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$\bullet$ (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 $D$ and the astronomical data in Appendix E.
a) $2.67 \times 10^{40} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$b) $7.07 \times 10^{33} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$
Physics 101 Mechanics
Chapter 10
Dynamics of Rotational Motion
Newton's Laws of Motion
Rotation of Rigid Bodies
Equilibrium and Elasticity
Cornell University
Rutgers, The State University of New Jersey
Simon Fraser University
Hope College
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huh? Problem 23 is an angular momentum problem. It is talking about our earth and our son. You will need to look at Appendix D and E in the back of your textbook to find some data about our earth. You'll need the mass of the earth radius of the earth and then our orbital radius around the sun. So party would like to know what is the angular momentum of the earth in a circular orbit around the sun. Okay, angular Momenta formula is I a maker. Now we will model the Earth as a particle. And that would be m r squared. I would be m r squared, Megan. Now, since the Earth is moving in a circle around the sun, we know that Omega can be calculated as two pi r over its period. So, using our data, we know we king simplify this two pi mass radius cube over the period. You know, we have two pi mass of the earth is 5.97 10 to the 24th race to the orbit of the earth around the sun now, period, we have to think How long did it take our earth to go around the sun that also comes in in your orbital data. In the back, your book and they list are it takes 365 point three days. We know we have to convert that into seconds. So every day is 24 hours and every hour is 3600 seconds. Gives are angular Momenta MME. Calculation to be 2.67 I'm Justin to the 40th kilogram. Leaders squared her second hey, part B to calculate the magnitude of the angular Momenta of the Earth due to its rotation around the axis. A little different still. I amega Now the earth is no longer a particle. It is a rotating sphere. So I become to fit some r squared times again. They tube high, are over the period of rotation. So that gives us 2/5. Sorry. Struck change, though two forfeits. Right? Is that page him going back to our data? Our mass of the earth is 5.97 times 10 24 radius of our earth. Now our period this time it takes us 24 hours to rotate in. Converting that two seconds you observe from the angular momentum kilograms made her squirt per second. Thank you for
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