Escape speed: What is the ratio of the escape speed of a rocket launched from sea level to the escape speed of one launched from Mt. Everest (an altitude of 8.85 km)? The radius of the earth is 6.38 × 10⁶ m. A) 1.0007 B) 1.0014 C) 1.0001 D) 0.9986 E) 0.9993

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Quizplus · Accepted Answer

The escape speed $v$ from a distance $r$ from the center of a mass (like Earth) is given by the formula $v = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. The escape speed from sea level (radius $r_1 = 6.38 \times 10^6$ m) and from the top of Mt. Everest (radius $r_2 = r_1 + 8.85 \times 10^3$ m) can be compared by taking the ratio of their escape speeds:$$\frac{v_1}{v_2} = \frac{\sqrt{\frac{2GM}{r_1}}}{\sqrt{\frac{2GM}{r_2}}} = \sqrt{\frac{r_2}{r_1}}$$Plugging in the values:$$\frac{v_1}{v_2} = \sqrt{\frac{6.38 \times 10^6 + 8.85 \times 10^3}{6.38 \times 10^6}} \approx \sqrt{1.001387} \approx 1.0007$$Therefore, the ratio of the escape speed from sea level to that from Mt. Everest is approximately 1.0007.

Escape Speed: What Is the Ratio of the Escape Speed