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, g h. Express the equation in terms of the radius R of the Earth, g, and h.

Question

Physics

Zachery Hayes

11 March, 07:59

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, g h. Express the equation in terms of the radius R of the Earth, g, and h.

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Answers (1)

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Averie Conner · Answer 1

g = g₀ [1 - 2 h / Re + 3 (h / Re) ²]

Explanation:

The law of universal gravitation is

F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

F = G Me / Re² m

We call gravity acceleration a

g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

R = Re + h

Therefore the attractive force is

F = G Me m / (Re + h) ²

Let's take Re's common factor

F = G Me / Re² m / (1 + h / Re) ²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

(1 + h / Re) ⁻² = 1 - 2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

F = G Me / Re² m [1 - 2 h / Re + 3 (h / Re) ²]

F = g₀ m [1 - 2 h / Re + 3 (h / Re) ²]

If we call the force of attraction at height

m g = g₀ m [1 - 2 h / Re + 3 (h / Re) ²]

g = g₀ [1 - 2 h / Re + 3 (h / Re) ²]