given-that-the-earths-mass-is-m-its-tangential-speed-as-it-revolves-around-the-sun-is-v-and-the-dist

Question

Given that the Earth's mass is $m,$ its tangential speed as it revolves around the Sun is $v$ , and the distance from the Sun to the Earth is $r,$ which of the following correctly describes the work done by the centripetal force, $W_{c}$ , in one year's time?
(A) $\mathrm{W}_{\mathrm{c}}>2 r\left(m v^{2} / r\right)$
(B) $\mathrm{W}_{\mathrm{c}}=2 r\left(m v^{2} / r\right)$
(C) $\mathrm{W}_{\mathrm{c}}<2 r\left(\mathrm{mv}^{2} / r\right)$
(D) Cannot be determined

Vishal Gupta · Accepted Answer

Hi in uh uniform circular motion as a century pittle force, which is radial in nature in direction. So it acts always for perpendicular to the direction of motion. Because it is tangential, the direction of motion is in general. Hence we can say Work done by centripetal force is always zero Means WC is zero. Hands only correct. Option seems to be correct. Option is that this work done by the centripetal force is less than to our multiplied by century pittle force mv square by uh, because options say W. C. Is more than this. No, it cannot be because it is zero. Option B says W. C. Is equal to to work into this car buyer, which is not correct Because W 30. So as W zero. So we can say WC is less than two are into invisible by art and option B. C. S. We cannot little mind this work. No, that is also not correct. So here we can say option. See is correct. Thank you.

Ze-Han Lee · Answer

All right, So, uh, in this problem, we know that Ah, the earth is moving along, Uh, the second copy of it around the around the cell. So this is the earth, and this is the South here and from the problem, we know that the mass of the earth is, um, and the potential velocity. So the message them, the tangential velocities read. Okay. And we also know that Ah, the distance from the sound to the earth is ah, are so These are so one to determine the the work down by the centripetal force W C in a year&#x27;s time. So a year means ah, company, period. Right. Just starting from this position and at the same position. So in a year time, actually, Ah, you can You can just ah, start from another direction. Seeing that because things are your time. There&#x27;s ah, we&#x27;ll come back to the original position and velocity actually will not change saying that says the sound. We&#x27;re not to do any work to the ah, tears. All right, so in that sense, we know that a central force centripetal force w c should people to zero. Okay, so we know that ah w c should be smaller than the given choices, which is to see times Ah, the squid time Sam of are all right.

Ze-Han Lee · Answer

All right, so in this problem, we have object, which is moving along The circular path Was Constance beefy? Okay, So based on this information, we know that the centripetal force is pointing to the sense of this off this off his path. All right, so this is the direction of the centripetal force. And as you can see, that to the Central Air Force is always from particular to the velocity. Right. So in that case, the ah, the force, the Central Force, there&#x27;s no work to the object. Okay, so the work is always zero.

Lisa Tarman · Answer

for number 28 were giving the radius of the orbit of the earth going around the sun. And we&#x27;re to find the velocity of the earth going around the sun. So velocities is it&#x27;s changing position per time. It goes the whole way around the circumference of the earth. So two pi r to pie are. And I was given my radius right by the time it takes one year for the earth to go around the sun. So that&#x27;s 365 days, and I&#x27;m just gonna put my conversions in here. If I&#x27;m gonna change days, two hours. I know there&#x27;s 24 hours in a day. Our worst a minute. They&#x27;re 60 minutes in an hour and then minutes two seconds, 60 seconds in a minute. I do a mess. Get three point. Oh, I should probably two significant figures. So 3.0 times. 10 to the fourth meters per second. Part B. We&#x27;re supposed to find the angular or who asked you? So that&#x27;s it&#x27;s change. Don&#x27;t fade A for time, right? I want to be in radiance. I know that one time around is two pi radiant. So it&#x27;s basically this without the are without the radius. So two pi over the time again, this whole conversion. So through 65 that I&#x27;m going to change hours, two minutes, two seconds. I get 2.0, times 10 to the negative Seventh radiance and then for port. See, I&#x27;m asked to find the centripetal acceleration. So centripetal acceleration picnic that linear velocity squared over the time. No, over the radius. So I just found when you were, lest he appear gonna square that divided by the radius And I was given the radius at the beginning. 1.5 times. 10 to the 11th. I get 0.6 times. No, that&#x27;s it. Point over six. And they&#x27;re being meters per second squared. I should probably keep two significant figures. So had a significant figure there.

Marshall Styczinski · Answer

So here we&#x27;re dealing with movement of the earth around the sun and were given that the period of a year in seconds is three point 16 times tend to the seven seconds the radius is 1.5 times tend to the 11 meters. And because they have conveniently given us these numbers and s I units, we won&#x27;t need to do any conversions and part, eh rest for the angular speed of the earth which is to pie over the period. And when you plug that into our calculator, we get 1.99 times 10 to the minus seven rads per second, and part B is finding tangential speed of the earth, which is the radius times. The angular speed, which will be 2.98 times 10 to the fore meters per second and part see, is finding the centripetal acceleration, which is the tangential velocity squared, divided by the radius. And since we just found the tangential velocity, we were given the radius. We have everything we need and when we plug in those values, we get 5.92 times 10 to the minus three meters per second squared

Mayukh Banik · Answer

are vital Acceleration is given by or by the speed squared over r and our bite a speech days GM over R So the Arctic that you know me that Mexico are square putting the numbers eight we get 6.67 times tend to the native 11 times mass of the sun is one for 989 time Stand to that 30 divided by radius warm for 9.6 times 10 to the nine square off that and that comes out to be 5.9 to 8 Parents gentle in 83 Made up for a second. Yeah. Now, for the second case, we have the orbit of the acceleration a prize or the centripetal acceleration A prize is given by on the guest square are off art Oh, my days to buy you Mardy So four by square, out of one d It was for price where times are these six foreign 37 times 10 to the six divided by the period is 23.93 hours times 3602nd. So that comes out to be 3.388 They&#x27;re sent to the native to made up by a second squared, so we find a prime over A. It was 3.388 times 10 to the native to divided by 5.92 a times 10 to the native. Three. It was five for a 715 So your pride centripetal acceleration at sequitur of the art he&#x27;s find by 715 times greater than odds acceleration.

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