Why has its magnetic field existed over at least 70% of geological time?
Why is it predominantly dipolar?
What determines its strength?
Why does its strength vary, but by so little?
Why does the magnetic compass needle point
approximately North?
Why does the averaged geomagnetic axis coincide
with the geographical axis?
Why does the polarity of the Earth’s field reverse?
What happens to the geomagnetic field during a
reversal and why?
Why does the frequency of reversals vary so greatly
over geological time?
Why is neither polarity of field favored over the
other?
What causes the slow secular change of the field?
What is the significance of the westward drift?
Can a single mechanism explain why other planets
and satellites are magnetic too?
The Earth's Magnetic field
Strength: 25 to 65 microteslas (0.25 to 0.65 gauss).
The north magnetic pole is in Greenland
It flips sometimes. (Last time was about 780,000 years ago.)
What does it do?
Van Allen Radiation Belts
Earth's shape
Precession
Precession
The North Pole Star changes! Just very slowly.
The Moon
Moon Structure
Formation
Giant-impact hypothesis
Earth's spin and moon's orbit have similar orientations
Moon samples indicate that the Moon's surface was once molten.
The Moon has a relatively small iron core.
The Moon has a lower density than Earth.
Evidence exists of similar collisions in other star systems (that result in debris disks).
The stable-isotope ratios of lunar and terrestrial rock are identical, implying a common origin.
Explorations
First Touch
Apollo Program
1961 to 1972
First manned flight: 1968
First landing: Apollo 11, July 1969
Last landing: Apollo 17, Dec 1972
Missions brought back 842 lbs of rocks
Left several science experiments
Installed Retro-Reflectors
Earth Moon System
Tides
Tidal Locking
Moon is leaving
3.8 cm/year
Tidal Braking
Over time, the tidal bulge will create a counter-torque to the rotation. This will cause the central object to rotate just a little bit slower. We can measure this:
$$\begin{equation}
\frac{dP_\textrm{rot}}{dt} = 0.0016 \; \textrm{s} / \; \textrm{century}
\end{equation}$$
But this implies a change in $L$! Where does it go?
Let's start with the angular momentum $L$ of the moon in orbit around the earth. It depends on the mass of the moon, its speed and its distance from the center of rotation.
$$\begin{equation}
L_\textrm{orbit} = M_\textrm{moon} v r = M_\textrm{moon}\left( \frac{G M_\textrm{E} }{r}\right)^{(1/2)}r = \left( G M_\textrm{E} M_\textrm{moon}^2 r\right)^{(1/2)}
\end{equation}$$
Consider the change in angular momentum as a function of time:
$$\begin{equation}
\frac{dL_\textrm{orbit}}{dt} = \left(G M_\textrm{E} M_\textrm{moon}^2\right)^{(1/2)}\frac{1}{2}r^{-1/2}\frac{dr}{dt}
\label{eq:moondLdt}
\end{equation}$$
If the earth and moon are considered the only two bodies in the system, then any change in ang. momentum of the moon will have to be balanced by a change in angular momentum of the earth, i.e. the earth's rotation speed.
$$\begin{equation}
\frac{dL_\textrm{orbit}}{dt} = -\frac{dL_\textrm{rot}}{dt}
\end{equation}$$
For a rotating body, the angular momentum is a product of the moment of inertia, $I$, and the angular velocity, $\omega$:
$$\begin{equation}
L_\textrm{rot} = I \omega
\end{equation}$$
Considering the earth a sphere:
$$\begin{equation}
I = \frac{2}{5}M_\textrm{E} R_\textrm{E}^2
\end{equation}$$
which leads to:
$$\begin{equation}
L_\textrm{rot} = \frac{2}{5}M_\textrm{E} R_\textrm{E}^2 \left( \frac{2 \pi }{P_\textrm{rot}}\right)
\end{equation}$$
where $P_\textrm{rot}$ is the rotational period. Considering the first time derivative:
$$\begin{equation}
\frac{d L_\textrm{rot}}{dt} = \frac{4 \pi M_\textrm{E} R_\textrm{E}^2}{5} \left(- \frac{1}{P_\textrm{rot}^2} \frac{dP_\textrm{rot}}{dt} \right)
\end{equation}$$
Using \eqref{eq:moondLdt}, and the conservation of angular momentum, we obtain:
$$\begin{equation}
\frac{\left(G M_\textrm{E} M_\textrm{moon}^2\right)^{1/2}}{2r^{1/2}}\frac{dr}{dt} = \frac{4 \pi M_\textrm{E} R_\textrm{E}^2}{5 P_\textrm{rot}^2} \frac{dP_\textrm{rot}}{dt}
\end{equation}$$
Rearranging for $dr/dt$:
$$\begin{equation}
\frac{dr}{dt}= \frac{8 \pi}{5} \left(\frac{M_\textrm{E} r}{G} \right)^{1/2}\frac{R_\textrm{E}^2}{M_\textrm{moon} P_\textrm{rot}^2} \frac{dP_\textrm{rot}}{dt}
\label{eq:drdtmoon}
\end{equation}$$
$dP/dt$ is measurable:
$$\begin{equation}
\frac{dP_\textrm{rot}}{dt} = 0.0016 \; \textrm{s} / \; \textrm{century}
\end{equation}$$
Putting this into \eqref{eq:drdtmoon} yields a rate of motion as:
$$\begin{equation}
\frac{dr}{dt} \approx 4 \; \textrm{cm} / \; \textrm{yr}
\end{equation}$$
Moon Phases
Mercury
The Messenger Mission
Mercury - stats
Atmosphere: None!
Temperature
Daytime: 700 K at equatorial noon
Nighttime: 100K at polar regions
Interior:
Little is known (hard to get to)
Probably has iron-rich interior
Probably has very thin silicate crust
Probably has some molten material in core – observed magnetic field is weak
Orbital and Rotational Periods
Orbital Period
87.969 Earth Days
Rotation Period
58.646 Earth Days
Venus
Physical Parameters
Mass
$4.867 \times 10^{24} \; \textrm{kg}$
$0.815 M_\textrm{E}$
Radius
$6,051.8 \; \textrm{km}$
$0.9499 M_\textrm{E}$
Year
224.701 days
0.615 years
Rotation
once every 243 Earth days
Clockwise!
Eccentricity
0.006772
(nearly circular)
Magnetic Field
None
(due to slow rotation)
Greenhouse Effect
How hot would a planet be, if it didn't have an atmosphere?
$a$ is the albedo, or how much energy is reflected.
(1 = all of it, 0 = none of it.)
The Luminosity of the sun is:
$$\begin{equation}
L_\textrm{Sun} = 4 \pi R_\textrm{Sun} \sigma_\textrm{SB} T_\textrm{Sun}^4
\end{equation}$$
Now consider a planet a distance $D$ away from the sun. The power received by the planet would just be the ratio of its cross section to a sphere of radius $D$.
The energy received per second will be
$$\begin{equation}
P_\textrm{in} = L_\textrm{Sun} (1-a)\left(\frac{\pi R_p^2}{4 \pi D^2} \right)
\end{equation}$$
If the planet radiates as a blackbody, then:
$$\begin{equation}
P_\textrm{out} = 4 \pi R_p^2 \sigma_\textrm{SB} T_p^4
\end{equation}$$
In thermodynamic equilibrium: $P_\textrm{out} = P_\textrm{in}$
Thus, we can say:
$$\begin{equation}
L_\textrm{Sun} (1-a)\left(\frac{\pi R_p^2}{4 \pi D^2} \right) = 4 \pi R_p^2 \sigma_\textrm{SB} T_p^4
\end{equation}$$
Solving for $T_p$
$$\begin{equation}
T_p = T_\textrm{Sun}(1-a)^{1/4} \sqrt{\frac{R_\textrm{Sun}}{2 D}}
\end{equation}$$
Putting number relevant to the Earth and Sun in to this equation, yields a temperature for the earth equal to 255K = -19° C=-1° F. Obviously, we're missing something from this analysis: the greenhouse effect.
Take away: the greenhouse effect is real.
Venus' Atmosphere
Top of clouds is 250 K, however, at the surface the temperature is 750 K, above the melting point of Lead.
More volcanos than any other object in the solar system. (>1500 major)
Unclear if they are active or not.
Mars
Physical Parameters
Mass
$6.42 \times 10^{23} \; \textrm{kg}$
$0.107 M_\textrm{E}$
Radius
$3397 \; \textrm{km}$
$0.533 M_\textrm{E}$
Year
686.9 days
1.8807 years
Rotation
24 h 37 min 22.6 s
Eccentricity
0.0934
Magnetic Field
Weak
Probably larger in the past
Mars' Atmosphere
The atmosphere of Mars is mainly composed of carbon dioxide (95 %). It contains only 2 % nitrogen and 0.1–0.4 % oxygen.
The air pressure is only 5–8 mbar. Much has been lost. No greenhouse effect.
Mars' Surface
It's red: Iron Oxide aka rust.
Polar Caps: water ice and CO2
Indications are that water has flowed on Mars in the past. Geologic features like river beds and other erosion indicators suggest this. No liquid water has been found on the surface. Too cold and the pressure is too low.