To solve this problem, we need to calculate the minimum energy required to launch a satellite into a circular orbit at an altitude of $2R$ from the surface of the Earth.$\\$Given:$\\$• Mass of the satellite $m \\$• Mass of the Earth $M \\$• Radius of the Earth $R \\$The altitude of the satellite's orbit is $2R$, so the total distance from the center of the Earth is $R + 2R = 3R. \\$The gravitational potential energy at the surface of the Earth is:$\\ U_{surface} = -\frac{GmM}{R} \\$The gravitational potential energy at the orbit (distance $3R$ from the center of the Earth) is:$\\ U_{orbit} = -\frac{GmM}{3R} \\$The kinetic energy required for a circular orbit at this altitude is given by:$\\ K_{orbit} = \frac{1}{2} \cdot \frac{GmM}{3R} = \frac{GmM}{6R} \\$The total energy at the orbit is the sum of kinetic and potential energy:$\\ E_{orbit} = U_{orbit} + K_{orbit} \\ E_{orbit} = -\frac{GmM}{3R} + \frac{GmM}{6R} \\ E_{orbit} = -\frac{GmM}{3R} + \frac{GmM}{6R} \\ E_{orbit} = -\frac{2GmM}{6R} + \frac{GmM}{6R} \\ E_{orbit} = -\frac{GmM}{6R} \\$The change in energy required to move the satellite from the surface to the orbit is:$\\ \Delta E = E_{orbit} - U_{surface} \\ \Delta E = \left(-\frac{GmM}{6R}\right) - \left(-\frac{GmM}{R}\right) \\ \Delta E = -\frac{GmM}{6R} + \frac{GmM}{R} \\ \Delta E = \frac{GmM}{R} - \frac{GmM}{6R} \\ \Delta E = \frac{6GmM}{6R} - \frac{GmM}{6R} \\ \Delta E = \frac{5GmM}{6R} \\$Therefore, the minimum energy required is $\frac{5GmM}{6R}$, which corresponds to Option 3.