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Chapter: 11th Physics : Gravitation

Solved Example Problems for Universal Law of Gravitation

Physics : Gravitation - Solved Example Problems for Universal Law of Gravitation

EXAMPLE 6.1

Consider two point masses mand m2 which are separated by a distance of 10 meter as shown in the following figure. Calculate the force of attraction between them and draw the directions of forces on each of them. Take m1= 1 kg and m2 = 2 kg


Solution

The force of attraction is given by


From the figure, r =10 m.

First, we can calculate the magnitude of the force


It is to be noted that this force is very small. This is the reason we do not feel the gravitational force of attraction between each other. The small value of G plays a very crucial role in deciding the strength of the force.

The force of attraction () experienced by the mass m2 due to m1 is in the negative ‘y’ direction ie., rˆ =−jˆ . According to Newton’s third law, the mass m2 also exerts equal and opposite force on m1. So the force of attraction () experienced by m1 due to m2 is in the direction of positive ‘y’ axis ie., rˆ = jˆ .


The direction of the force is shown in the figure,


Gravitational force of attraction between m1 and m2

=  which confirms Newton’s third law.

 

EXAMPLE 6.2

Moon and an apple are accelerated by the same gravitational force due to Earth. Compare the acceleration of the two.

The gravitational force experienced by the apple due to Earth


Here M– Mass of the apple, ME– Mass of the Earth and R – Radius of the Earth.

Equating the above equation with Newton’s second law,


Simplifying the above equation we get,


Here aA is the acceleration of apple that is equal to ‘g’.

Similarly the force experienced by Moon due to Earth is given by


Here Rm- distance of the Moon from the Earth, Mm – Mass of the Moon

The acceleration experienced by the Moon is given by


The ratio between the apple’s acceleration to Moon’s acceleration is given by


From the Hipparchrus measurement, the distance to the Moon is 60 times that of Earth radius. Rm = 60R.


The apple’s acceleration is 3600 times the acceleration of the Moon.

The same result was obtained by Newton using his gravitational formula. The apple’s acceleration is measured easily and it is 9.8 m s−2 . Moon orbits the Earth once in 27.3 days and by using the centripetal acceleration formula, (Refer unit 3).


which is exactly what he got through his law of gravitation.


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