1169061310Xuj

1169061310Xuj - Alvarado, Patrick Homework 10 Due: Nov 16...

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Alvarado, Patrick – Homework 10 – Due: Nov 16 2006, 10:00 am – Inst: Andrei Sirenko 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Calculate the average speed oF the Moon around the Earth. The Moon has a period oF revolution oF 27.3 days and an average dis- tance From the Earth oF 3 . 84 × 10 8 m. 1. 61374 . 2 m / s 2. 1022 . 9 m / s correct 3. 3 . 68245 × 10 6 m / s 4. 3688.5 m/s 5. 2045.8 m/s 6. 8 . 83789 × 10 7 m / s 7. 2054.6 m/s 8. 584000 m/s 9. None oF these Explanation: Let : T = 27 . 3 d and R = 3 . 84 × 10 8 m . T = 27 . 3 d × 24 h d × 3600 s h = 2 . 35872 × 10 6 s . v = D T = 2 π R T = 2 π (3 . 84 × 10 8 m) 2 . 35872 × 10 6 s = 1022 . 9 m / s . keywords: 002 (part 1 oF 1) 10 points Calculate the centripetal Force exerted on the Earth by the Sun. Assume that the period oF revolution For the Earth is 365 . 25 days, the average distance is 1 . 5 × 10 8 km and the Earth’s mass is 6 × 10 24 kg. 1. 2 . 66331 × 10 32 N 2. 3 . 56775 × 10 19 N 3. 4 . 6238 × 10 29 N 4. None oF these 5. 1 . 28439 × 10 26 N 6. 7 . 24562 × 10 22 N 7. 1 . 62932 × 10 21 N 8. 7 . 24562 × 10 20 N 9. 3 . 56775 × 10 22 N correct Explanation: Let : T = 365 . 25 d and R = 1 . 5 × 10 8 km = 1 . 5 × 10 11 m . The period is T = 365 . 25 d × 24 h d × 3600 s h = 3 . 15576 × 10 7 s , and the velocity is v = D T = 2 π R T = 2 π (1 . 5 × 10 11 m) 3 . 15576 × 10 7 s = 29865 . 3 m / s , so the centripetal Force is F = m v 2 R = (6 × 10 24 kg) (29865 . 3 m / s) 2 1 . 5 × 10 11 m = 3 . 56775 × 10 22 N .
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Alvarado, Patrick – Homework 10 – Due: Nov 16 2006, 10:00 am – Inst: Andrei Sirenko 2 keywords: 003 (part 1 of 2) 10 points On the way to the moon, the Apollo astro- nauts reach a point where the Moon’s gravi- tational pull is stronger than that of Earth’s. Determine the distance of this point from the center of the Earth. The masses of the Earth and the Moon are respectively 5 . 33 × 10 24 kg and 7 . 36 × 10 22 kg. The distance from the Earth to the Moon is 3 . 9 × 10 8 m. Correct answer: 3 . 4899 × 10 8 m. Explanation: If r e is the distance from this point to the center of the Earth and r m is the distance from this point to the center of the Moon, then from the formula GmM e r 2 e = GmM m r 2 m we obtain q = r m r e = r M m M e = s 7 . 36 × 10 22 kg 5 . 33 × 10 24 kg = 0 . 11751 . On the other hand, r e + r m = R . Eliminating r m from the last two equalities, we obtain r e = R q + 1 = 3 . 9 × 10 8 m 0 . 11751 + 1 = 3 . 4899 × 10 8 m . 004 (part 2 of 2) 10 points What is the acceleration due to the Earth’s gravity at this point? The universal gravita- tional constant G is 6 . 672 × 10 - 11 N m 2 / kg 2 . Correct answer: 0
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This note was uploaded on 09/04/2008 for the course PHYS 106 taught by Professor Opyt during the Spring '08 term at NJIT.

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1169061310Xuj - Alvarado, Patrick Homework 10 Due: Nov 16...

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