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EXAMPLE 12-3 THE DEPENDENCE OF GRAVITY ON ALTITUDE RWP If you climb to the top of Mt. Everest, you will be about 5.50 mi above

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EXAMPLE 12-3 THE DEPENDENCE OF GRAVITY ON ALTITUDE "RWP If you climb to the top of Mt. Everest, you will be about 5.50 mi above sea level. What is the acceleration due to gravity at this altitude? PICTURE THE PROBLEM At the top of the mountain, your distance from the center of the Earth is r - Re + h, where h - 5.50 mi is the altitude. Ar = 5,50 ml REASONING AND STRATEGY First, use F - GmM /r to find the force due to gravity on the - RE + h mountaintop. Then, set F = my to find the acceleration & at the height A. Known Height of Mt. Everest, h = 5.50 mi. Unknown Acceleration due to gravity at height , &; - ? SOLUTION 1. Calculate the force F due to gravity at a height h F - G- mME above the Earth's surface: (RE + h) 2. Set Fequal to my, and solve for & F - G- ( RE + h)3 - ME & - G.. ( Re + 1) 3. Factor out Re/ from the denominator, iMe and use the fact that GM,/R;] - g (1 + 2 ) (1 + 4 ) 9.81 m/'s2 4. Substitute numerical values, with 8850 m - 9.78 m/s h - 5.50 mi - (5.50 ml) (1609 m/ml) = 8850m and Re - 6.37 x 10* m: 6.37 x 10* m INSIGHT As expected, the acceleration due to gravity is less as one moves farther from the center of the Earth. Thus, if you were to climb to the top of Mt. Everest, you would lose weight-not only because of the physical exertion required for the climb, but also because of the reduced gravity. In particular, a person with a mass of 60 kg (about 130 lb) would lose about half a pound of weight just by standing on the summit of the mountain. A plot of & as a function of h is shown in FIGURE 12-7 (a]. The plot indicates the altitude of Mt. Everest and the orbit of the International Space Station. FIGURE 12-7 (b) shows & out to the orbits of communications and weather satellites, which are at an altitude of roughly 22,200 mi. PRACTICE PROBLEM Find the acceleration due to gravity at the altitude of the International Space Station's orbit, 322 km above the Earth's surface. [Answer: f, - 8.89 m/s', a reduction of 9.38% compared to the acceleration of gravity on the surface of the Earth] Some related homework problems: Problem 14, Problem Is CONTINUED "Real World Physics applications are denoted by the acronym EWRQuestion 16 (1 point) Retake question What is the acceleration of gravity, in m/s , at 349 km from the surface of Earth? This question is identical to Example 12-3. Add the distance given to the radius of the Earth and convert to meters. Plug this value into the expression for g, keeping in mind that the distance to the center of the Earth is not the radius (the object is not at the surface!)

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