# What is acceleration of free fall ?

## Can it be in simple words please

Jul 21, 2017

$g = 9.80665$ ${\text{m/s}}^{2}$ (see below)

#### Explanation:

In situations where a particle is in free-fall, the only force acting on the object is the downward pull due to earth's gravitational field.

Since all forces produce an acceleration (Newton's second law of motion), we expect objects to accelerate toward earth's surface due to this gravitational attraction.

This acceleration due to gravity near Earth's surface (symbol "$g$") is the same for all objects near Earth's surface (that aren't affected by any other forces which can easily dominate this gravitational force, such as subatomic particles and their electromagnetic interactions).

The value of $g$ is standardized as a constant:

• color(blue)(g = 9.80665 color(blue)("m/s"^2

However, there are a lot of factors that can affect this value depending on where the object is located, so approximations are almost always used in calculations (most commonly $10$ ${\text{m/s}}^{2}$, $9.8$ ${\text{m/s}}^{2}$, or $9.81$ ${\text{m/s}}^{2}$).

ul(bb("Extra Info":

This value of $g$ was both experimentally determined and determined via Newton's law of gravitation, which states

• ${F}_{\text{grav}} = \frac{G {m}_{1} {m}_{2}}{{r}^{2}}$

where

• ${F}_{\text{grav}}$ is the gravitational force experienced between two objects

• $G$ is the gravitational constant (don't confuse this with $g$!), defined as $6.674 \times {10}^{-} 11 \left({\text{N"·"m"^2)/("kg}}^{2}\right)$

• ${m}_{1}$ and ${m}_{2}$ are the masses of the two objects in kilograms, in no particular order

• $r$ is the distance between them, in meters

If the object is near Earth's surface, the distance between Earth and the object is essentially the radius us the earth (${r}_{\text{E}}$). Also, one of the objects' masses is earth's mass ${m}_{\text{E}}$, so we then have

• color(green)(F_"grav" = (Gm_"E"m)/((r_"E")^2)

What we can do to find the value of $g$ is...

First recognize Newton's second law, which if the acceleration is $g$ is

${F}_{\text{grav}} = m g$

we can substitute this value in for ${F}_{\text{grav}}$ in the above equation, to yield

mg = (Gm_"E"m)/((r_"E")^2)

Dividing both sides by $m$:

• $\textcolor{red}{g = {\left(G {m}_{\text{E")/((r_"E}}\right)}^{2}}$

What does this equation tell us?

Notice how the value of the object $m$ is no longer a part of this equation...this proves that $g$ is completely independent of the object's mass.

The value of $g$ can thus be found using the gravitational constant, the mass of Earth ($5.9722 \times {10}^{24}$ $\text{kg}$) and Earth's radius ($6371008$ "m"):

g = ((6.674xx10^-11("N"·"m"^2)/("kg"^2))(5.9722xx10^24cancel("kg")))/((6371008color(white)(l)"m")^2)

= color(blue)(9.8 color(blue)("m/s"^2#