Compute the ratio of the magnitudes of the Earth\'s orbital angular momentum and

Question

Compute the ratio of the magnitudes of the Earth's orbital angular momentum and its rotational angular momentum.

We know that the Angular momentum L = m*r*? = m*r*2p/T

Let r1 be the orbital radius, r1 = 1.50x10^11 m and T1 = 365day*24*3600 = 3.154x10^7 s
Let r2 be the Earth's radius, r2 = 6.38x10^6 m and T2 = 24*3600 = 86400 s

So L1/L2 = (r1/T1)/(r2/T2)
= r1*T2/(r2*T1)
= 1.50x10^11*86400/(6.38x10^6*3.154x10^7) =
= 64.4

** This is not the correct answer, and no matter how I try to change it I can not see to come up with the right one. Please help!

Explanation / Answer

The orbital angular momentum of the Earth is given by Lo = mr^2.w m is the mass of the Earth r is the orbital radius of the Earth =1.50x10^11 m The rotational angular momentum of the Earth is given by Lr = Iw Lr = 2/5 mR^2.w R is the radius of the Earth =6.38x10^6 m The ratio of the magnitudes of the Earth's orbital angular momentum and its rotational angular momentum is given by: Lo/Lr = (mr^2.w)/(2/5 mR^2.w) Lo/Lr = 5r^2/2R^2 Lo/Lr = 5*(1.50x10^11 m)^2/2*(6.38x10^6 m)^2 Lo/Lr = 1.38*10^9

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.