Escape speed: A satellite is in circular orbit at an altitude of 2300 km above the surface of a nonrotating asteroid with an orbital speed of 5.9 km/s. The minimum speed needed to escape from the surface of the asteroid is 14.6 km/s, and G = 6.67 × 10^-11 N ∙ m²/kg². The mass of the asteroid is closest to A) 1.8 × 10²⁴_kg. B) 8.9 × 10²³_kg. C) 1.3 × 10²⁴_kg. D) 2.7 × 10²⁴_kg. E) 3.6 × 10²⁴_kg.

Question

Quizplus · Accepted Answer

The correct answer is A because the escape speed from the surface of the asteroid is given by:$$v_{esc} = \sqrt{\frac{2GM}{r}}$$where:* $v_{esc}$ is the escape speed* $G$ is the gravitational constant* $M$ is the mass of the asteroid* $r$ is the radius of the asteroidWe can use the given information to solve for the mass of the asteroid:$$v_{esc} = 14.6 	ext{ km/s}$$$$G = 6.67 	imes 10^{-11} 	ext{ N} \cdot 	ext{m}^2/	ext{kg}^2$$$$r = 2300 	ext{ km} + 6371 	ext{ km} = 8671 	ext{ km}$$Substituting these values into the equation, we get:$$14.6 	ext{ km/s} = \sqrt{\frac{2(6.67 	imes 10^{-11} 	ext{ N} \cdot 	ext{m}^2/	ext{kg}^2)M}{(8671 	ext{ km})(1000 	ext{ m/km})^2}}$$Solving for $M$, we get:$$M = 1.8 	imes 10^{24} 	ext{ kg}$$

Escape Speed: a Satellite Is in Circular Orbit at an Altitude