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 Earth's surface due to the increase...

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 Earth's 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 𝑅 of the Earth, 𝑔, and ℎ .

gh=

Suppose a 91.75 kg hiker has ascended to a height of 1802 m above sea level in the process of climbing Mt. Washington. By what percent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use 𝑔=9.807 m/s2 and 𝑅=6.371×106 m. Enter your answer as a positive value.

weight change in %=

Answer 1

Above earth's surface, the force on an object is proportional to 1/(R+h)², where R radius of the earth, and h is the height above earth. Why so? Because the gravitational force is G*M*m/(R+h)². Be careful to understand that the denominator is always the total distance from center of earth to the center of the object. Usually, this is denoted simply by "R", but in the given question, R is the radius of the earth, which doesn't change. So the formula should use R+h instead, which is the total distance between the centers of the earth and the hiker.

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So weight W at any height = G* M / (R+h)²

and weight at surface = m* g = G* M / R²

Combining these two equations, we get the weight at at any height

W = m * g * R²/ (R+h)²

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Now let

W₀ = m* g

The percentage change can be calculated by

Weight change in % = 100 * (W₀ - W)/ W₀

Substitute the W and W₀ values from the previous equations to get the answer.