Problem Set 12

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SUNY Empire State College *

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1200

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Astronomy

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Feb 20, 2024

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docx

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Problem Set # 12- Problems 7, 20, 30, 36, 47 7. At new moon, the Earth, Moon, and Sun are in a line. Find the direction and magnitude of the net   gravitational force exerted on   (a)   the Earth,   (b)   the Moon, and   (c)   the Sun. To find the direction and magnitude of the net gravitational force of sun, moon, and earth. It would follow as: (6.67x 10^-11N*m/kg2) (5.97x10^24kg) ((2x10^30kg/((1.50x10^11m)2) + (7.35x 10^22kg) ((3.84x 10^8m) ^2) ________ = 3.56 x 10^22 N (direction towards sun) ________ (6.67 x 10^-11N*m/kg2) (7.35 x 10^22kg) ((2 x 10^30kg/ ((1.50 x 10^11m)2)-3.84 x 10^8kg) -(5.97 x 10^24kg)/ (3.84 x 10^8m) ^2) ________ =2.40 x 10^20 N (direction towards sun) ________ (6.67x10^-11N*m/kg2)(2x10^30kg)((7.35x10^22kg/((1.50x10^11m)^2- 3.84x10^8m)^2+(5.97 x 10^24kg)((1.50 x 10^11m)^2) ________ = 3.58 x 10^22 N (direction towards earth, moon)
20. At some point along the direct path from the center of the Earth to the center of the Moon, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth.   (a)   How far from the center of the Earth does this occur?   (b)   At this location, how far is the spacecraft from the surface of the Moon? How far is it from the surface of the Earth? a) r1 = 9.0139 (D -r1) r1 (1+ 9.0139) = 9.0139 D r1 = 9.0139 / 10.0139 D r1 = 0.9001 3.84 10^8 r1 = 3.46 x 10^8 m b) Given: Radius of moon = 1.74 x 10^6m Distance of moon = r2 = D-r1 3.84 x 10^8 - 3.46 x 10^8 = 3.8 x 10^7m Now the distance measured from surface for the spacecraft is: 3.8 x 10^7 – 1.74 x 10^6 = 3.6 x 10^7m c) now for the distance of earth’s surface: 3.46 x 10^8m – 6.37 x 10^6m = 3.40 x 10^8m
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