Question

The Hubble Space Telescope was released from the Space Shuttle, which was in a circular orbit at 400km earth altitude. The relative velocity (from the Space Shuttle bay)of the ejection is vxo= ‒ 0.25m/s, vyo= ‒ 0.31m/s and vzo= 0.09m/s. Determine the distance between the two orbiting objects after 5 and 60 minutes.

Answer 1

Answer:

d = (75 i ^ + 93 j ^ + 27 k ^) m ,  d2 = (900 i ^ + 1116 j ^ + 324 k ^) m

Explanation:

The two objects are in circular orbit together, therefore with the same angular velocity, after the launch they move with the relative velocity, so we can use the kinematic relation

       

         v = d / t

         d = v t

Reduce time to units SI

         t = 5 min (60 s / 1 min) = 300 s

X axis

         x = vₓ t

         x = 0.25 300

         x = 75 m

Y axis  

        y = $v_{y}$ t

        y = 0.31 300

       y = 93 m

Z axis

        z=  $v_{z}$ t

        z = 0.09 300

       z = 27 m

       d = (75 i ^ + 93 j ^ + 27 k ^) m

For the time of 1 h

       t2 = 1 h (3600s / 1 h) = 3600

       x2 = 900 m

       y2 = 1116 m

       z2 = 324 m

      d2 = (900 i ^ + 1116 j ^ + 324 k ^) m