A small space telescope at the end of a tether line of length L moves at linear speed v about a central space station. What will be the linear speed of the telescope if the length of the line is changed to x*L? x = 2.8; v = 2 m/s?

The answer to the question is

The linear speed of the telescope will be 5.6 m/s if the length of the line is changed to x*L where x = 2.8; and initial velocity v = 2 m/s

Explanation:

Speed = v₁ = ωL = 2 m/s

When the line is changed to x*L where x = 2.8 the linear speed will be

v₂ = 2.8 * L * ω = 2.8 * 2 = 5.6 m/s

The linear speed varies with the angular speed following the relation v/r = ω where

ω = angular speed

v = linear speed and

r = radius of the path of travel of the object at the vertex

v' = 0.714 m/s

Explanation:

Solution:

- Assuming no external torque is acting on the system then the angular momentum is conserved for the system.

- The initial momentum angular Mi and final angular momentum Mf are as follows:

Mi = Mf

m*L*v = m*x*L*v'

Where,

m : mass of the telescope

L : Length of teether line

v: Initial speed

v' : Changed speed.

- Then we have:

L*v = x*L*v'

v' = v / x

v' = 2 / 2.8

v' = 0.714 m/s