\[ \text{Total energy} = \frac{-GMm}{2(2R)} \] If energy \( \frac{10^4R}{6} \) is added, then \[ \frac{-GMm}{4R} + \frac{10^4R}{6} = \frac{-GMm}{2r} \] where \( r \) is the new radius of revolution and \( g = \frac{GM}{R^2} \). \[ -\frac{mgR}{4} + \frac{10^4R}{6} = -\frac{mgR^2}{2r} \quad (m = 10^3\, \text{kg}) \] \[ -\frac{10^3 \times 10 \times R}{4} + \frac{10^4R}{6} = -\frac{10^3 \times 10 \times R^2}{2r} \] \[ -\frac{1}{4} + \frac{1}{6} = -\frac{R}{2r} \] \[ r = 6R \]
The total mechanical energy of a satellite in a circular orbit is given by:
\[ E = -\frac{GMm}{2r}, \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( r \) is the radius of the orbit.
When energy is supplied to the satellite, the total energy increases, which results in an increase in the radius of the orbit. Let the initial radius of the orbit be \( 2R \), and the new radius be \( r' \).
Using the conservation of energy:
\[ E_{\text{initial}} + \text{Energy supplied} = E_{\text{final}}, \]
\[ -\frac{GMm}{2(2R)} + \frac{10^4R}{6} = -\frac{GMm}{2r'}. \]
Simplifying, we find that the new radius \( r' \) is:
\[ r' = 6R. \]
Thus, the correct answer is \( r' = 6R \), and the correct answer is Option (4).
A sportsman runs around a circular track of radius $ r $ such that he traverses the path ABAB. The distance travelled and displacement, respectively, are:

A body of mass $100 \;g$ is moving in a circular path of radius $2\; m$ on a vertical plane as shown in the figure. The velocity of the body at point A is $10 m/s.$ The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as $10 m/s^2$)


Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: