Answer:
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]
