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If we are in a free fall which implies we are accelerating at 9.8 m/sec every sec. And let's say that we are falling into a pit that has enormous depth. So isn't this be possible that we may accelerate and surpass the magnitude of speed of light?

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marked as duplicate by John Rennie, Carl Witthoft, Kyle Kanos, Qmechanic Mar 28 '15 at 13:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You make two wrong assumptions in your question, namely that if an object is accelerating the velocity would keep increasing ad infinitum without limit, and that the acceleration due to gravity on earth is always $9.8 m/s^2$ these are both not the case.

First of. The theory of relativity doesn't allow for objects that have mass to go faster than the speed of light. You can (informally) see this by the following equation.

\begin{equation} m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \end{equation}

Where $m_0$ is the mass the object for an observer at rest with respect to the object, $v$ is the velocity of the object and $c$ is the speed of light in vacuum. When the object would reach the speed of light $c$ the numerator becomes zero and the mass would become infinitely big, and it would therefore require an infinite amount of force to accelerate it. It would also have an infinite amount of kinetic energy ($1/2mv^2$), which I hope you will understand is not possible.

Secondly the rate at which object accelerate toward the earth, or fall towards any object in general is not a constant. The classical formula for the gravitational acceleration $g$ is the following. \begin{equation} g = \frac{GM}{r^2} \end{equation} Where $M$ is the mass of the earth, $G$ is the gravitational constant and $r$ is the distance between the centre of the earth and the object. The reason people say the gravitational acceleration is $9.8m/s^2$ is that the change in $g$ is negligible if we are talking about object falling over distances on the order of several meters: for the derivation of this fact see: https://physics.stackexchange.com/a/35880/76430.

I hope this somewhat answers your question.

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Without invoking relativity, let's look at energy. The force of gravity between two objects goes as

$$F= \frac{GMm}{r^2}$$

Which implies that the force is weaker when you are far away and stronger when you get closer. potential energy is the integral of force, and we know the sum of potential and kinetic energy is constant. Putting potential energy =0 at infinity, we can solve for the velocity of an object falling from far away to a distance $r$:

$$v = \sqrt{\frac{2GM}{r}}$$

If you now put $\frac{GM}{r^2}=g$ and $r=6400 km$ you get the velocity of an object that falls to earth "all the way from infinity" - it is

$$v_{max}=\sqrt{2gr}\approx 11.2 km/s$$

Which is the escape velocity of earth. Classically that is the fastest an object can fall due to the gravity of earth - well below the speed of light. The effects of relativity are quite small at these speeds.

How much more massive would an object have to be to break the light speed barrier? Using the above equation, if we could make the earth denser you could keep falling; effectively $g$ on the surface would be greater. Now to get to the speed of light you would have to shrink the earth quite a lot:

$r_{new}=\left(\frac{v_escape}{c}\right)^2 r_e = 9 mm$

If you could compress all matter of the earth into a sphere of 9 mm radius and fell to this sphere from a great height you would have to consider relativistic effects. But I think we are safe in the classical realm for another day.

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The short answer is: in ultra-deep gravity wells one has to bring relativistic considerations into the picture. The result of this is that an 'event horizon' forms in ultra-deep gravity wells such that an observer looking from a distance into the gravity well can only look till a finite depth. In loose terms, that specific finite depth corresponds to the depth at which infalling objects reach the light speed.

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