Ncert Exercises & Solutions

Q. 32 An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy?

Hint: The potential energy of the object increases as it goes away from the earth.

Step 1: Find the initial potential energy of the object.

Consider the diagram where an object of mass m is raised from the surface of the earth to a distance (height) equal to the radius of the earth (A).

The potential energy of the object at the surface of the earth =

- \frac{GMm}{R}

Step 2: Find the final potential energy of the object.

PE of the object at a height equal to the radius of the earth =

- \frac{GMm}{2 R}

Step 3: Find the gain in the potential energy of the object.

The gain in PE of the object = \frac{- GMm}{2 R} - (- \frac{GMm}{R}) = \frac{- GMm + 2 GMm}{2 R} = + \frac{GMm}{2 R} = \frac{{gR}^{2} \times m}{2 R} = \frac{1}{2} mgR (∵ GM = {gR}^{2})

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