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.

tried gh=g*(1-2h/R) and was wrong?

and then A 91.75 kg hiker has ascended to a height of 1867 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

Acceleration due to gravity is given as = Field strength of Earth's gravitational field at that point,

so gh = G M / ( h + Re)^2

and g is gh when h = 0

g = G M / Re^2

so gh / g = ( G M / ( h + Re)^2 ) / ( G M / Re^2 )

gh = ( Re / (Re +h))^2 g .........Ans

at 1867 m,

gh = ( 6.371 x 10^6 / ( 6.371x10^6 + 1867))^2 g

gh = 0.999707 g

%change = ( g - 0.999707g) / g   x 100 = 0.029 %

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