Celestial Mechanics

PHYS390 (Astrophysics)

Professor Lee Carkner

Lecture 5

Questions

1) Can we achieve enormous resolving power by dramatically increasing N?

Answer: NoExplain: The more slits, the larger the grating has to be.

Large gratings require the light to be more spread out, need higher intensity.

2) Can we achieve enormous resolving power by using very high orders?

Answer: NoExplain: Increase n, decrease brightness, need higher

intensity to use higher orders.

Test #1

Next Wednesday, Dec 1 Multiple choice (25%)

Like in-class questions

Problems (75%) Like PALs and homework

Sample equation sheet on web page Just need pencil and calculator

Kepler’s Laws In the 1600’s Johannes Kepler

used Tycho Brahe’s data to find the laws of planetary motion

Kepler’s First Law

Half of the longest (major) axis is the semi-major axis called a

Half the distance between the foci is called c

e = c/a

e = 0 is circle (a=r) e = 1 is line

c a

Kepler’s Third Law

There is a relationship between the orbital size and period

Only in the solar system

and if P is in years and a is in AU

P2=[42/G(m1+m2)] a3

Where m1 and m2 are the masses of the two objects

Reduced Mass Consider two masses m1 and m2

located distances of r1 and r2 from the center of mass of the system

We can define the reduced mass,

= m1m2/(m1+m2)

r1 = -(/m1)rr2 = (/m2)r

Can think of as the reduced mass () orbiting the total mass (M)

Angular Momentum

Orbiting objects have angular momentum In general:

e.g., a single mass, m in a circular orbit of

radius, r:

so for two bodies, total L is:

L = m1r1v1 + m2r2v2

L = [GMa(1-e2)]½

Velocity

Energy of reduced mass

The total system energy is then:

E = -(Gm1m2/2a)

v2 = GM[(2/r)-(1/a)]

Visual Binaries Consider a binary star composed

of two masses in two elliptical orbits around the center of mass (cm)

m1/m2 = a2/a1

1 = a1/d

m1/m2 = 2/1

even if we don’t know the distance, if we can measure ’s we can find the mass ratio

cm

Finding Mass

If we do know the distance, we can get a1 and a2

P2 = (42a3)/[G(m1+m2)]

Have to use m1+m2 and m1/m2 to pull out m’s separately

Inclination

The observed subtended angle is now

ά = cos i then a = d = άd/cos i

m1+m2 = (42/G)(d/cos i)3(ά3/P2)

Next Time

Read: 7.2-7.3 Homework: 2.6, 2.14, 7.4, 7.7, 7.12,