A space rocket is launched such that very far from Earth it has a speed of 20 km/s.What was its initial launch speed from the surface of Earth? (R_E = 6.38 $\times$10⁶ m,M_E = 5.98 $\times$ 10²⁴ kg,G = 6.67 $\times$ 10^-11 N·m²/kg²) A)22.9 km/s B)20.0 km/s C)45.8 km/s D)21.5 km/s E)25.5 km/s

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Quizplus · Accepted Answer

To find the initial launch speed from the surface of Earth, we can use the formula for escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, where $G = 6.67 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2$, $M = 5.98 \times 10^{24} \, \text{kg}$ (mass of Earth), and $R = 6.38 \times 10^6 \, \text{m}$ (radius of Earth). The given speed of 20 km/s is less relevant for calculating the initial speed directly but indicates the rocket achieved escape velocity plus additional kinetic energy. Calculating escape velocity gives us the minimum speed needed to escape Earth's gravity without considering air resistance or Earth's rotation, which is approximately 11.2 km/s. However, the question seems to imply a need for the initial speed given a final speed far from Earth, which would typically involve energy considerations. Without additional details, the escape velocity calculation provides a baseline, but the initial speed would be higher to achieve a final speed of 20 km/s. The correct answer, based on the information provided and typical physics principles, would be an initial speed higher than escape velocity, but the exact value cannot be directly calculated from the given information. The choice provided closest to an expected higher initial speed due to the need to overcome Earth's gravity and achieve the final speed is A) 22.9 km/s, assuming some energy is converted to potential energy and some remains as kinetic energy at 20 km/s far from Earth.

A Space Rocket Is Launched Such That Very Far from Earth