Answer:
a
The orbital speed is [tex]v= 2.6*10^{3} m/s[/tex]
b
The escape velocity of the rocket is [tex]v_e= 3.72 *10^3 m/s[/tex]
Explanation:
Generally angular velocity is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
Where T is the period which is given as 1.6 days = [tex]1.6 *24 *60*60 = 138240 sec[/tex]
Substituting the value
[tex]w = \frac{2 \pi}{138240}[/tex]
[tex]= 4.54*10^ {-5} rad /sec[/tex]
At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as
[tex]\frac{GMm}{r^2} = mr w^2[/tex]
Where G is the universal gravitational constant with a value [tex]G = 6.67*10^{-11}[/tex]
M is the mass of the earth with a constant value of [tex]M = 5.98*10^{24}kg[/tex]
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
[tex]r = \sqrt[3]{\frac{GM}{w^2} }[/tex]
[tex]= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }[/tex]
[tex]= 5.78 *10^7 m[/tex]
The orbital speed is represented mathematically as
[tex]v=wr[/tex]
Substituting value
[tex]v= (5.78*10^7)(4.54*10^{-5})[/tex]
[tex]v= 2.6*10^{3} m/s[/tex]
The escape velocity is mathematically represented as
[tex]v_e = \sqrt{\frac{2GM}{r} }[/tex]
Substituting values
[tex]= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }[/tex]
[tex]v_e= 3.72 *10^3 m/s[/tex]
