You are a visitor aboard the New International Space Station, which is in a circ

Question

You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital speed of v = 1.91 km/s. The station is equipped with a High Velocity Projectile Launcher, which can be used to launch small projectiles in various directions at high speeds. Most of the time, the projectiles either enter new orbits around the Earth or else eventually fall down and hit the Earth. However, as you know from your physics courses at the Academy, projectiles launched with a great enough initial speed can travel away from the Earth indefinitely, always slowing down but never falling back to Earth. With what minimum total speed, relative to the Earth, would projectiles need to be launched from the station in order to "escape" in this way? For reference, recall that the radius of the Earth is R = 6.37 * 10^6 m, the mass of the Earth is M = 5.98 * 10^24 kg, the acceleration due to gravity on the surface of the Earth is g = 9.81 m/s^2 and the universal gravitational constant is G = 6.67 * 10-11 N*m2/kg2.

Explanation / Answer

From the orbital speed we get the distance from the center of the earth (r = R + h ) to the space station so V = sqrt [ GM / r ]

R is radius of the earth, h is the heigt of the space station from earth surface

so 1.91x10^3 = sqrt [ (5.98x10^24)(6.67x10^-11)/ r ] so r = 1.0933x10^8 m

Escape velocity from the space station is

V= sqrt [ 2GM/r] = sqrt [2 (5.98x10^24)(6.67x10^-11)/ 1.0933x10^8 m ] = 2701m/s or 2.7km/s

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