To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,
[tex]F_g = F_c[/tex]
[tex]\frac{GmM}{r^2} = \frac{mv^2}{r}[/tex]
Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity
[tex]\frac{GM}{r^2} = \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r} = v^2[/tex]
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,
[tex]v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}[/tex]
[tex]v = 4564.42m/s[/tex]
From the speed it is possible to use find the formula, so
[tex]T = \frac{2\pi r}{v}[/tex]
[tex]T = \frac{2\pi (6.371*10^6)}{4564.42}[/tex]
[tex]T = 8770.05s\approx 146min\approx 2.4hour[/tex]
Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.
PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is
[tex]KE = \frac{1}{2} mv^2[/tex]
[tex]KE = \frac{1}{2} (100)(4564.42)^2[/tex]
[tex]KE = 1.0416*10^9J[/tex]
Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.
