An $8000 \mathrm{~kg}$ satellite is launched from the surface of the Earth into outer space. What initial kinetic energy is needed by the satellite in order to reach a great (i.e., infinite) distance from the Earth, neglecting the effects of air resistance in the atmosphere? $\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}, \mathrm{M}_{\mathrm{E}}=5.97 \times 10^{24} \mathrm{~kg}, \mathrm{R}_{\mathrm{E}}=6.37 \times 10^{6} \mathrm{~m}\right.$.)

Question

Quizplus · Accepted Answer

The correct answer is A because the initial kinetic energy needed by the satellite is equal to the change in its gravitational potential energy, which is given by:$$\Delta U = -\frac{GM_EM_s}{r_f} - (-\frac{GM_EM_s}{r_i}) = \frac{GM_EM_s}{r_i}$$where $G$ is the gravitational constant, $M_E$ is the mass of the Earth, $M_s$ is the mass of the satellite, $r_i$ is the initial distance of the satellite from the center of the Earth, and $r_f$ is the final distance of the satellite from the center of the Earth (which is infinity). Substituting the given values into this equation, we get:$$\Delta U = \frac{(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2})(5.97 \times 10^{24} \mathrm{~kg})(8000 \mathrm{~kg})}{6.37 \times 10^{6} \mathrm{~m}} = 5.00 \times 10^{11} \mathrm{~J}$$

An 8000 kg8000 \mathrm{~kg}8000 kg Satellite Is Launched from the Surface of the Earth into the Earth

An $8000 \mathrm{~kg}$ Satellite Is Launched from the Surface of the Earth into the Earth