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Download Comparing Earth, Sun and Jupiter
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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 1 arcmin = 1/60 degree 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 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