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

To determine the distance between the two orbiting objects after 5 and 60 minutes, calculate the initial position and use the laws of motion to find the subsequent positions.

Step-by-step explanation:

To determine the distance between the two orbiting objects after 5 and 60 minutes, we can start by finding their initial position. Since the ejection velocity is given as vxo = ‒ 0.25 m/s, vyo = ‒ 0.31 m/s, and vzo = 0.09 m/s, we can use these values to find the initial position of the Hubble Space Telescope relative to the Space Shuttle. Given that the Space Shuttle is in a circular orbit at 400 km altitude, we can calculate the initial distance between the two objects using Pythagoras' theorem.

After finding the initial distance, we can use the laws of motion to calculate the subsequent position of the Hubble Space Telescope after 5 and 60 minutes. Since the Space Shuttle is in orbit, its velocity remains constant, so we can use this velocity to determine the displacement of the Hubble Space Telescope relative to the Space Shuttle over the given time intervals.

By adding the displacement to the initial distance, we can find the final distance between the two orbiting objects after 5 and 60 minutes.

Answer 2

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

Step-by-step 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