Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by,
- $-\frac{GMm}{R + h}$
- $\frac{GMmh}{R\left(R + h\right)}$
- mgh
- $\frac{GMm}{R + h}$
The Correct Option is B
Solution and Explanation
$ = - GMm[\frac{1}{R+h} - \frac{1}{R}] $
$= \frac{GMmh}{R(R+h)}$
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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.