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 = 2.45 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 RE = 6.37

Explanation / Answer

We know that
the escape velocity can be derived as

the kinetic energy of the satellite is equal to the potential

that is

   0.5*m*v^2 = G*M*m/R

solving for v

   ve = sqrt(2*G*m/R)

and we know that the orbital vlocity is

centripetal force = gravitational force

   mv^2/R = G*m*M/R^2

solving for v , vo = G*M/R

so the relation between escape velocity and orbital velocity is

   ve = sqrt(2) *v0

and we know the value of sqrt(2) = 1.414

so Ve = 1.414*vo

Vo is the orbital velocity of an object and Ve is the escape velocity from any point in that orbit

          now the escape velocty of the object with Vo = 2.45*10^3m/s is

           Ve= 1.414*2.45*10^3m/s

               = 3464.3 m/s

              = 3464.3 Km/s

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