Consider the following: The Earth has a mass of M. = 6.1 x 1024kg , and a radius
ID: 1789044 • Letter: C
Question
Consider the following: The Earth has a mass of M. = 6.1 x 1024kg , and a radius of R. = 6.4 x 106m. The moon, meanwhile, has a mass of M ) = 7.3 x 1022kg and a radius of R = 1.74 × 106m . Also, the distance from the Earth to the moon is 3.8 × 108m . If the moon orbits the Earth once every 28 days, and is tidelocked (le the same side of the moon always faces the Earth), what is the energy and angular momentum of a) b) c) The Earths rotation The moons rotation Total for the Earth-moon system (this will not be the sum of a and b)Explanation / Answer
given Me = 6.1*10^24 kg
Mm = 7.3*10^22 kg
Re = 6.4*10^6 m
Rm = 1.74*10^6 m
d = 3.8*10^8 m
a. energy of rotation of earth = 0.5*(2MeRe^2/5)*w^2 = Ee
w for earth = 2 pi rad/24 hours = 2* pi rad / 24*60*60 s
hence Ee = 2.64272*10^29 J
angular momentum of rotating earth = (2 Me Re^2 / 5) * w
Ie = 9.41934*10^40 kg m^2 / s
b. energy of rotation of moon = 0.5*(2Mm*Rm^2/5)*w^2 = Em
w for moon = 2*pi/28 days = 2*pi/28*24*60*60 rad/s
hence
Em = 2.981724*10^23 J
angular momentum, Im = (2Mm*Rm^2/5)w = 2.296*10^29 kg m^2 / s
c) total energy = 2.642722981724*10^29 J
now energy of the revolution of moon around earth has to be included too
E = 0.5(Mmd^2)*w^2 ( w for moon)
E = 3.55529*10^28 J
hence total energy = 2.998251981724*10^29 J
total angular momentum = 9.41934*10^40 + 2.296*10^29 = 9.41934*10^40 KG M^2/S ( MOON HAS INSIGNIFICANT CONTRIBUTION to angular moemntum)
bit this does not include angular momentum of moons rotation around earth
I = Mmd^2*w = 2.73777*10^34 kg m^2/s
hence
total angular moemntum = 9.4193427377*10^40 kg m^2/s