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

Question

Quizplus · Accepted Answer

The escape speed from a planet's surface is given by:
Ve = sqrt((2GM)/r)
Where G is the gravitational constant, M is the mass of the planet and r is the distance from the planet's center. Since the question asks for the ratio of the escape speeds, we can cancel out these common factors and only consider the distance from the center of the Earth:
Ve1/Ve2 = sqrt(r2/r1)
Where Ve1 is the escape speed from sea level and Ve2 is the escape speed from the altitude of Mt. Everest. Plugging in the values given, we get:
Ve1/Ve2 = sqrt((r + h)/r) = sqrt((6.38×10^6 + 8.85×10^3)/6.38×10^6) = 1.0007 (rounded to four decimal places)

Therefore, the ratio of the escape speed of a rocket launched from sea level to the escape speed of one launched from Mt. Everest is 1.0007, which is closest to choice A.

What Is the Ratio of the Escape Speed of a Rocket