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If Moon'$ orbital time around the Earth is 27.322 days, predict the angular ' speed of Moon about the Earth in (4) rudls and (b) "Iday:...

Question

If Moon'$ orbital time around the Earth is 27.322 days, predict the angular ' speed of Moon about the Earth in (4) rudls and (b) "Iday:

If Moon'$ orbital time around the Earth is 27.322 days, predict the angular ' speed of Moon about the Earth in (4) rudls and (b) "Iday:



Answers

Calculate the angular speed of the Moon as it orbits Earth. The Moon completes one orbit about Earth in $27.4$ days and the Earth-Moon distance is $3.84 \times 108 \mathrm{~m} 3.84 \times 10^{8} \mathrm{~m} .$ SSM $\underline{\text { Example } 8-1}$

Hello. In this problem we are studying the orbital motion of the moon around planet Earth. So that's the Earth and that's the moon. And we assume in this problem that the orbital is perfectly circular and we know the fact that the radius of this orbital is 3.84 Times 10 to the power of eight. So that's the radius from the center of the earth, that's the radius. And we have also that it takes 27.3 days for the moon to complete a full cycle around the Earth. We're asking in part 8 to calculate the mean velocity for the moon around the Earth. Okay, so we know that velocity is basically the distance crossed over the time that it took to cross this distance and dealing with this. We need to get the circumference of this orbital and the circumference of a perfect circle is basically two pi multiplied by our and we need to divide by time. So let's blogging numbers now now and block this on the calculator to know the answer. So we have to Pi multiplied by 3.84 times 10 to the power of eight. That's in meters. And we need the answer to be in meters per second. So here we have 27.3 days. Each day contains 24 hours, each hour contains 16 minutes and each minute contains 60 seconds. Blackness on your calculator. You will know that the mean velocity of the moon around the Earth is 1022.9 m per second. So that's the answer of part eight. In Part B. Were asked to calculate the centripetal exploration of the moon around the Earth. The centripetal acceleration A. R. Can be calculated directly from the law. The square divided by our and this basically squaring the velocity And divided by the Radius which is 3.8, 4 times 10 to the power of eight. And blocking this on the calculator. We know that the answer is 2.72 times 10 to the power of negative, 3m/s square. So that's the answer for part B. And that's the exploration of the moon around the Earth.

If the moon orbited at four times the present distance of the Earth, what would be the orbiting period? So we have that this centripetal force M one V squared over R. Is equal to the gravitational force. G. M one I am too over R squared. We also have that the period is equal to two pi R over the velocity. So solving the first equation for the velocity, plugging it into the second, Your two pi the square of R cubed over G times M. We know the mass of the Earth is 5.97110 to the 24th kilograms. We know the current distance is 384.410 to the 6th meters, and we need the race to be four times that distance. So solving for the period And we get one 90 10 to the 7th seconds.

All right, So this question is about the moon. And so let's get down the givens, and then we can discuss the questions. Um, so it rotates on a tax is 27.3 days. So then that means it's period is 27.3 days, and, um, the radius is 1.74 times 10 to the six meters Look, and then we want the period in seconds. And so, for a what we need to do is convert this two seconds. So I still do this, Um, in the same way, I learned how to do in high school chemistry. So maybe this is the way you do it, too. But basically, we kind of make this little Charton, so we want to get it into second. So, um, one day is, um 24 hours and then I put the 24 hours up here is because that's what I want as a unit and then day down here cause it cancels with this day, and then I'm left with hours. And then there's 60 seconds and in our 60 minutes per hour. So then there's 3600 seconds per hour, and so um, then so that now the hour cancels out and I'll cancel the day like I talked about. So now we just need to multiply 27.3 times. 24 times serving 600. And that I got, um, Let's see. 123 2.36 times 10 to the sex. Actually, I could just copy and paste the number. Um, seconds. So homeless, right? I have a great patient. 1234 by six. I like to just double check on these things. So that's two point 36 So just keeping three sink fixes. The original number had six sync figs, um, seconds. And we want to get the frequency and radiance for second. And so what to do that? What you want to do is use the formula that period is to pry over Omei guy, and then you can get the Omega is therefore two pi over tea. But you could also kind of start with this equation, just keeping in mind that Omega is radiance for second. So in one full rotation, that takes this amount of time, and that's that's how maney radiance, right? There's two pi radiance in a revolution, so we really just need to do to pie divided by, um t thats number that we just got a few times pride right by that? Yes, Route one. We'll copy and paste the number, I guess again, Um, Rezian's per second. So, um, yeah, you could do the scientific notation conversion if you want. And the linear speed of a rock. Um, just due to the earth rotation. So what you want to do for that is V is equal to our omega. So we just want to take our own makeup that we got before and multiply it by r and R is 1.74 times 10 to the six with that. Oh, wow. It's actually a non astronomical number, 4.64 meters per second. And next, um, we want to get o compare it with a speedboat person on the earth or creator. All right, so basically, we have to kind of read all of this. I'm just gonna go through this sort of fast. So v is our omega, and then I make goes to pi over tea, and then t of the earth is, um so it's 24 hours. So then that's, um we talked about. So then that's 24 times 3600 20 seconds seconds per hour, right? 24 hours times 3600 seconds per hour. And then we want to multiply by the radius of the earth. So it's like that. Yeah. Um, six point 6.37 times 10 of the six imminence again times two pi. So let's go live all of this into a calculator. 6.37 times. 10 to the six times, Two times by and then divided by 24 times. 30. 600. I got 463 meters per second. Ah, double check my work. You know, 27.3 to 4. There's 100. Okay, great. So on earth, you're going to be going a lot faster. And it's at the end of the problem. Yeah,

So Kepler's third Law is pretty easy to understand when we're saying that the length of the period of a or object in orbit over the length of its semi major axis in sequel dues of constant value and this constant values usually defined as four pi squared over GM. So when we get everything into the right units, this for price wearing over GM is equal to one. So the right units for everything R t needs to be in years. You're a or the lensing to be in astronomical units and you're masses need to be in solar units. So national kuna is the average length from the earth to the sun. The solar units is the mass of the sun in kilograms, which is 1.99 times 10 to the power 30th and yeah, that's pretty much the units you need. So when we have this and we assume that everything is in the right units, we can say that T squared over a cube is equal to one. So the right units for tea would be years. So we are given 3 55.256 as the amount of days something one turn that into years we can say that's one year. So one squared equals one over a cube where a is the length of the the distance from that, the orbital distance of the planet in question, which is the earth. And they would also be in astronomical units. So what would they have to be? Well, it's pretty intuitive that they would have to equal warm, which makes sense when we're defining one Astrodome unit as the average distance from the Earth to the sun and one astronomical unit is about 1400. 1005 150000000 So it's a lot of kilometers and he was right down to it's 1.45 times 10 to the power eighth. So this is 1.45 times 10 to the power eight kilometers


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