Download Comparing Earth, Sun and Jupiter

The solar system
• Mass dominated by the Sun: an average star
• Planets are solid, almost spherical bodies
orbiting the Sun:
¾ Mercury: small, very close to Sun. Heavily
cratered, no detectable atmosphere
¾ Venus: most Earthlike in terms of size,
composition. Covered in thick clouds.
¾ Earth: only planet with liquid surface
water
¾ Mars: red surface due to high iron
content. Thin, dry atmosphere
¾ Jupiter, Saturn, Neptune, Uranus: Gas
giants with many small moons
• Smaller bodies found in the asteroid belt
between Mars and Jupiter.
¾ Often nonspherical, with orbits close to
the ecliptic plane
¾ Almost certainly pieces left over from
formation of SS.
• Most of the planets rotate in the same
direction as they orbit the Sun, and in the
same plane
¾ Exceptions are Venus, Uranus, (Pluto)
•
•
•
•
The Sun
Mostly Hydrogen and Helium
High luminosity due to central fusion reactions
Sunspots: cooler regions where magnetic field
is strongest
More features visible at other wavelengths
The Milky Way
• Our galaxy is characterized by a relatively thin disk of
stars and dust. This appears as a roughly linear
feature on the sky
¾ The dust obscures most of the light from these
stars. If there were no dust, the centre of the galaxy
would be as bright as the full moon
¾ Looking in infrared light, which is less affected by
dust, we can see the full structure of our Galaxy.
• Young stars and open clusters, in roughly circular
orbits about the centre, lie in the disk.
¾ Globular clusters and older stars form the halo,
a roughly spherical collection of stars, much
larger than the disk, on high velocity, radial
orbits.
Basic Geometry: the skinny triangle
• Most celestial objects have sizes of about 1“ (1
arcsecond) or less
ƒ 1 arcmin = 1/60 degree
ƒ 1 arcsec = 1/60 arcmin = 1/3600 deg
• An arc on the circle
subtends an angle θ.
• These are clearly
related by: s = θ
2πr
360
where θ is in degrees.
• If θ is expressed in degrees, the equation is
written as:
s
θ
=
r 57.3
• The chord D made by drawing a straight line
between the ends of the arc, s, produces an
isosceles triangle with 2 long sides, r.
r
D
θ
r
• In astronomy the distances are so large and
the angular diameters so small that s≈D.
• And then we have:
D
θ
=
2πr 360
Size and Shape of Earth
• Evidence that the Earth is round:
¾ Land at sea level disappears before hills
ƒ Ship hulls vanish before their masts
¾ The altitude of stars depends on latitude
¾ The shadow of a lunar eclipse is always
circular
• Eratosthenes calculated the size of the Earth
using a simple geometric argument, comparing
the height of the Sun at two different
locations, at the same time.
¾ At Alexandria the Sun was 7.2° north of
overhead
¾ Using basic geometry he related the
distance on the spherical Earth (5000
stadia) to the angular distance around the
circumference (7.2°) and found a radius of
~39000 stadia.
¾ We don’t know exactly what this unit is in
our terms, but best estimates suggest 1
stadium = 157 m; with this conversion
Eratosthenes’ method gives a radius for
Earth of ~6250km. This is very close to
the modern value of 6378km.
Distance to the Moon
• Eclipses show Moon is between Earth/ Sun
• Lunar eclipses can be used to determine the
distance to the moon
¾ Since we know the diameter of the Earth
(13000 km), and the angular size of the
Sun (0.53 deg) we can work out the size of
Earth’s shadow.
¾ Comparison with the angular size of the
shadow on the moon gives the distance to
the moon, 350,000 km.
Distance to the Sun:
• Aristarchos used another simple geometric
argument, based on the phases of the Moon
¾ When moon is in quarter phase, EarthMoon-Sun must form a right-triangle
¾ By measuring the angle between the Moon
and Sun, as viewed from Earth, you can
solve the triangle
¾ He measured this angle to be 87 degrees.
The modern value is 89.75 degrees. This
angle is very difficult to measure
accurately.
¾ The distance to the Sun is then
350,000/cos(89.75). Since the angle is
close to 90 deg, the cosine is close to zero.
A small error in measurement leads to a
very wrong answer.
Exercises
1.
At what distance would a loonie subtend
an angle of 1”?
The equation relating physical diameter, D, to
the subtended angle θ and the distance r is:
θ
D
=
2πr 360
The diameter of a loonie is D=26.55 mm.
Therefore the distance D is given by:
(
0.02655m )
r=
2π
= 5476.3m
360
(1 / 3600)
2.
What is the maximum elongation of
Mercury?
Mercury’s orbit is 0.387 times that of Earth.
Therefore
sin θ = 0.387
θ = 22.7 degrees
3. At
a distance of 350,000 km, and
subtending an angle of 0.5 degrees, how
large is the moon?
Again we use the equation:
θ
D
=
2πr 360
with r=350,000 and θ=0.5 we get:
0.5
D = 2π (350,000km )
360
= 3054.3km