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

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, g_h. Express the equation in terms of the radius of the Earth R_E, g and h. A 74.35 kg hiker has ascended to a height of 1841 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 R_E = 6.371 Times 10^6 m.

Explanation / Answer

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

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

The first eq. can be used to express GM in terms of g and substituted in the second eq;
From first eq;
GM = gR2

Sub. into second eq;
gh = gR2 / (R+h)2

Weight change is inversely proportional to distance from center of earth ratio squared.
This is an example of the so called 'inverse square law of a point source'.

Radius of Earth = 6.371E6 m
Radius of Earth + 1841 m = 6.371E6 + 0.001841E6 = 6.372841E6
Inverse ratio = 6.371/6.372841 = 0.9997111
Inverse ratio squared = 0.9994223
reduced "g" at height of 1841 m = (9.807)(0.9994223) = 9.801335 9.801
% difference in "g-value" is the same as % difference in weight, and equals:
(9.807 - 9.801)/9.807 = 0.006/9.807 = 0.0612 0.06%
ANS: Hiker's weight changes by 0.06%

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.