Calculate the current angular momentum of the Earth-Moon system as an isolated s

Question

Calculate the current angular momentum of the Earth-Moon system as an isolated system. For simplification, we will assume the total momentum of the system involves only the rotation of the Earth about its axis and the orbit of the moon around the Earth. Also, we will assume all motion is with respect to the axis of the Earth, the axis of Earth's rotation and the Moon's orbit are parallel, the Moon's orbit about the Earth is circular, and that the motion is with respect to the center of the Earth (not the center of mass of the Earth-Moon system). Since current estimates have the Moon receding from the Earth at approximately 3-4 cm per year, what factors are changed in the system to conserve angular momentum by having the distance change?

Explanation / Answer

mass of earth = 5.9736×1024 kg ; radius = 6,371.0 km = 6.371*10^6 m

mass of moon = 7.3477 × 1022 kg ; moon does 1 rotation in 27.32 days

radius of orbit around earth (semi major axis of ellipse is taken) = 384,399 km = 3.844*10^8 m

I of earth = 2/5 mr2 = 0.4 * (5.9736×1024) * (6.371*10^6)^2 = 9.698 * 10^37 kg/m2

of earth = (2)/86400 = 7.272 *10^-5 rad/s

Ang momentum of earth = I = 9.698 * 10^37 kg/m2 * 7.272 *10^-5 rad/s = 7.053 * 10^33 (ans)

Ang momentum of moon about earth :---

I of moon abt earths axis ; taking moon as pt mass

I = (7.3477 × 1022 kg) * (3.844*10^8 m)^2 = 1.0857 *10^40 kg/m2

of moon = (2)/(27.32 * 86400) rad/s = 2.66 *10^-6rad/s

Ang momentum of moon = I = (1.0857 *10^40 kg/m2) * (2.66 *10^-6 rad/s) = 2.89 * 10^34

Total ang momentum of moon earth system is thus = 3.595 * 10^34 (ans)

According to this model, the variable of distance between earth & moon increasing can be offset by decreasing the rotation speed of moon about earth i.e. moon will then complete a revolution about earth in more than 27.32 days (ans)

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