Answer:
t = √ ( 2h R² / GM)
This is the correct expression so we see the equation found is not correct
Explanation:
To answer this question, let's look for an expression for the time and fall of the body, using kinematics
v = vo + a t
v² = v₀² + 2 a h
Suppose the body is released, so its initial velocity is zero, when they reach the ground they have the speed
(at)² = 2 a h
t = √ 2h / a
To find acceleration, let's use Newton's second law, where force is the force of gravitation
F = m a
G m M / (R + h)² = m a
a = G M / (R + h)²
In general, if R >> h we can approximate the denominator expressions to R
a = G M / R²
Substituting
t = √ ( 2h R² / GM)
This is the correct expression so we see the equation found is not correct
Since the found equation is not correct, its dependence does not give units of time