Activity 10.3 Class 9 Science Gravitation

Activity 10.3 Class 9 Science Gravitation Chapter 10

Activity 10.3 Class 9 Science gravitation chapter 10

What activity asks?

Activity 10.3 asks us to drop a piece of paper and stone from some height and see which reaches the ground first. The activity also asks us to do the same thing in a vacuumed glass.

 

Observation:

The stone reaches fast while paper arrives at the bottom late.

Explanation:

We all know that earth attracts all the object with the force proportional to the mass of the object.

F= GMm/R2

(Where G is the universal gravitational constant; M is the mass of the earth; m is mass of the object, and R is the radius of the earth.)

⇒ma=GMm/R2

⇒a=GM/R2

 

Thus we see that all the body falls with the same acceleration. We call this Acceleration due to gravity (g). Its value is 9.8m/s2.

But this does not happen in reality. Other forces come into play in a free fall. For example Friction due to air. Light objects like paper have a higher surface area. They offer higher resistance to the air; while the stone is heavy with a small source area. Stone offer lower friction with the air. As a result, when we fall paper and stone at the same time, stone reaches the ground fast.

A glass from which air is drawn out does not offer any resistance whether it is a stone or a paper or something else. As a result, all the objects fall at the same speed and reach the bottom simultaneously.

 

Next: Activity 10.4 Class 9 Science Gravitation.

See also: Activity 10.2 Class 9 Science Gravitation.

 

Ref: Chapter 10.

Activity 10.3 Class 9 Science Gravitation Chapter 10

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