Astronomy Problem Set I

Problem set I Problem 1 The distance from Earth to Moon is a lot (less than) the distance from the Earth to the Sun. So, it an observer on the Moon sees the Earth & Sun in more or less the same direction. The configuration looks like this: to day I night the Sun Earth moon a) The observer on the Moon sees mostly the nighttime part of the Earth: low. nearly new Earth, crescent Earth, or thin crescent Ear b) en observer on the Earth soos m-sth the daylit part of the moon, lo s nearly full moon c) If Earth is nearly noon, so the Configuration was to the the the Sun Landon is cuer A Person in England cannot see the m-th it is "undergrand" (a the Earth is The way A Person Gagulant → man England can enly see these Livectrang Is harib Faland ← Th M d) to see the perfect you must go to the opposite side of the Earth from England: to the to the < England to the M- The moon is above the horizon from these locations e) These locations are in the night-time part of the earth Problem 2 a) it light trevels straight down, it moves along any affect light rays↓ The czer S there is no she dou (b) H-L T H Icn = FI T think The triangle: H-L L clearly tan a = L=H-L tan s H-L L R= (H-2) (c) Ris defined & be 5 H L tans ) 1-(1/4) H tan 7 (d) tans = tan 30° = 0.577 H = 1/2 cm: R = (no shadows!) H = 1.5 cm: R = 1 - 1 / 15 / 0.577 = 0.577 H = 2 cm: R = 1 - 1 / 2 / 0,577 = 0,8667 H = 10 cm: R = 1 - 1 / 10 / 0.577 = 1.56 H = 100 cm: R = 1 - 1 / 100 / 0.577 = 1.72 H = 10000 cm: R = 1 - 1 / 1000 / 0.577 = 1.73 (e) the measurements did not prace that light travels in straigt lines, becerse nobody studied the length of the shaden cest by very short hats (3) Eratosthenes Cakulation (4) Eratosthenes assamer that the Earth is perfectly spherical