# The value of $g$ in free fall motion on earth [closed]

When we release a heavy body from a height to earth. We get the value of $$g=9.8 \ ms^{-2}$$. Now, I'm confused about what it means. For example, does it mean that the body's speed increases to $$9.8$$ every second? Or, does it mean that the speed of the body is $$9.8 \ m/s$$?

It means the speed increases by $$9.8$$ m/s every second. At the beginning (when you release the body) its speed is $$0$$. After $$1$$ second the speed is $$9.8$$ m/s, after $$2$$ seconds the speed is $$19.6$$ m/s, and so on.

It means the speed of the falling body increases with 9.8 m/s each second.

The other guys here (@Thomas Fritsch and @AWanderingMind) are perfectly right, and just to see that: g is an acceleration, and acceleration is change of velocity with time, or velocity per time. Like velocity itself is distance per time.

You can see that $$g$$ has units of acceleration, namely $$\frac {m}{s^2}$$ or $$\frac {m}{s} \left (\frac 1s \right )$$. Last form gives an easy interpretation,- speed change per 1 second. Additionally, there are couple of assumptions in use:

• Earth is assumed an ideal sphere, otherwise $$g$$ would depend on radius $$R$$ at exact $$\varphi, \lambda$$ (Latitude, Longitude) coordinates on surface.
• Free-fall has to be in pure vacuum, otherwise gravitational acceleration would be opposed by air resistance (air drag force), which would make net acceleration dependent on such parameters like falling body cross-section, velocity, etc.
• Finally, mass distribution in Earth along axis parallel to the surface normal is assumed to be uniform, namely- density variation function $$\rho(R)$$ has same profile along any coordinates $$\varphi, \lambda$$. Otherwise, you would need to account for local mass distributions in $$g$$ calculations.

$$F=-G\frac{Mm}{r^2}$$

$$\frac{F}{m}=-G\frac{M}{r^2}$$

$$a=-G\frac{M}{r^2}$$

Force divided by mass is by definition acceleration. This is denoted as g, as in, the acceleration due to gravity

For the surface of the earth this value is around $$-9.81$$.

As such the standard definition of acceleration applies. Acceleration is the rate at which velocity changes, a constant acceleration of x, means that every second, the velocity of the body increases by x, or in a formula: $$v=at$$

Look at the units $$\frac{m}{s} \frac{1}{s}$$

Distance traveled per second, per second