Escape speed: A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.2 km/s. The minimum speed needed to escape from the surface of the planet is 14.9 km/s, and G = 6.67 × 10^-11 N ∙ m²/kg². The orbital period of the satellite is closest to A) 72 min. B) 65 min. C) 58 min. D) 51 min. E) 44 min.

Question

Quizplus · Accepted Answer

The orbital period can be found using the formula for the period of a satellite orbiting a planet: $T = 2\pi \sqrt{\frac{r^3}{GM}}$, where $r$ is the distance from the center of the planet to the satellite, $G$ is the gravitational constant, and $M$ is the mass of the planet. Given the altitude of the satellite and the radius of the planet derived from the escape speed, we can calculate the orbital period. The closest answer given the calculation is approximately 72 minutes.

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