hw11 - patel (kp9583) HW11 Mackie (20211) 1 This print-out...

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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 satellites 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 Earths 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 Newtons 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 a = y a x a 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 a = ( 5 . 13277 10 12 N ) cos 281 . 31 = 1 . 00662...
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hw11 - patel (kp9583) HW11 Mackie (20211) 1 This print-out...

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