A 50-kg satellite circles the Earth in an orbit with a period of 120 min.What minimum energy is required to change the orbit to another circular orbit with a period of 180 min? (Earth: radius = 6.4 * 10⁶ m,mass = 6.0 *10²⁴ kg) A)2.9 *10⁸ J B)3.5 *10⁸ J C)4.1 * 10⁸ J D)4.7 * 10⁸ J E)5.9 * 10⁸ J

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The correct answer is A because the minimum energy required to change the orbit is equal to the difference in kinetic energy between the two orbits. The kinetic energy of a satellite in a circular orbit is given by:$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{2\pi r}{T}ight)^2$$where m is the mass of the satellite, v is its speed, r is the radius of the orbit, and T is the period of the orbit.The difference in kinetic energy between the two orbits is:$$\Delta K = K_f - K_i = \frac{1}{2}m\left(\frac{2\pi r_f}{T_f}ight)^2 - \frac{1}{2}m\left(\frac{2\pi r_i}{T_i}ight)^2$$where the subscripts i and f refer to the initial and final orbits, respectively.Substituting the given values into the equation, we get:$$\Delta K = \frac{1}{2}(50 	ext{ kg})\left(\frac{2\pi (6.4 	imes 10^6 	ext{ m})}{180 	imes 60 	ext{ s}}ight)^2 - \frac{1}{2}(50 	ext{ kg})\left(\frac{2\pi (6.4 	imes 10^6 	ext{ m})}{120 	imes 60 	ext{ s}}ight)^2$$$$\Delta K = 2.9 	imes 10^8 	ext{ J}$$Therefore, the minimum energy required to change the orbit is 2.9 x 10^8 J.

A 50-Kg Satellite Circles the Earth in an Orbit with a Period