Acceleration Due to Gravity

When an object is dropped from the top of a hill or even a tree, the body moves and increases in velocity until it touches the ground with a velocity of finite value. Such movement is influenced by the earth’s gravitational field. The increase in velocity is therefore due to acceleration due to gravity which is usually represented by ‘g’. The motion of such body under gravity is always described as motion under free fall.

However, when two bodies of different masses are released from a height above the ground level, they do hit the ground at the same time. This is because acceleration due to gravity at a location is the same for all bodies irrespective of their masses and thus reach the ground at the same time.

This constant acceleration is called acceleration due to gravity and has a value of  or .

When a body is released from a height so that it falls towards the centre of the earth, ‘g’ is positive; but when a body is thrown upward, it goes against ‘g’ thereby decreasing in velocity until it momentarily comes to rest at the maximum height. For upward movement, ‘g’ is negative.

The equations connecting acceleration due to gravity, ‘g’ are as follows:

For downward movement, v2=u2+2gs and  s=ut+12gt2

For upward movement, v2=u2–2gs and s=ut−12gt2

When a body is released from rest at a certain height so that it falls towards the centre of the earth,

For upward movement, s=ut+12gt2

Since,  u = 0, 2s = gt²

∴ t2=2sg

Hence, t=2sg−−√

This equation shows that the time to reach the ground does not depend on the mass of the object.

Click on the slides to learn more about Gravitation:

Determination of Acceleration Due to Gravity

The value of ‘g’ could be determined using:

1. Formula method: A body is released from a height ‘s’  and the time t is taken; then use s=12gt2to get the value of ‘g’.
2. Simple Pendulum Experiment method: The value of ‘g’ could also be determined using this experiment.

The period T for the oscillation is given by: T=2πlg−−√

By linearizing this formular, we have T2=4π2(lg)

When T² is plotted against l, the equation is T2=(4π2g)l

1. Hence, the slope for such graph is 4π2g

When l is plotted against T², the equation is l=(g4π2)T2

And the slope for such graph is g4π2

In any case, from the slope, you get the value of ‘g’.

(NB: Educator should carry out the two experiments with the students.)

EVALUATION

1. What is the value of acceleration due to gravity?
2. What is the mathematical relationship between the period of oscillation T and the length of the string used l in a simple pendulum experiment?