Rotational velocity of Earth

by Patrick Ford
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All right. So here we have an example, recalculating the rotational velocity of the earth, um, as into different situations as it rotates on itself and rotates around the sun. So let's look at the first one earth around itself. Okay, so, actually, I want w of Earth around self. Okay, well, W is either delta data over Delta T or two pi over tea or two pi f now, as the earth rotates around itself. What's the Delta Fada? What's the time? I'm not telling how many degrees it's going through. I'm not telling you this either. So one of the things you can do is you can then use some information you know about the earth. And you know how long the period of the earth is. How long the er it takes to make a complete revolution around itself. If it's a complete revolution around itself, that takes 24 hours, right? So it takes 24 hours. So what I can do is I can just write w equals two pi divided by 24 hours. Obviously, I can't just leave. It is 24 hours. I have to convert. I have to convert that into seconds. Okay, so it's gonna be two pi times, 60 time 60 and you multiply all of this. You get that seven. That the W for the earth is 7.27 times 10 to the negative fifth radiance per second. It's a very small number. That's because, um, the earth takes a very long time to spin around itself, right relative to like a DVD or something. All right. Um, be how long, Or rather, what is the rotational velocity of the earth rotating around the sun? Very similar problem. W equals. I can use any of these equations. In this case. What I'm going to use is again two pi over tea and the period of the earth going around the sun is days, okay? And you don't have to worry about be peers or any of that stuff just simplified by doing this. So it's two pi 365. So it's basically all of this. This is one day, right? 24 hours. So it's all of this times 24 times 3 65. So I'm gonna repeat the bottom there, and this is gonna be an even smaller number. I get 1.99 times to the I have it here to the negative seven radiance per second. Okay, then The last thing I wanna point out to you guys is that if the Earth is rotating around itself in this case, we're talking about a We're talking about a rigid body that's spinning around itself because it's just sort of a sphere. Not exactly, but you can approximate the sphere that spins around itself. Right? Um, the earth rotating around the sun actually acts as a tiny point object. Why? Because relative to the size of the sun and relative to the distance between the two, the earth is so tiny that you would consider it a point mass going around the circle. Here you have a rigid body spinning around itself. So that's an important distinction in these two examples. A knob checked can, depending on the situation, be thought of as a rigid body that spins around itself or a tiny point mass that spins around, uh, a circular path. Okay, that's it for this one. Hopefully make sense. Let me know if you guys have any questions