Question

a) What is the angular momentum of the Moon around the Earth? The Moon mass is 7.40e22 kg and it orbits 3.80e8 meters from the Earth.

b) And what is the angular momentum of the Earth around the Sun? Mass of the Earth is 5.98e24 kg and it orbits 1.50el 1 m from the Sun.

Answer 1

Given:

m = 7.4E+22 kg
r = 3.8E+8 m

Angular momentum:

L = r * P

P = momentum
r = position from origin

P = m * v

So

L = r * m * v

We have r and m, solve for v using circumference of orbit and period of orbit

C = 2 * π * r
C = 2 * π * (3.8E+8 m)
C = 2,387,608,400 m

T = 27.321582 days (a known value)
T = 2,360,585 s (converted to SI unit)

Velocity = C/T
v = (2,387,608,400 m) / (2,360,585 s)
v = 1,011 m/s

Now solve for angular momentum

L = (3.8E+8 m) * (7.4E+22 kg) * (1,011 m/s)
L = 2.84E+34 kg-m^2/s

.......................................

Angualar moment L = I ω.

For orbitting earth
I = [2/5] mr^2 + m R^2 where R is the orbital radius.
R =1.496e11m
R/r = 1.496e11/ 6.378e6 = 23455.
Hence the term 2/5 mr^2 can be ignored as small.
I = m R^2
.
T = 1 year = 3.1536e+7 s
ω = 2π/T = 2π/T = 2π/3.1536e+7s = 1.99e-7rad /s
.
L = I ω. = 5.98e24*1.50e11*1.496e11*1.99e-7
.
L = 2.66e+40 kg.m^2S^-1.