Sapling Learning Map A The acceleration due to gravity, g, is constant at sea le

Question

Sapling Learning Map A The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, gh. Express the equation in terms of the radius of the Earth RE, gand h. A 78.70 kg hiker has ascended to a height of 1815 m in the process of climbing Mt. Washington. By what percent has the hiker's weight changed by climbing to this elevation? Use g 9.807 m/s2 and RE 6.371 x10 m. (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.) click to edit Number w change O Previous Check Answer Next Exit

Explanation / Answer

Acceleration due to gravity on the surface of earth g = GM/R 2   -----------(1)

Accleration due to gravity at a height h from the surface of the earth g ' = GM /(R+h) 2   ------------(2)

From equation (1) and ( 2) , g ' / g = [ GM /(R+h) 2 ]/[GM/R 2 ]

= R 2 /(R+h) 2

Divide numerator and denominator by R 2 you get ,

g ' / g = 1/[1+(h/R)] 2

= [1+(h/R)] -2

Neglect higher order trems you get ,

g ' / g = 1-(2h/R)

g ' = g [1-(2h/R)]

(b). Mass m = 78.7kg

Weight at bottom of the mountain W = mg = 78.7 x9.807 = 771.8109 N

Accleration due to gravity at height h is g ' = g[1-(2x1815)/(6.371 x10 6)]

= 9.807[1-(5.6976x10 -4)

= 9.80141 m/s 2

Weight at height h is W ' = mg '

= 78.7 x 9.80141 = 771.37114 N

Required percentage = [(W'-W)/W]x100

= -0.056976 %

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