sol12 - The center of every circular satellite orbit must...

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PHYSICS 103 Chapter 5 -- HW #12 Questions 20. We know that the force of gravity decreases as the square of the distance from the center of the Earth. The altitude of 300 km represents a small increase in the distance compared to the Earth’s radius of 6370 km. Therefore, the force of gravity will be a little less that it is on the Earth’s surface. To actually calculate the decrease in the force, we need to square the ratio of the distances; (6370/6670) 2 = 0.91. Therefore, the force of gravity on the astronaut in orbit is 91% of what it is on the ground. 28. We cannot determine the mass of the Moon from its orbit because the properties of its orbit do not depend on its mass. The Moon would have the same orbit if its mass were half as large. To determine the Moon’s mass we could send a satellite to orbit the Moon and measure the satellite's orbital radius and period. 34. It is not possible for a satellite to remain stationary over Iraq, because a circular orbit that would keep the satellite directly above Iraq does not have its center at the center of Earth.
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Unformatted text preview: The center of every circular satellite orbit must be at the center of Earth because the gravitational force points toward Earth’s center. The gravitational force provides the centripetal force necessary for circular motion and the centripetal force for uniform circular motion always points toward the center of the circle. Exercise 12. We can do this problem two ways. We know that the force of gravity decreases as the square of the distance between the satellite and the center of Earth. Therefore, ( 29 ( 29 2 2 2 2 6.6 1 600 kg 10 m/s 138 N 6.6 old E new old new E r R F F mg r R = = ÷ ÷ = = ÷ Note that this is quite a bit smaller than the satellite’s weight of 5500 N on the ground. We can also solve this problem using the law of universal gravitation. ( 29 ( 29 ( 29 ( 29 1 2 2 2 24 2 11 2 2 6 6.6 600 kg 5.98 10 kg m 6.67 10 N 135 N kg 6.6 6.37 10 E E m m mM F G G r R-= = × = × × = ÷ × ×...
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