Question

A very large number of particles of same mass m are kept in XY plane at points (1,1),(2,2),(4,4),(8,8) and so on. The total gravitation field at the origin has the magnitude? ​

Answer 1

Answer:

The total gravitation field at the origin has the magnitude equal to

$E = \frac{2Gm}{3}$

Explanation:

Gravitational field due to a point mass at "r" distance from it is given by formula

$E = \frac{Gm}{r^2}$

now we know that gravitational field due to more than one masses is vector sum of the field due to each mass

so we will have

$E = \frac{Gm}{1^2 + 1^2} + \frac{Gm}{2^2 + 2^2} + \frac{Gm}{4^2 + 4^2} + \frac{Gm}{8^2 + 8^2}......................$

now we have

$E = \frac{Gm}{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64}.............)$

now its a Geometric Progression series of infinite terms so sum of all terms is given as

$E = \frac{Gm}{2}(\frac{1}{1 - \frac{1}{4}})$

$E = \frac{2Gm}{3}$

#Learn

Topic : Gravitational field intensity

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