Question

what is the size of the increase in velocity per second during free fall?

Answer 1

During free fall, force of gravity is the only force acting on a body.

By Newton's law, it is given by

F = GMm/R^2
G = universal gravitational constant.
M = mass of the earth
m = mass of the body
R = radius of the earth

Also, the force is given as F = ma.

On equating the two we get

ma = GMm/R^2

a = GM/R^2

a is the acceleration. To be specific, it's called acceleration due to gravity. Every free falling body is acted upon by this. To find it, we substitute all the values in the above relation (they are all constants, so the final value always comes the same)

a = (6.67x10^-11) x (6x10^24) / (6.4x10^6)^2

a = g = 9.8 m/s^2

It also means g = 9.8 (m/s)/s.

Which also means that every second, the speed of a freely falling body increases by 9.8 m/s. Hence, the answer to your question.

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