(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix E and the astronomical data in Appendix F.

Solution 38E The angular momentum can be calculated using the relation, L = mvr, where r is the perpendicular distance between the object and axis of rotation, m is the mass of the object and v is the linear velocity of the particle. When the earth orbits around the sun 7 Time period T = 365.3 days = 365.3 × 24 × 60 × 60 s = 3.15 × 10 s 2 Angular speed of the earth = T = 2×3.147rad/s 3.15×10 7 = 2 × 10 rad/s 11 Now, the orbital radius of earth is r0= 1.50 × 10 m 11 7 Linear speed of the earth v = r 0 = 1.50 × 10 m × 2 × 10 rad/s v = 3 × 10 m/s Now, mass of the earth m = 5.97 × 10 24kg (a) Now, substituting the values of m, v and r in th0 equation, L = mvr , we can calculate the angular momentum. L = 5.97 × 10 24kg × 3 × 10 m/s × 1.50 × 10 11m 40 2 L = 2.68 × 10 kg.m /s This is the magnitude of earth’s angular momentum in its circular orbit around the sun. Yes, it is reasonable to model the earth as a particle as the earth’s radius is much smaller than the orbital radius. (b) The earth takes 24 hours to make one rotation around its axis. Therefore, time period T = 24 hrs = 24 × 60 × 60 s = 86,400 s 1 Angular speed = 2 rad/s = 7.26 × 10 5 rad/s 1 86,400 If the earth is modeled as a perfect sphere, its moment of inertia is given as, I = 2 × M ×eR e2 5 2 24 6 2 2 I = 5 × 5.97 × 10 kg × (6.37 × 10 ) m 37 2 I = 9.7 × 10 kg.m Now, substituting the values of I and in, 1 37 2 5 L =2I = 917 × 10 kg.m × 7.26 × 10 rad/s 33 2 L =27.04 × 10 kg.m /s This is the required approximate value of angular momentum.