Question:

There are $2$ bodies of masses $100\, kg$ and $1000\,kg$ separated by a distance $1\,m$. At what distance from smaller body, the intensity of gravitational field will be zero

Updated On: Jul 7, 2022
  • $\frac{1}{9}m$
  • $\frac{1}{10}m$
  • $\frac{1}{11}m$
  • $\frac{10}{11}m$
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The Correct Option is C

Solution and Explanation

Here $\frac{G \times 1000}{(1-x)^2} = \frac{G \times 100}{x^2}$ i.e. $\frac{1}{x} = \frac{10}{1 - x} $ i.e. $1 - x = 10x$ i.e. $x = \frac{1}{11} m $
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Concepts Used:

Gravitational Potential Energy

The work which a body needs to do, against the force of gravity, in order to bring that body into a particular space is called Gravitational potential energy. The stored is the result of the gravitational attraction of the Earth for the object. The GPE of the massive ball of a demolition machine depends on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between GPE and the mass of an object. More massive objects have greater GPE. Also, there is a direct relation between GPE and the height of an object. The higher that an object is elevated, the greater the GPE. The relationship is expressed in the following manner:

PEgrav = mass x g x height

PEgrav = m x g x h

Where,

m is the mass of the object,

h is the height of the object

g is the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.