet comes to the star is .6 au and the
farthest is 2.7 au. If you add .6 2.7 3.3
this is the peri , and aphe . Divide this number
by 2 you get 1.65 . Our answer for a was 1.69 ,
not bad for a hand calculator. They probably used
more accurate data..
On the internet I found the mass of 16 Cygni to
be 1 solar mass. I need a better accurate number
than 2.2 years.( some more decimals)
13
(No Transcript)