hw11 - patel(kp9583 – HW11 – Mackie –(20211 1 This...

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Unformatted text preview: patel (kp9583) – HW11 – Mackie – (20211) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A satellite moves in a circular orbit around the Earth at a speed of 5 . 3 km / s. Determine the satellite’s altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius 6370 km and mass 5 . 98 × 10 24 kg. The value of the universal gravitational constant is 6 . 67259 × 10 − 11 N · m 2 / kg 2 . Correct answer: 7835 . 09 km. Explanation: Let : v = 5 . 3 km / s , R e = 6370 km , M e = 5 . 98 × 10 24 kg , and G = 6 . 67259 × 10 − 11 N · m 2 / kg 2 . The gravitational force provides the cen- tripetal acceleration, so GmM e r 2 = mv 2 r r = GM e v 2 = (6 . 67259 × 10 − 11 N · m 2 / kg 2 ) × 5 . 98 × 10 24 kg (5 . 3 km / s) 2 parenleftbigg 1 km 1000 m parenrightbigg 3 = 14205 . 1 km , and the height of the satellite above the Earth’s surface is h = r- R e = 14205 . 1 km- 6370 km = 7835 . 09 km . 002 10.0 points If our Sun were four times as massive as it is, how many times faster or slower should the Earth move in order to remain in the same orbit? Correct answer: 2. Explanation: Let : M ′ = 4 M . The gravitational force supplies the cen- tripetal force: F = mv 2 r = GM m r 2 . v = radicalbigg GM r ∝ √ M for the same distance, so v ′ v = √ 4 M √ M = √ 4 = 2 . v ∝ √ M tells us that more massive bodies have higher speeds for a given orbit. 003 (part 1 of 2) 10.0 points Three masses are arranged in the ( x, y ) plane as shown. y [m]- 5- 3- 1 1 3 5 x [m]- 5- 3- 1 1 3 5 2 kg 1 kg 5 kg What is the magnitude of the result- ing force on the 2 kg mass at the ori- gin? The universal gravitational constant is 6 . 6726 × 10 − 11 N · m 2 / kg 2 . Correct answer: 3 . 07069 × 10 − 11 N. Explanation: patel (kp9583) – HW11 – Mackie – (20211) 2 Let: m o = 2 kg , ( x o , y o ) = (0 m , 0 m) , m a = 1 kg , ( x a , y a ) = (1 m ,- 5 m) , and m b = 5 kg , ( x b , y b ) = (4 m ,- 3 m) . Applying Newton’s universal gravitational law for m o and, F ao = G m o m a ( x a- x o ) 2 + ( y a- y o ) 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) × (2 kg) (1 kg) (1 m) 2 + (- 5 m) 2 = 5 . 13277 × 10 − 12 N , where tan θ a = y a x a θ a = arctan parenleftbigg y a x a parenrightbigg = arctan parenleftbigg- 5 m 1 m parenrightbigg = 281 . 31 ◦ , so the components of this force are F a x = F a cos θ a = ( 5 . 13277 × 10 − 12 N ) cos 281 . 31 ◦ = 1 . 00662 ×...
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This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.

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hw11 - patel(kp9583 – HW11 – Mackie –(20211 1 This...

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