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G49wP80U84m3_1210016271_jwk572

# G49wP80U84m3_1210016271_jwk572 - homework 08 – KIM JI –...

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Unformatted text preview: homework 08 – KIM, JI – Due: Oct 24 2007, 4:00 am 1 Question 1, chap 9, sect 1. part 1 of 1 10 points On the way from a planet to a moon astro- nauts reach a point where that moon’s gravi- tational pull is as strong as that of the planet. The masses of the planet and the moon are, respectively, 6 . 13 × 10 24 kg and 7 . 36 × 10 22 kg. The distance from the center of the planet to the center of the moon is 3 . 7 × 10 8 m. Determine the distance of this point from the center of the planet. Correct answer: 3 . 33461 × 10 8 m (tolerance ± 1 %). Explanation: If r p is the distance from this point to the center of the planet and r m is the distance from this point to the center of the moon, then from the formula GmM p r 2 p = GmM m r 2 m , we obtain q = r m r p = radicalBigg M m M p = radicalBigg 7 . 36 × 10 22 kg 6 . 13 × 10 24 kg = 0 . 109574 . On the other hand, r p + r m = R. Eliminating r m from the last two equalities, we obtain r p = R q + 1 = 3 . 7 × 10 8 m (0 . 109574) + 1 = 3 . 33461 × 10 8 m . Question 2, chap 9, sect 1. part 1 of 1 10 points A satellite moves in a circular orbit around the Earth at a speed of 6 . 9 km / s. Determine the satellite’s altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius R earth = 6370 km and mass M earth = 5 . 98 × 10 24 kg . You will need G = 6 . 67259 × 10- 11 N m 2 / kg 2 Correct answer: 2011 . 03 km (tolerance ± 1 %). Explanation: The gravitational force provides the cen- tripetal acceleration. GmM r 2 = mv 2 r , where M is the mass of the Earth and m is the mass of the satellite. Solving for r yields r = GM v 2 = 8 . 38103 × 10 6 m Then the height of the satellite above the Earth’s surface is h = r − R earth = 2011 . 03 km Question 3, chap 9, sect 1. part 1 of 2 10 points Given: G = 6 . 67259 × 10- 11 N · m 2 / kg 2 . A 3 . 5 kg mass weighs 34 . 3 N on the surface of a planet similar to Earth. The radius of this planet is roughly 7 . 2 × 10 6 m. Calculate the mass of of this planet. Correct answer: 7 . 61372 × 10 24 kg (tolerance ± 1 %). Explanation: By Newton’s Law of Universal Gravitation, W = G mM planet r 2 , so M planet = W r 2 Gm = (7 . 2 × 10 6 m) 2 (6 . 67259 × 10- 11 N · m 2 / kg 2 ) × 34 . 3 N 3 . 5 kg = 7 . 61372 × 10 24 kg . homework 08 – KIM, JI – Due: Oct 24 2007, 4:00 am 2 Question 4, chap 9, sect 1. part 2 of 2 10 points Calculate the average density of this planet. Correct answer: 4869 . 79 kg / m 3 (tolerance ± 1 %). Explanation: The volume of the planet is V = 4 3 π r 3 so its average density is ρ = M planet V = 3 M planet 4 π r 3 = 3 (7 . 61372 × 10 24 kg) 4 π (7 . 2 × 10 6 m) 3 = 4869 . 79 kg / m 3 ....
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G49wP80U84m3_1210016271_jwk572 - homework 08 – KIM JI –...

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