A rocket is launched straight up from the surface of Earth with a speed v,such that the rocket reaches a maximum height of R_{E above} the surface.Find the launch speed of the rocket.Neglect air resistance.(R_E = 6.38 $\times$ 10⁶ m,M_E = 5.98 $\times$ 10²⁴ kg,G = 6.67 $\times$ 10^-11 N·m²/kg²) A)2.81 km/s B)5.59 km/s C)11.2 km/s D)7.91 km/s E)None of the above

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Quizplus · Accepted Answer

The launch speed of the rocket can be found using energy conservation principles. At the maximum height, the kinetic energy of the rocket is zero, and all the launch kinetic energy has been converted into gravitational potential energy. The formula to calculate the initial speed $v$ is derived from equating the initial kinetic energy to the potential energy at the maximum height: $$ \frac{1}{2}mv^2 = \frac{G M_{\text{Earth}} m}{R_{\text{Earth}} + h} $$ where $h$ is the maximum height above the Earth's surface, $R_{\text{Earth}}$ is the radius of the Earth, $M_{\text{Earth}}$ is the mass of the Earth, and $G$ is the gravitational constant. Given $R_{\text{Earth}} = 6.38 \times 10^6 \, \text{m}$, $M_{\text{Earth}} = 5.98 \times 10^{24} \, \text{kg}$, $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$, and $h = R_{\text{Earth}}$, solving this equation for $v$ gives a value close to 7.91 km/s, making option D the correct answer.

A Rocket Is Launched Straight Up from the Surface of Earth