Acceleration due to Gravity

The acceleration due to gravity, commonly denoted by \( g \), is the rate at which an object accelerates when falling freely under the influence of gravity. On Earth, this value is approximately: \[ g \approx 9.8 \, \text{m/s}^2 \] This means any object in free fall near Earth's surface (ignoring air resistance) will increase its velocity by \( 9.8 \, \text{m/s} \) every second.

Free Fall and Equations of Motion

When an object is falling freely:

The only force acting on it is gravity.
It experiences a uniform acceleration \( g \).

The motion can be analyzed using kinematic equations: \[ v = u + gt \] \[ s = ut + \frac{1}{2}gt^2 \] \[ v^2 = u^2 + 2gs \] where:

\( u \): initial velocity (m/s)
\( v \): final velocity (m/s)
\( t \): time (s)
\( s \): displacement (m)

Variation of \( g \)

The value of \( g \) is not constant everywhere. It varies slightly with:

Altitude: \( g \) decreases as you go higher above sea level.
Latitude: Due to Earth's rotation and shape (bulging at the equator), \( g \) is slightly lower at the equator than at the poles.
Depth: \( g \) decreases as you go deeper underground.

Relation Between \( g \) and Gravitational Force

The acceleration due to gravity is derived from Newton’s Law of Universal Gravitation: \[ g = \frac{GM}{r^2} \] where:

\( G \): gravitational constant
\( M \): mass of the Earth
\( r \): distance from the center of the Earth

A stone is dropped from rest from a height of 20 m. What is its velocity just before hitting the ground? (Assume \( g = 9.8 \, \text{m/s}^2 \))

Solution:

Given: \( u = 0 \), \( s = 20 \, \text{m} \)
Using: \( v^2 = u^2 + 2gs \)
\[ v^2 = 0 + 2 \times 9.8 \times 20 = 392 \Rightarrow v = \sqrt{392} \approx 19.8 \, \text{m/s} \]

The stone hits the ground with a speed of approximately \( 19.8 \, \text{m/s} \).

An object is dropped from a height of 45 m. How long does it take to reach the ground?

Solution:

Given: \( u = 0 \), \( s = 45 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \)
Using: \( s = ut + \frac{1}{2}gt^2 \)
\[ 45 = 0 + \frac{1}{2} \times 9.8 \times t^2 \Rightarrow t^2 = \frac{90}{9.8} \approx 9.18 \Rightarrow t \approx 3.03 \, \text{seconds} \]

The object reaches the ground in about 3.03 seconds.

Written by Thenura Dilruk