The acceleration due to gravity, g, is constant at sea level on the Earth\\\'s s

Question

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, g and h. A 78.70 kg hiker has ascended to a height of 1893 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 × 106 m. (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.)

Explanation / Answer

A. The formula for the acceleration due to gravity "g" at the surface of the Earth is;
g = GM/R^2

GM = gR^2
The formula for the acceleration due to gravity at a height "h" above the Earth's surface is;
g' = GM/(R+h)^2
using above relation:
g' = gR^2/(R+h)^2

B. radius of earth = 6.371*10^6 m

at hight h radius will be = 6.371*10^6 + 1893 = 6372893 m

R^2/(R+h)^2 = [(6.371*10^6)/(6372893)]^2 = 0.99940601

reduced "g" at height h

g' = 9.807*0.99940601 = 9.80117474

weight ratio = (9.807 - 9.80117474)/9.807 = 0.00059339

percent change = 0.059339% = 0.0593%

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