Answer:
a) v² = G M R³, b) T = 2π /[tex]\sqrt{GMR}[/tex], c) [tex]m \sqrt{GMR^5 }[/tex]
Explanation:
a) The kinetic energy is
K = ½ m v²
to find the velocity let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / R
force is the universal force of attraction
F = G m M / r²
we substitute
G m M R² = m v² R
v² = G M R³
the kinetic energy is
K = ½ m G M R³
b) angular and linear velocity are related
v = w R
w = v / R
w = [tex]\frac{\sqrt{GMR^3 }}{R}[/tex]
w = [tex]\sqrt{GMR}[/tex]
the angular velocity is related to the period
w = 2π / T
T = 2π / w
we substitute
T = 2π /[tex]\sqrt{GMR}[/tex]
c) the angular moeomto is
L = m v r
L = m RA G M R³ R
L = [tex]m \sqrt{GMR^5 }[/tex]