What is minimum energy required to launch a satellite of mass m from the surface of a planet of mass m and radius r in a circular orbit at an altitude 2R:

Explanation

To solve this problem, we can use the concept of gravitational potential energy. The potential energy of an object due to gravity is given by:PE = -G * M * m / rwhere:G is the gravitational constantM is the mass of the planetm is the mass of the satelliter is the distance between the center of the planet and the satelliteIn this scenario, we need to consider the initial and final potential energies of the satellite:Initial potential energy (PE_i): When the satellite is at the surface of the planet (r = R), its potential energy is:PE_i = -G * M * m / RFinal potential energy (PE_f): When the satellite reaches the circular orbit at an altitude of 2R (r = 3R), its potential energy is:PE_f = -G * M * m / (3R)Escape velocity: The minimum energy required to launch the satellite and overcome the gravitational pull of the planet is known as the escape velocity (v_e). It can be calculated using the following equation:1/2 * m * v_e^2 = -G * M * m / RSolving for v_e, we get:v_e = sqrt( 2 * G * M / R )Kinetic energy (KE): Once the satellite reaches the circular orbit, it needs to have enough kinetic energy to maintain its orbital motion. This kinetic energy can be calculated using the following equation:KE = 1/2 * m * v^2where v is the orbital velocity of the satellite.Total energy (E): The minimum total energy required (E) is the sum of the escape velocity and the kinetic energy needed for the orbit:E = KE + PE_f = 1/2 * m * v^2 - G * M * m / (3R)Relating orbital velocity and escape velocity: In a circular orbit, the orbital velocity (v) is related to the escape velocity (v_e) by:v = v_e / sqrt(2)Substituting this relationship into the equation for total energy:E = 1/2 * m * (v_e^2 / 2) - G * M * m / (3R)Simplifying the equation:E = (1/4) * G * M * m / R - G * M * m / (3R)Combining terms:E = ( -3G * M * m + G * M * m ) / (12R)This simplifies to:E = 5GmM / 6RTherefore, the minimum energy required to launch the satellite (5GmM/6R) corresponds to option d in the question