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.

Question

Quizplus · Accepted Answer

The escape speed $v$ from a distance $r$ from the center of a mass (like Earth) is given by $v = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. The escape speed from sea level ($r = R$, the radius of the Earth) is $v_1 = \sqrt{\frac{2GM}{R}}$, and from Mt. Everest ($r = R + h$, where $h$ is the height of Mt. Everest) is $v_2 = \sqrt{\frac{2GM}{R + h}}$. The ratio of the escape speeds is $\frac{v_1}{v_2} = \sqrt{\frac{R + h}{R}}$. Substituting $R = 6.38 \times 10^6 m$ and $h = 8.85 \times 10^3 m$, the ratio is very close to 1, with a slight increase due to the additional altitude of Mt. Everest, making the correct answer 1.0007.

What Is the Ratio of the Escape Speed of a Rocket