(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? Consult Appendix E and the astronomical data in Appendix F

Question

Accepted Answer

Hey everyone welcome back in this problem. We are asked to determine the angular momentum magnitude. Okay. For mars revolving around the Sun assuming a circular orbit. Okay. And we're given some information about mars its mass, the radius and its orbit radius and period. Okay, so the mass we'll call it M that we're given is 6. times 10 to the 23 kg. The radius Is equal to 3.39 times 10 to the six m. The radius of the orbit R 002, eight Times 10 to the 11 m. And finally the period T. is equal to 687 days. Alright, We're looking for angular momentum. The magnitude. Let's recall what is angular momentum, angular momentum. L is given by i omega where i is the moment of inertia and omega is the angular speed. Alright, so we don't have omega but we do have the period T. So let's think about how we can relate period to angular speed or angular velocity omega. When we know that t the period is going to be equal to two pi over omega. And so omega, It's gonna be equal to two pi over tea, Which is gonna be two pi over 687 days. Okay. And if we want to put this into our fundamental unit, we want this in seconds. Okay. Where you are going to have to multiply in the denominator. Okay, we're going to get 1.585 times 10 to the minus seven. That's going to be ratings per second. Okay. Alright. So now we have omega are angular speed or angular velocity. The only thing left in order to find l the angular momentum we're looking for is this i the moment of inertia. We're told to treat mars as a particle. Okay. So when we hear particle we think point mass. Now let's recall moment of inertia for a point mass. Using you can use the table in your textbook or provided by your professor. Okay. And that's going to be equal to M R squared. Okay, So we just have M R squared omega. And in this case when we're thinking about our and let me write this as big are until we decide which are we're going to use am R squared omega When we're thinking about are the radius that we're using because we're told to treat this as a particle. We don't need to concern ourselves with the radius of the planet itself. We just want to concern ourselves with the radius of the orbit. Okay, Because we're asked about the angular momentum for it revolving around the sun. Okay, so the center the axis is the sun. And so the radius we're interested in is the radius of that orbit. Okay, If we weren't treating it as a particle, we would need to consider the other radius as well. Okay, But in this case we're treating it as a particle. We just care about the radius of the orbit. So this is going to be m times the radius of the orbit squared times omega, which is equal to 6.4, 2 times 10 to the 23 kg. They are not squared 2.28 times 10 to the m squared Times, 8, 5 times 10 to the -7 radiance per second. Alright. And what we're left with oh, is equal 23. times 10 to the 39 kg meter squared per second. And so that angular momentum is 3.53 times 10 to the 39 kg meter squared per second. That's going to correspond with answer D. That's it for this one. Thanks everyone for watching. See you in the next video.

(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? Consult Appendix E and the astronomical data in Appendix F

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Key Concepts

Angular Momentum

Moment of Inertia

Modeling as a Particle