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Homework 7 - Solutions

# Homework 7 - Solutions - homework 07 FRENNEA KYLE Due Mar 7...

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homework 07 – FRENNEA, KYLE – Due: Mar 7 2007, 4:00 am 1 Question 1 part 1 of 1 10 points On the way from a planet to a moon as- tronauts reach a point where that moon’s gravitational pull is stronger than that of the planet. The masses of the planet and the moon are, respectively, 5 . 38 × 10 24 kg and 7 . 36 × 10 22 kg. The distance from the cen- ter of the planet to the center of the moon is 3 . 29 × 10 8 m. Determine the distance of this point from the center of the planet. Correct answer: 2 . 94549 × 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 G m M p r 2 p = G m M m r 2 m , we obtain q = r m r p = radicalBigg M m M p = radicalBigg 7 . 36 × 10 22 kg 5 . 38 × 10 24 kg = 0 . 116963 . 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 . 29 × 10 8 m (0 . 116963) + 1 = 2 . 94549 × 10 8 m . Question 2 part 1 of 1 10 points A satellite moves in a circular orbit around the Earth at a speed of 6 . 7 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: 2518 . 86 km (tolerance ± 1 %). Explanation: The gravitational force provides the cen- tripetal acceleration. G m M r 2 = m v 2 r , where M is the mass of the Earth and m is the mass of the satellite. Solving for r yields r = G M v 2 = 8 . 88886 × 10 6 m Then the height of the satellite above the Earth’s surface is h = r - R earth = 2518 . 86 km Question 3 part 1 of 2 10 points Given: G = 6 . 67259 × 10 11 N · m 2 / kg 2 . A 1 . 9 kg mass weighs 18 . 62 N on the surface of a planet similar to Earth. The radius of this planet is roughly 5 . 6 × 10 6 m. Calculate the mass of of this planet. Correct answer: 4 . 60583 × 10 24 kg (tolerance ± 1 %). Explanation: By Newton’s Law of Universal Gravitation, W = G m M planet r 2 , so M planet = W r 2 G m = (5 . 6 × 10 6 m) 2 (6 . 67259 × 10 11 N · m 2 / kg 2 ) × 18 . 62 N 1 . 9 kg = 4 . 60583 × 10 24 kg .

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homework 07 – FRENNEA, KYLE – Due: Mar 7 2007, 4:00 am 2 Question 4 part 2 of 2 10 points Calculate the average density of this planet. Correct answer: 6261 . 16 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 (4 . 60583 × 10 24 kg) 4 π (5 . 6 × 10 6 m) 3 = 6261 . 16 kg / m 3 . Question 5 part 1 of 1 10 points Given: G = 6 . 673 × 10 11 N · m 2 kg 2 Mars has a mass of about 6 . 40 × 10 23 kg, and its moon Phobos has a mass of about 9 . 11 × 10 15 kg. If the magnitude of the gravitational force between the two bodies is 4.33 × 10 15 N, how far apart are Mars and Phobos? Correct answer: 9 . 47907 × 10 6 m (tolerance ± 1 %). Explanation: Basic Concept: F g = G m 1 m 2 r 2 Given: m 1 = 6 . 40 × 10 23 kg m 2 = 9 . 11 × 10 15 kg F g = 4 . 33 × 10 15 N G = 6 . 673 × 10 11 N · m 2 kg 2 Solution: r 2 = G m 1 m 2 F g = parenleftbigg 6 . 673 × 10 11 N · m 2 kg 2 parenrightbigg · ( 6 . 4 × 10 23 kg ) ( 9 . 11 × 10 15 kg ) 4 . 33 × 10 15 N = 8 . 98528 × 10 13 m 2 Thus r = radicalbig 8 . 98528 × 10 13 m 2 = 9 . 47907 × 10 6 m Question 6 part 1 of 1 10 points
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Homework 7 - Solutions - homework 07 FRENNEA KYLE Due Mar 7...

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