Question

One way to provide artificial gravity (i.e., a feeling of weight) on long space voyages is to separate a spacecraft into two parts at the ends of a long cable, and set them rotating around each other. A craft has been separated into two parts with a mass of 70600 kg each, at the ends of a cable with their centers of mass 155 m apart, rotating around the center point of the cable with a period of 385.3 seconds.
1) If the cable is reeled in so that the the centers of the two pieces are now only 119.35 m apart, what will the new period be?
2) What happens to the angular momentum L and kinetic energy K of the system consisting of both pieces of the space ship when the two pieces are pulled closer to the center?

Answer 1

Answer:

1)  T = 649.86 s, 2)   L₀ = L_f,   $\frac{K_o}{K_f}$ = 4.8

Explanation:

1) As the system of the two bodies is isolated, its angular momentum is conserved

             

initial instant.  r₀ = 155 m, T₀= 385.3 s

      L₀ = I₀ w₀

final instant. r = 119.35 m

      L_f = I w

      L₀ = L_f

      I₀ w₀ = I w

      w = $\frac{I_o}{I} \ w_o$

let's consider each object as punctual

      I = m r²

at angle velocity and period are related

         w = 2pi / T

     

we substitute

      $\frac{2\pi }{T} = \frac{m r^2}{m _o^2 } \ \frac{2\pi }{T_o}$

      $\frac{1}{T} = ( \frac{r}{r_o} )^2 \ \frac{1}{T_o}$

       T = $(\frac{r_o}{r} )^2 \ T_o$

let's calculate

       T = $( \frac{155}{119.35} )^2 \ 385.3$

        T = 649.86 s    

2) The angular momentum is conserved because the system is isolated.

Let's look for kinetic energy

        K_total = 2 K = 2 (½ I w²)

        K_total = I 4π² / T²

        K_total = 2m r² 4 π² / T²

 

for r = 155 m

         K₀ = 8π² m r₀² / T₀²

for r = 119.35 m

          K_f = 8π² m r² / T²

the relationship is

        $\frac{K_o}{K_f} = ( \frac{r_o \ T}{ r \ \ T_o} )^2$

        $\frac{K_o}{K} = ( \frac{ 155 \ \ \ 649.86}{ 119.35 \ 385.3})^2$

       $\frac{K_o}{K_f}$ = 4.8