Celestial Mechanics

Mul Apin tablet

http://www.mesopotamia.co.uk/astronomer/explore/exp_set.html

Kepler 1571-1627

Note that all of the discussion below is in terms of the solar system, but all results apply to stellar systems

Corresponded with Brahe and acquired records after his death. (i.e. refused to give them up to his heirs)

Keplers laws :

Planets move in ellipses, with one focus at the sun

Keplers laws :

Ellipses: a circle is a point moving so that its distance from one point is constant. An ellipse is a point which moves so that the sum of its distance from two points is constant

Relations for ellipse

\color{red}{ r_1 + r_2 = 2a}
a is semi-major axis
b is semi-minor axis
(why?)
Eccentricity e measures deviation from a circle
\color{red}{ b = a\left( {1 - e^2 } \right)^{1/2} }
(e = 0 means a circle)
Equation is $$\color{red}{ r = \frac{{a\left( {1 - e^2 } \right)}}{{1 + e\cos \left( \theta \right)}} }$$

Keplers laws :

Second Law

A vector drawn from the planet to the sun will sweep out equal areas in equal times
A = B = C Means "Planet moves faster closer to sun"

Third Law

The period (P) and the semi-major axis (a) are related by

\color{red}{ \frac{{P^2 }}{{a^3 }} = Const}

Note that these are empirical laws: there is no indication of where they come from...

Newton

1642-1727 (born the day of Galileo's death)

Newton (in Principia) showed that the motion of the planets can be undestood in terms of the gravitational force at the earth's surface.

Universal Gravitation + Mechanics ⇒

Keplers 3rd law
Keplers 2nd law
Keplers 1st law
Energy and Virial Theorem
Tidal forces and Roche's limit
Stability of orbits

\color{red}{ \vec F = m_{} \vec a_{} = - \frac{{GMm_{} }}{{\left| {r_{} } \right|^2 }}\hat r_{} }

The fundamental relations:

But now we want to have two bodies moving under mutual gravitation, so that the force of m₂ on m₁ is

\color{red}{ \vec F_{21} = m_2 \vec a_2 = - \frac{{Gm_1 m_2 }}{{\left| {r_{12} } \right|^2 }}\hat r_{12} }

with the changing velocity and acceleration of m₁
due to m₂ shown thus:

Centre of Mass

Define Relative and Centre of Mass variables.
\color{red}{ \vec R = \frac{{m_1 \vec r_1 + m_2 \vec r_2 }}{{m_1 + m_2 }},\vec r = \vec r_1 - \vec r_2 }
where R is the vector for the centre of mass, and r is the vector describing the relative displacement of one body from the other.
Then by differentiating these, we can get the C.of M. and relative velocities
\color{red}{ \vec V = \frac{{m_1 \vec v_1 + m_2 \vec v_2 }}{{m_1 + m_2 }},\vec v = \vec v_1 - \vec v_2 }
and accelerations.
\color{red}{ \vec A = \frac{{m_1 \vec a_1 + m_2 \vec a_2 }}{{m_1 + m_2 }},\vec a = \vec a_1 - \vec a_2 }

The C.of M. motion can be eliminated immediately, so that A = 0 (no C.of M. acceleration.)., and the motion of the bodies follows the pattern shown. with the motion of the centre of mass shown by the series of X's (ie constant motion, no acceleration).

The Relative motion contains all of Kepler's laws:

Kepler's Third Law.

It is theoretically justified for circular orbits via:

\color{red}{ \frac{{d^2 \vec r}}{{dt^2 }} = \vec a = \vec a_1 - \vec a_2 = - \frac{{Gm_1 }}{{\left| {r_{} } \right|^2 }}\hat r_{} - \frac{{Gm_2 }}{{\left| {r_{} } \right|^2 }}\hat r_{} = - \frac{{GM}}{{\left| {r_{} } \right|^2 }}\hat r}

so
\color{red}{ \frac{{mv^2 }}{r} = \frac{{GMm}}{{\left| {r_{} } \right|^2 }},v = \frac{{2\pi r}}{P} \Rightarrow P^2 = \frac{{4\pi ^2 }}{{GM}}r^3 }
Two changes from Kepler
1. We know the constant.
2. The mass is the total mass of the system (not just the sun): VERY important, since it will let us deduce mass of stars

Second Law

\color{red}{ \vec F = m_{} \vec a_{} = - \frac{{GMm_{} }}{{\left| {r_{} } \right|^2 }}\hat r_{} }

Take × product with r
i.e. Angular momentum doesn't change: L=const

\color{red}{ \begin{array}{l} \vec L = m\vec r \times \vec v \Rightarrow \\ \frac{{d\vec L}}{{dt}} = m\left( {\frac{{d\vec r}}{{dt}} \times \vec v + \vec r \times \frac{{d\vec v}}{{dt}}} \right) \equiv 0 \\ \end{array}}

What does this have to do with 2nd Law? Area of triangle

\color{red}{ A_\Delta = \frac{1}{2}\left| {\vec r \times \vec vdt} \right|}

so if dt is fixed,

\color{red}{ A_\Delta = \frac{1}{{2m}}\left| {\vec L} \right|}

is constant

1st Law:

MUCH HARDER

Split into tangential and radial components

\color{red}{ \vec v = v_r \hat r + v_\theta \hat \theta = \frac{{dr}}{{dt}}\hat r + r\frac{{d\theta }}{{dt}}\hat \theta }

\color{red}{ E = \frac{1}{2}m\vec v.\vec v = \frac{1}{2}m\left( {\left( {\frac{{dr}}{{dt}}} \right)^2 + r^2 \left( {\frac{{d\theta }}{{dt}}} \right)^2 } \right)}

\color{red}{ \vec L = m\vec r \times \vec v = r^2 \frac{{d\theta }}{{dt}}\hat \theta }

So we can write the energy and ang mom. conservation laws in terms of the new variables.

Conservation of energy:

\color{red}{ E = \frac{1}{2}mv^2 - \frac{{GMm}}{r}}

\color{red}{ \frac{{dE}}{{dt}} = m\left( {\vec v.\frac{{d\vec v}}{{dt}} - GM\frac{d}{{dt}}\left( {\frac{1}{{\left| {\vec r} \right|}}} \right)} \right)}

Take dot product with v:
\color{red}{ \vec v.\frac{{d\vec v}}{{dt}} = \vec v.\frac{{d^2 \vec r}}{{dt^2 }} = - \frac{{GM}}{{\left| {r_{} } \right|^2 }}\vec v.\hat r}
and
\color{red}{ \frac{d}{{dt}}\left( {\frac{1}{{\left| {\vec r} \right|}}} \right) = - \frac{{\vec v.\hat r}}{{\left| {r_{} } \right|^2 }}}

Can combine these to give...

\color{red}{ \frac{{dE}}{{dt}} = 0}

\color{red}{ \left| {\vec L} \right| = m\left| {\vec r \times \vec v} \right| \Rightarrow mr^2 \frac{{d\theta }}{{dt}}}

Energy is a conserved quantity. define

\color{red}{ J = \frac{{\left| {\vec L} \right|}}{m}}

We now need to combine the last 2 equations and energy conservation to give us...

\color{red}{ E = \frac{1}{2}mv^2 - \frac{{GMm}}{r} = \frac{1}{2}m\left( {\frac{{dr}}{{dt}}} \right)^2 + \frac{{L^2 }}{{r^2 }} - \frac{{GMm}}{r}}

(look familiar?) so

\color{red}{ \begin{array}{l} \frac{{dr}}{{dt}} = \sqrt {2E - \frac{{L^2 }}{{r^2 }} + \frac{{2GM}}{r}} \\ r^2 \frac{{d\theta }}{{dt}} = J \\ \end{array}}

This is correct, but not very interesting: to get the shape of the orbit, we need to divide the last two equations.
\color{red}{ \frac{1}{{r^2 }}\frac{{dr}}{{d\theta }} = \sqrt {\frac{{2E}}{{J^2 }} - \frac{1}{{r^2 }} + \frac{{2GM}}{{rJ^2 }}} }
To solve this is a mess: you need to make the following substitution.
\color{red}{ u = \frac{1}{r},u_0 = \frac{{GM}}{{J^2 }},e = \left( {1 + \frac{{2EJ^2 }}{{G^2 M^2 }}} \right)^{1/2} }

and this gives the solution...
\color{red}{ \frac{1}{r} = \frac{1}{{r_0 }}\left( {1 + e\cos \left( \theta \right)} \right)}
(Wasn't Newton a smart cookie!)

Comments: The fundamental quantities are J, the angular momentum, and E, the total energy.

All the conic sections are hidden in this:

\color{red}{e = 0} ⇒ circle ⇐ \color{red}{E = - \frac{{G^2 M^2 }}{{2J^2 }}}
\color{red}{0 < e < 1} ⇒ ellipse ⇐ \color{red}{0 > E > - \frac{{G^2 M^2 }}{{2J^2 }}}
\color{red}{e = 1} ⇒ parabola ⇐ \color{red}{E = 0} (many comets are very close to this: cannot distinguish closed elliptical orbits with P > 1000000 years from parabola)
\color{red}{1 < e} ⇒ hyperbola ⇐ \color{red}{0 < E}. Note that if E > 0, then the system is unbound.

Energy and the Virial theorem

Total energy of the moon in orbit = K.E. + P.E.

\color{red}{ K.E. = \frac{1}{2}mv^2 ,P.E. = - \frac{{GMm}}{r}}

and

\color{red}{ \frac{{mv^2 }}{r} = \frac{{GMm}}{{r^2 }}}

so
\color{red}{ K.E. = - \frac{1}{2}P.E.}
This result is much more general than the "derivation" suggests. For any bound system interacting via gravitational forces
\color{red}{ \left\langle {K.E.} \right\rangle = - \frac{1}{2}\left\langle {P.E.} \right\rangle }
e.g. how much energy would you get by changing the orbit of Mars to that of Earth?
e.g. If a star shrinks, (so that -\color{red}{\left\langle {P.E.} \right\rangle } increases) it will heat up (so that \color{red}{\left\langle {K.E.} \right\rangle } increases) and ¹/₂ the energy must be radiated into space.

Tidal Effects

Newton thought of

water on the near side of earth as being pulled away by the moon,
the earth being pulled away from the water on the far side,
gives tides on opposite sides of the earth.

More accurately, there is a force on every point on the earth, and one must subtract the force on the centre from this.Note that the effect is really that the water at ± 90⁰ (with the moon to the right at 0⁰) is pulled flat!
The size of the effect is given by

\color{red}{ \delta \vec F = \frac{{GMm_{} }}{{\left| {r_{} } \right|^2 }}\hat r_{} - \frac{{GMm_{} }}{{\left| {r_{} + \delta r} \right|^2 }}\hat r \Rightarrow \left| {\delta \vec F} \right| = - - \frac{{2GMm_{} }}{{\left| {r_{} } \right|^3 }}\delta r}

Note that δr is the radius of the object, r is the distance from the mass. Also note that the Moon is more important than the sun because of the 1/r³

Babylonians Observed total Eclipse 15 April 136 BC.
But they shouldn't have!

Earth's rotation has slowed down, by 1/100 sec/century, because of tidal effects! i.e. earth isn't a very good time-keeper
Long term effects of tidal friction: Friction extracts energy from the system, so the rotational speed of the earth decreases.

What is the energy loss?
\color{red}{ \omega = \frac{{2\pi }}{P} \approx 7.27 \times 10^{ - 5} }
so angular acceleration
\color{red}{ \frac{{d\omega }}{{dt}} = - \frac{{2\pi }}{{P^2 }}\frac{{dP}}{{dt}} = 2.67 \times 10^{ - 21} rad/s^2 }
- Hence the power consumption:
  \color{red}{ P = \frac{{dE}}{{dt}} = \frac{{d\left( {\frac{1}{2}I\omega ^2 } \right)}}{{dt}} = I\omega \frac{{d\omega }}{{dt}} \approx 2.5 \times 10^{17} W}
  (note: Power consumption of human race ~10¹⁵ W!)
Angular momentum must remain constant (no external torques): how can this happen?
- For the earth-moon system
  \color{red}{ L = mvr + I\omega }
  and so if ω decreases, vr must increase.
- How much?
- \color{red}{ \frac{{mv^2 }}{r} = \frac{{GMm}}{{\left| {r_{} } \right|^2 }} \Rightarrow v = \left( {\frac{{GMm}}{{\left| {r_{} } \right|}}} \right)^{1/2} }
  so
  \color{red}{ L = m\left( {GMr} \right)^{1/2} + I\omega }
- Hence r must increase: i.e the moon moves away from the earth. How much?
- \color{red}{ \frac{{dL}}{{dt}} = 0 = m\left( {GMr} \right)^{1/2} \frac{{dr}}{{dt}} + I\frac{{d\omega }}{{dt}}}
- so v_r~7 nm/s ~ 4 cm/yr.
- Note this is a subtle effect: if the earth was NOT rotating, the moon would be dragged towards the earth
Tidal effects and solid bodies:
e.g. LEP at CERN
It was noticed that the energy of the accelerator was changing on a 12 hour basis (ΔE/E ≈ 10^-4)
This corresponds to a change of ΔR ≈ 600 µm (to be compared with 6.5 km for the ring radius.)

Tidal Forces and Roche's limit
If the difference between the sides of the moon is larger than the cohesive forces holding it together, it will fall apart.
- Small objects are held together by inter-molecular forces: these are very strong, but short-range (i.e. they don't accumulate).
- Roches limit: e.g. assume the "moon" is made of two spheres:
- Assume the cohesive gravitational force holding one sphere together = the differential gravitational force on that sphere due to the other.
  Hence the minimum distance before the moon breaks up is given by...
  
  δr is the radius of the moon,
  
  R be the radius of the earth
  
  d is the minimum distance before breakup occurs
- The cohesive force must exceed the tidal force, so to find when breakup occurs, assume they are equal.
- \color{red}{ \begin{array}{l} \delta \vec F = - \frac{{G\frac{m}{2}\frac{m}{2}}}{{\left| {\delta r_{} } \right|^2 }} - \frac{{2GMm_{} }}{{\left| d \right|^3 }}\delta r \\ d = \left( {\frac{{2M}}{m}} \right)^{1/3} \delta r \approx 2.5R\left( {\frac{{\rho _{earth} }}{{\rho _{moon} }}} \right)^{1/3} \\ \end{array}}
- Note Saturn's rings lie inside the Roche limit for a moon : probably not a moon that was pulled apart but one that could never form.
Note the cohesive force is given by internal forces until the moon is about 100 km in radius.
Small moons are irregular, large one are round
Eros Iapetus
The numerical constant depends on exactly what assumptions we make for the shape of the moon: obviously ours isn't very sensible. (but only change answer by about 50% )
Density is crucial: see later on. A high density star (white-dwarf or neutron star) can rip apart much heavier low density companion!

Instability

Similar idea: if a satellite moves too far away, then it can be dragged away from the planet.

We can consider this as a tidal problem again: we need that the tidal force dragging the moon away from the earth > the attractive gravitational force.
The cohesive force must exceed the tidal force, so to find when breakup occurs, assume they are equal. (D is the earth-sun distance, d is the minimum earth-moon distance before breakup occurs).
\color{red}{ d = \left( {\frac{{M_{earth} }}{{M_{sun} }}} \right)^{1/3} D}
What is the maximum distance for the moon?
d = 1.7 × 10⁹ m (≈ 4 × current distance)
When does this happen (assuming, wrongly, that energy loss from the earth remains constant)?
The total distance ≈ 1.2 × 10⁹ m at .23 m/yr => 5.2 × 10⁹ yrs (≈ the age of the solar system)

Tidal locking

A (semi-)stable configuration will occur when tidal forces increase the orbital period of the moon and decrease the rotational period until the two are the same. e.g.

Earth-Moon, Jupiter-Io
It needn't be locked in 1:1 e.g. Mercury is in 1.5 (rotational) : 1.0 (orbital) lock.

Yet another effect of tides (and frictional forces in general): Orbits become more circular, since tides and hence energy oss are largest at closest approach

Resonances:

Usually attractions of other planets/moons is random and cancels out.

However if the period of (e.g) two moons are in a simple ratio, the force will be systematically applied. e.g.

\color{red}{ \frac{{d_1 }}{{d_2 }} = 1.58 \Rightarrow \frac{{P_1 }}{{P_2 }} = 2}

so (assuming M₁ is the lighter) it can be kicked out of its orbit.

We can see this (particularly) in Saturn's rings. Note no material in Cassini's division: This corresponds to a 2:1 resonance with Mimas (innermost large moon of Saturn).

Precession:

The moon exerts a torque on the bulge of the earth, causing it to precess in the same way as a top (calculation is messy: result is 26000 years).
The sun exerts torque on the earth-moon system, causing the plane of the moon's orbit to precess with a period of 18.6 yrs

General comments

Ideal Newtonian 2-body systems are totally stable
3-body systems are almost always unstable
Tidal effects and (later) General Relativity cause even 2-body systems to change

Now we take a first look at stars

Celestial Mechanics

Kepler 1571-1627

Keplers laws :

Keplers laws :

Keplers laws :

Second Law

Third Law

Newton

Universal Gravitation + Mechanics ⇒

The fundamental relations:

Centre of Mass

Kepler's Third Law.

Second Law

1st Law:

Conservation of energy:

Energy and the Virial theorem

Tidal Effects

Tidal effects and solid bodies:

Tidal Forces and Roche's limit

Instability

Tidal locking

Resonances:

Precession: