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

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 91.75 kg hiker has ascended to a height of 1802 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/s^2 and RE = 6.371 x 10^6 m. (Hint: Keep 4 significant figures in your weight calculation to find the percent difference.)

Explanation / Answer

a)
on the surface of the earth, g = G*Me/Re^2

at altitude h,

g_h = G*Me/(Re+h)^2

b)

g_h/g = Re^2/(Re+h)^2

= (6.371*10^6)^2/(6.371*10^6 + 1802)^2

= 0.9994

g_h = g*0.9994

g_h = g*(1 - 0.0006)

= g - 0.0006*g

= g - 0.06% of g

so,

we know weight = m*g

so, decrease in weight 0.06% <<<<<<-------Answer

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