hw8_solutions - Homework#8 Solutions Astro 10 spring 2003...

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Homework #8: Solutions Astro 10, spring 2003 General Notes to Graders: Please take a point off the total if there is no section number on the homework. If numerical answers are roughly correct, do not take marks off. Award part marks if the student has made some progress with the question. Be generous. 1. [10 points] a) Alright, before we begin we need to know the true diameters of the Sun and the Earth. Noting that the diameter is simply twice the radius, we find from the textbook that: 2 R Sun = 1 . 392 × 10 11 cm 2 R Earth = 1 . 28 × 10 9 cm Note that the ratio of the Earth’s diameter relative to the Sun’s diameter is simply 2 R Earth / 2 R Sun = 0 . 0092 . Therefore, if we are modeling the Sun as being 30 cm in diameter, then the Earth at the same scale will be 30 cm × 0 . 0092 = 0 . 28 cm in diameter (roughly the size of a small pebble). b) The Earth-Sun distance is given as: D E - S = 1 . 49 × 10 13 cm The ratio of the Earth-Sun distance to the diameter of the Sun is simply D E - S / 2 R Sun = 107 . Therefore, if we are modeling the Sun as being 30 cm in diameter and the Earth as 0.276 cm in diameter, then we must place them apart at a (rough) distance of 30 cm × 107 = 3120 cm , or 31.2 m (which is huge - imagine a pebble in orbit around a basketball at a distance about equal to the length of a basketball court! ). c) Proxima Centauri has a parallax of 0.77”, so it’s distance is d = 1 0 . 77” = 1 . 3 pc There are 3.09 × 10 18 cm in a parsec, so the distance to Proxima Centauri is 4 . 02 × 10 18 cm. The ratio of this distance to the Sun’s diameter is 4 . 02 × 10 18 / 2 R Sun = 2 . 9 × 10 7 . Therefore, if the Sun’s diameter is 30 cm, the distance to Proxima Centauri is 30 cm × 2 . 9 × 10 7 = 8 . 7 × 10
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