A schematic drawing of the major sections inside the earth
Region | Thickness | Constituents |
---|---|---|
Inner core | 1300 km | Solid Iron & Nickel |
Outer core | 2250 km | Liquid Iron & Nickel |
Mantle | 2900 km | Rocky (heavy elements) |
Crust | 8-40 km | Rocky (lighter elements) |
Atmosphere | 16 - 480 km | Gasses |
Fraction (by mass) | Fraction (by number) | Species |
---|---|---|
75.5 % | 78.1 % | N2 |
23.1 % | 20.9 % | O2 |
1.3 % | 0.93 % | Ar |
0.05 % | 0.04 % | CO2 |
trace | trace | Ne, He, CH4, Neon, He, Methane, Krypton |
The Atmosphere of the Earth
Simulation of the geo-dynamo fields
Contemporary Physics, 1997, 38 4, Glatzmaier, G. and Roberts, P. http://dx.doi.org/10.1080/001075197182351
The Dynamo Theory: Moving charges.
Magnetic field flips
USGS
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.)
Illustration of Earth's magnetic field
ESA/C. T. Russel
Van Allen Probes
Johns Hopkins University Applied Physics Laboratory
Van Allen Probe - Science Overview
Van Allen Probe - Instruments
The spheroid allows for a torque, which leads to a precession.
The North Celestial Pole
https://in-the-sky.org/article.php?term=precession_of_the_equinoxes
The North Pole Star changes! Just very slowly.
Lunar internal structure
Moon Formation
By NASA/JPL-Caltech - http://www.nasa.gov/multimedia/imagegallery/image_feature_1454.html, Public Domain, https://commons.wikimedia.org/w/index.php?curid=8626942
Luna 2: Soviet Moon Probe, September 12, 1959
By NASA - http://nssdc.gsfc.nasa.gov/database/MasterCatalog?sc=1959-014A, Public Domain, https://commons.wikimedia.org/w/index.php?curid=16509292
On the lunar surface
NASA: flickr album
Scale Image of Earth and Moon
A torque will arise if the two angular velocities are different, due to the tidal bulge.
Lunar Laser Ranging Experiment with the stereo camera in the background (NASA image number AS11-40-5952). This Retroreflector was left on the Moon by astronauts on the Apollo 11 mission. Astronomers all over the world have reflected laser light off the reflectors to measure precisely the Earth-Moon distance.
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: $$\begin{equation} \frac{dr}{dt} \approx 4 \; \textrm{cm} / \; \textrm{yr} \end{equation}$$The moon in orbit showing its phases.
The phases of the moon in 2024
This high-resolution mosaic of WAC images shows Mercury as it appeared to MESSENGER as the spacecraft departed the planet following the mission's first flyby of Mercury.
Credit: NASA/Johns Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington
Mariner 10 spacecraft - launched November 3, 1973, Mercury 1st flyby March 1974
By NASA / Jet Propulsion Laboratory - http://solarsystem.nasa.gov/multimedia/display.cfm?IM_ID=1563http://www.nasa.gov/multimedia/imagegallery/image_feature_1483.html, Public Domain, https://commons.wikimedia.org/w/index.php?curid=1182348
The complicated trajectory to get to Mercury.
The 1st science orbit
By NASA, Jet Propulsion Laboratory, Applied Physics Laboratory - MESSENGER Website, Public Domain, https://commons.wikimedia.org/w/index.php?curid=15235525
MESSENGER has provided multiple lines of evidence that Mercury’s polar regions host water ice. The permanently shadowed craters near Mercury’s north pole have thermal environments that allow water ice to be stable in these craters either at the surface or a few tens of centimeters below the surface.
NASA/Johns Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington
Mercury rotates 3 times during 2 orbits.
Orbital and Rotational Periods | |
---|---|
Orbital Period | 87.969 Earth Days |
Rotation Period | 58.646 Earth Days |
NASA/Goddard Space Flight Center Scientific Visualization Studio.
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) |
A planet is located a distance $D$ away from an energy source.
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.)
A planet is located a distance $D$ away from an energy source.
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.
https://www.weather.gov/jetstream/absorb
The Soviet Union's Venera 14 probe captured two color panoramas of Venus's surface in 1982. This panorama came from the rear camera. Russian Academy of Sciences / Ted Stryk
https://www.planetary.org/space-images/venus-surface-panorama-from-venera-14-camera-2
Venera Mission
Mars, from space. Mosaic of the Valles Marineris hemisphere of Mars projected into point perspective, a view similar to that which one would see from a spacecraft. The distance is 2500 kilometers from the surface of the planet, with the scale being .6km/pixel.
Credit: NASA/JPL-Caltech
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 |
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.
It's red: Iron Oxide aka rust.
Polar Caps: water ice and CO2
Mars may have had oceans
NASA's Goddard Space Flight Center
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.
Mars' Larger moon Phobos
European Mars Express Mission
The small moon Deimos
European Mars Express Mission
Two relatively small moons: Phobos and Deimos