Question

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its orbital speed (in m/s)? m/s (b) If we want to propel a portion of the rocket to infinity (in the direction tangential to the circular orbit), what's the escape speed from there (in m/s)? m/s

Answer 1

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is  $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

            $w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

       Substituting the value

         $w = \frac{2 \pi}{138240}$

             $= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              $\frac{GMm}{r^2} = mr w^2$

Where  G is the universal gravitational constant with a value  $G = 6.67*10^{-11}$

            M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     $r = \sqrt[3]{\frac{GM}{w^2} }$

                        $= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

                        $= 5.78 *10^7 m$

The orbital speed is represented mathematically as

                   $v=wr$

Substituting value

                  $v= (5.78*10^7)(4.54*10^{-5})$

                     $v= 2.6*10^{3} m/s$    

The escape velocity is mathematically represented as

                            $v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

                             $= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

                             $v_e= 3.72 *10^3 m/s$