The orbital period increases if the orbital distance is increased.
Let's cut through the weeds and the trash
and get down to the real situation:
A stone is tossed straight up at 5.89 m/s .
Ignore air resistance.
Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for
(5.89 m/s / 9.8 m/s²) = 0.6 seconds.
At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.
His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .
Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at
(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.
With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.
(0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.
Answer:
Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest.
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,


Where,
m = Mass of spacecraft
M = Mass of Earth
r = Radius (Orbit)
G = Gravitational Universal Music
v = Velocity
Re-arrange to find the velocity



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,


From the speed it is possible to use find the formula, so



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



Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.