When a falling meteor is at a distance above the Earth's surface of 3.34 times the Earth's radius, what is its free-fall acceleration due to the gravitational force exerted on it?

To calculate the free-fall acceleration experienced by a falling meteor, we can use the law of universal gravitation. The formula for the gravitational force between two objects is:

F = G * (m1 * m2) / r^2

Where:
F is the force of gravity
G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
m1 and m2 are the masses of the two objects
r is the distance between the centers of the two objects

In this case, the mass of the meteor is irrelevant since we are only concerned with its free-fall acceleration. The mass of the Earth can be considered constant.

The free-fall acceleration experienced by the meteor is equal to the gravitational force divided by the mass of the meteor:

a = F / m

However, we know that the gravitational force is given by:

F = m * g

Where m is the mass of the meteor and g is the acceleration due to gravity. Therefore, we can rewrite the equation for free-fall acceleration as:

a = F / m = (m * g) / m = g

In other words, the free-fall acceleration of the meteor is equal to the acceleration due to gravity, which is constant near the Earth's surface.

Hence, the free-fall acceleration of the meteor, regardless of its distance above the Earth's surface, is approximately 9.8 m/s^2.