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# ASTA01H3 Study Guide - Midterm Guide: Angular Diameter, Light-Year, Celestial Equator

Department
Astronomy
Course Code
ASTA01H3
Professor
Kristen Menou
Study Guide
Midterm

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An asteroid has an aphelion distance of 4.6 A.U. and a perihelion distance of 1.8 A.U. Calculate the orbital semi major axis, eccentricity, and
period. a = semi-major axis e = eccentricity T = period q = perihelion Q = aphelion q = a x (1 - e) Q = a x (1 + e) q + Q = 2 x a T^2 =a^3 (if a in AU and
T in Earth years)
An asteroid has a perihelion distance of 2.0 AU and an aphelion distance of 4.0 AU. Calculate its orbital semi-major axis, eccentricity and
period.
rp =Perihelion = a(1-e)ra = Aphelion = a(1+e) So rp + ra = a -ea + a + ea = 2a
a = (2.0 AU + 4.0 AU)/2 = 3.0 AU
rp = a(1-e)
e = 1 - rp/a
e = 1 - 2.0/3.0 = 1/3 = 0.333
Kepler's 3rd Law: P(yrs)2 = a(AU)3
P = a3/2 = (3.0)3/2 = 5.20 years
Halley's comet has a perihelion of 0.6 AU and an orbital period of 76 years. What is its aphelion distance from the Sun? The Period gives the
semi-major axis of the orbit a3 = P2
a = P2/3 = (76 AU)2/3 = 17.9 AU,ra + rp = 2a so ra = 2a - rp = 2(17.9) - 0.6 = 35.2 AU
[Halley's comet reaches maximum distance from the Sun between the orbits of Neptune and Pluto]
What would be the orbital period at a distance of 5 light-year
For a period of 5 million years, you need to solve: (5,000,000)^2 = T^3 You should get T = 29,240 AU, which is about 0.462 light-years.
At a distance of 5 light-years, that is 316,200 AU. So you need to solve:
P^2 = (316,200 AU)^ 3
Suppose you traveled to a planet with 4 times the mass of the Earth, and 4 times the
diameter of the Earth. Would you weigh more or less on that planet than on Earth? By
what factor will your weight change?
What are the minimum and maximum times of travel of radio messages
One AU is 1.496 Â· 1011 m, a distance the light covers in (1.496 Â· 1011)/(2.99792 Â· 108
) s = 499 s, or8.316 minutes.To travel with the speed of light the minimum distance d = 28.66
AU, one needs t = d/c = 28.66âˆ—499s = 14300 s = 238 minutes = 3.97 hours.To travel with the
speed of light a distance d = 50.30 AU, one needs t = d/c = 50.30 âˆ—499 s = 25100s = 418
minutes = 6.97 hours.
the circumference of Earth by observing the altitude of the sun above horizon. He measured that on the day of summer solstice the sun at
noon was = 82 140Î± â—¦ above horizon, i.e., not far from the zenith. Knowing that theEarth equator is tilted to the ecliptic by angle = 23 270Î¸ â—¦
what is the geographical latitude of Alexandria?How many degrees from the zenith is the star Polaris?A. During the summer solstice the height
of the sun above horizon Î±, the geographical latitude andÏ† the Earthâ€™s axis tilt angle = 23.5Î¸ â—¦are connected via = 90 +Î± â—¦âˆ’Ï† Î¸The easiest way to see
this is to think of a situation during equinox, when Î¸doesnâ€™t play a role (effectively Î¸= 0). In this situation, at the geographical north pole where = 90 Ï† â—¦
we would see the sun at Î±= 0;whereas at the equator ( Ï†= 0) weâ€™d see the sun at = 90Î± â—¦at noon. So the correct formula for all intermediate Ï†(at
equinox!) is = 90 Î± â—¦âˆ’Ï†. The only difference in midsummer is that the sun is higher above the horizon by Î¸, which results in the above formula.These
relationships follow from a close look at the diagram on p. 24 and 25 of our textbook, and have been extensively discussed in the tutorials.In our case,
we obtain latitude of Alexandria = 90 + = 31 130Ï† â—¦âˆ’Î± Î¸ â—¦ B. Polarisâ€™ distance from zenith is 90 = = 58â—¦âˆ’Ï† Î± âˆ’Î¸ â—¦490, because the altitude above horizon
of Polaris (north celestial pole around which the stellar sphere revolves) is always equal .Ï†
What is the gravitational acceleration (in units of Earthâ€™s acceleration, g = 9.81 m/s2) on the surface ofTitan, the largest moon of Saturn? It
has a mass equal M = 1.345 Â· 1023 kg and radius equals R = 2576km. Letâ€™s call the surface gravity on Titan gT . Carefully sticking to the SI system
of units, we have gT =GM/R2 =6.67259e-11*1.345e23/(2576e3)2 m/s = 1.35 m/s2. (2576e3 here is R expressed in meters; we
need to convert all units to m, kg, s; a.k.a. the international system of units SI). Thatâ€™s (1.35/9.81) g, i.e. 0.137g (0.137 of Earthâ€™s surface gravity).
â€”-> WEB ASSIGN QUESTIONS <â€”â€”â€”-
How long to cross diameter of milk way?: 8000 The stars seem to move across the sky together, maintaining their positions relative to one
another. Why?: All the stars are so far from Earth that we cannot perceive their individual motions For an observer at the North Pole, what would
the star trails look like? A dome of concentric circles, that appears as straight horizontal lines at the horizon. Stars: All stars are part of some
constellation. A star with an ancient Arabic name is probably relatively bright. The Î±star in a constellation is usually brighter than the star.Î² Magnitude
System: An object of magnitude -2 is very bright., A star must be brighter than sixth magnitude to be seen with the unaided eye. If two stars'
brightness is separated by one magnitude, the brighter star is about 2.5 times brighter than the dimmer one. Motions of the sky: The celestial sphere
is still useful even though it is no longer accepted as physical reality. The apparent rotation of the sphere is caused by Earth's rotation. Precession is
literally a wobbling of Earth's rotational axis.Modern Constellations: They are 88 well defined regions on the celestial sphere.Apparent Visual
Magnitude: Light output and distance cannot be determined from a star's apparent visual magnitude alone Celestial equator: Earth's rotation on its
axis is responsible for the observed motion of a star 23.5 DEGREES TO 0 what celestial circles would coincide: the Celestial Equator and the
Ecliptic Motion of the moon relative to stars: The moon moves Eastward relative to the stars about 13 degrees per day Gibbous phase has:
âˆ«duration of about one week Total lunar eclipse: moon is a reddish colour and in earths umbra July 1st- Yemeni is the constellation that the sun is on
Solar Eclipses: Because the moon's angular diameter in the sky as viewed from Earth changes, the moon passing between Earth and the sun may
not entirely block the sun from view during a solar eclipse. The sun's corona becomes visible during a solar eclipse. During a solar eclipse, the moon
casts a shadow on Earth.SunRise: new moon (eastern horizon), third qt (southern sky), Wanning present (in southeastern sky), Wanning gibbous
(south-western) First-quater moon to new moon: 21 days Retrograde motion: Saturn, Pluto, Neptune, Uranus, Jupiter. Present phase: saturn,
uranus, pluto, Jupiter, neptune,Ancient astronomy: The moon are perfect, earth is corrupt Nicolaus Copernicus: Inner planets orbit the sun faster
and pass outer planets as they orbit around the sun. Copernicus : It was a more elegant explanation of retrograde motion. New ellipse: is a circle
Force: change in motion is acceleration, change in motion requires force, forces always occur in pairs, the presence of more mass means more gravity
Kepler's Law: Planet moves fastest in its orbit when its closest to the sun, Orbits of planets are the shape of ellipses, in order to sweep out equal
areas in equal times, a plant must change its speed throughout its orbit. Constant Speed: The acceleration is not zero, and there must be a force
acting on the object in the same direction as the acceleration Newtons LawThe acceleration is not zero, and there must be a force acting on the
object in the same direction as the acceleration: The amount of force exerted on the sun by Earth is the same as the amount of force exerted on
Earth by the sun
How many suns would it take, laid side by side, to reach the nearest star? Use powers-of-ten notation. The diameter of the sun is Dsun = 1.4 x
1011 cm = 1.4 x 109 m. The distance to the nearest star is d* = 4.2 ly x 9.46 x 1015 m/ly = 4.0 x 1016 m. Hence the number of suns, laid side by
side, required to reach to the nearest star would be -> n = d* /Dsun = 4.0 x 1016 m / 1.4 x 109 m = 2.8 x 107 suns, i.e., about 28 million