Correct Answer - Option 1 :
\(\frac{GMm}{12R}\)
The correct answer is option 1) i.e. \(\frac{GMm}{12R}\)
CONCEPT:
-
Gravitational Potential Energy: It is the energy possessed by a body at a certain point when work is done by the force of gravity in bringing the object from infinity to that point.
- The gravitational potential energy between two masses m1 and m2 separated by a distance r is given by
\(U = -\frac{Gm_1m_2}{r}\)
EXPLANATION:
- The gravitational potential energy of the satellite, \(U = -\frac{GMm}{R}\) (where M is the mass of earth)
The kinetic energy of the satellite, \(K =\frac{1}{2}mv^2\)
The centripetal force of the satellite is provided by the gravitational force exerted by the earth:
\(\Rightarrow \frac{mv^2}{R} = \frac{GMm}{R^2}\)
\(\Rightarrow v^2 = \frac{GM}{R}\) ----(1)
The total energy of the satellite, TE = K + U
\(\Rightarrow TE =\frac{1}{2}mv^2 +( -\frac{GMm}{R})\) ----(2)
Substituting (1) in (2),
\(\Rightarrow TE= \frac{1}{2}m(\frac{GM}{R}) +( -\frac{GMm}{R})\)
\(\Rightarrow TE =- \frac{GMm}{2R}\)
The work done or energy required to move from 2R to 3R will be:
\(\Rightarrow W =- \frac{GMm}{2}\times [\frac{1}{3R} - \frac{1}{2R}]\)
\(\Rightarrow W = \frac{GMm}{12R}\)
