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Problem 38 Easy Difficulty

(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.

Answer

a. $2.67 \times 10^{40} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$
b. $7.07 \times 10^{33} \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$

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Video Transcript

in this question, we are asked to calculate the magnitude of the angular momentum off the earth in a circular orbit around the sun. And we are asked to calculate the magnitude of the angular momentum of the Earth due to its rotation around its access through the north and south poles. So this is two different types of angular momentum in thief first part part A. We're going to be thinking of the sun as a point particle. Um, and the reason why it is reasonable to do so. It's because the earth is very, very small in comparison to its orbit around the sun. So if we're looking at the sun or system, the earth basically appears as a point particle, and we can sort of ignore, um, the fact that it has actual dimension. So to calculate the angular momentum for a point particle, um, we're gonna use em V cross our And essentially, um, what this boils down to for a circular orbit is just NPR. So v cross our would technically be VR signed data. Um, but V and R are at 90 degrees to one another for a circular orbit, and so the angle between them is 90 and so sign of 90 is going to equal one, and we don't need to include it. So we're just going to use angular momentum is equal Thio NPR and we don't really know what the speed of the earth is. But what we can dio is rewrite rewrite V as two pi r over the period. So that's the distance that the earth travels around the sun. Um, that's the circumference of its path. Two pi r divided by the period or the time it takes thio do a full, um, circular orbit around the sun. So overall we get the following formula. So we're gonna just plug in our variables. So the M will be the mass of the earth. The R is going to be the radius of the orbit or the distance between the sun and the earth. So that's gonna be no. 149.6 times 10 to the 9 m squared. And then that's going to be over the period. We know that the period is 365 days. Takes a year for the earth to complete one orbit around the sun. Um but we're gonna convert that into seconds because that's our standard unit. So 365 days times 24 hours in a day. Times 60 minutes in an hour, Times 60 seconds in a minute. Um, that will get us our period and seconds. So once we accurately enter all that into our calculator, that's going to give us 2.66 times 10 to the 40 kilograms meters squared per second. So this is the angular momentum of the Earth due to its rotation about the sun or its orbit about the sun. Okay, okay, let's move on to part B. So in this part, whereas to calculate the magnitude of the angular momentum of the earth due to its rotation about it's axis So this is going to be a little bit different. Obviously, we're not going to be able to consider. Um the earth is just a point particle this time, because the Earth is the system, in this case is the only thing we're considering. So it obviously has quite a large, um, breath to it. So it's not a point particle eso in order to calculate the angular momentum for something like that and extended shape. We have to use I times omega. I is what we call the moment of inertia. So it's basically an objects resistance to rotation, Kind of like mass is an objects resistance to movement and for different shapes. There's going to be different formulas for the moment of inertia. So for a sphere such as the Earth, it's going to be 2/5 m times a squared, a being the radius of the sphere. So let's just go ahead and plug in our values for the earth here. And the radius is 6371 kilometers or six million meters. So the moment of inertia for the earth is 9.69 times 10 to the 37 kg meter squared. And then what we're going to do is we're going to plug that in to the formula now. A mega is the angular speed. Um, and we can rewrite that as two pi over t. So two pi over t t being the period of the rotation. So very similarly Ah, to the previous calculation for the period, we have to make sure that that is in fact, in seconds. So, of course, the rotation of the earth takes 24 hours. We're gonna multiply by 60 twice to get that into seconds. And once we do that, we get a value of 7.5 times 10 to the 33 kilogram meter squared per second. So that is our angular momentum of the Earth. Due to the rotation about its access, you can see that the angler mo mentum in Part B is much smaller than the angular momentum in part A. And that's part partly the reason why it's OK. Thio consider the Earth as just a point particle in party, so the Serra final answer.

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