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 vo = 2.72 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 sufficiently great 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 RE = 6.37 × 106 m, the mass of the Earth is ME = 5.98 × 1024 kg, the acceleration due to gravity on the surface of the Earth is g = 9.81 m/s2 and the universal gravitational constant is G = 6.67 × 10-11 N·m2/kg2.

You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital speed of Vo 2.72 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 sufficiently great 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 RE = 6.37 x 106 m, the mass of the Earth is ME 5.98 x 1024 kg, the acceleration due to gravity on the surface of the Earth is g 9.81 m/s2 and the universal gravitational constant is G 661011N m2lkg2 Number

Explanation / Answer

Escape velocity: Ve= sqrt (2GM/r)

where G is the universal gravitational constant (G = 6.67×1011 m3 kg1 s2),M the mass of the body to be escaped

and r the distance from the center of mass of the mass M to the object.

Ve = sqrt(2*6.67 × 10-11 N·m2/kg2*5.98 × 10^24 kg /(6.37 × 10^6 m))

Ve = 11190.73m/s

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