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Problem 6. At the radius of the Earth' s orbit (1.50 x 1011 m), the Sun's radiation has an in- tensity of 1.36 kW /m? From this fact, find the rate at whi...

Question

Problem 6. At the radius of the Earth' s orbit (1.50 x 1011 m), the Sun's radiation has an in- tensity of 1.36 kW /m? From this fact, find the rate at which the mass of the Sun is decreasing At this rate, how long until the Sun is gone?

Problem 6. At the radius of the Earth' s orbit (1.50 x 1011 m), the Sun's radiation has an in- tensity of 1.36 kW /m? From this fact, find the rate at which the mass of the Sun is decreasing At this rate, how long until the Sun is gone?



Answers

The Sun radiates energy at a rate of about 4 $\times 10^{26}W$. ($a$) At what rate is the Sun's mass decreasing? ($b$) How long does it take for the Sun to lose a mass equal to that of Earth? ($c$) Estimate how long the Sun could last if it radiated constantly at this rate.

Let's do part, eh? Well, using E is equal to M C Square and this kangaroo Ginny's Delta V is equal to, uh, Delta M Times C Square and dividing Delta T on both sides. We have dealt any divided by Delta T. It's equal to, um one divided by C square into Delta E divided by Delta uh, T All right, since ah Delta M Times C squared is Dante and further, um, Delta M divided by Delta T Delta AM, divided by Delta T, is equal to four. Multiply by 10 to the power 26 Jule Bert second, divided by 3.0 multiplied by 10 to the power feed a meter per second old square. And this Delta hem divided by Delta T, is equal to 4.44 Multiply by 10 to the power nine kilogram for a second and which is approximately called too, is approximately called to four. Multiplied by tend to the bomber mine kilogram for a second. Now let's to Part B well again. Using a Delta T is equal to muscle earth divided by Delta M divided by Delta T. And let's plug in the values Aah! Delta T is equal to 5.98 Multiply by 10 to the power. 24 kilogram is ah, mass of the earth divided by Okay for point for four. Most of my bite into the bollard nine kilogram per second. Multiply by 3.156 Multiply by 10 to the power seven seconds for a year on. Simply find this Delta T is approximately gold, too, for multiply by 10 to the bomber. Seven years now let's do park si the list. I'm Wittig Morse off son. Time is equal to muscle son, divided by DealTime divided by Delta T with muscle Sonny's one point fine. Nine. Multiply by 10 to the power tardy kilogram divided by, uh, change in Mass is 4.4 four. Most of lie by 10 to the Power Mind Kilogram or second multiply by, UM, three point one 56 multiplied by 10 to the dollar. Seven. Second spur here and therefore time is a cool too one point for two. Multiply by 10 to the power 13 years, which is approximately equal to one multiplied by 10 to the power 13 years

Okay, First of all, let me summarize the current problem. So we imagine the oath and a son. This is the sun, and this is here. The sun is radiating with some unknown power per unit area hs um And when the sun's radiation reaches the earth, the powerful you'd error is estimated to be 1.5 times 1.5 kilowatt. Amita squared. We know that the distance between the earth and the sun is 1.5 times 10 to the 11 meters, and the radius of the sun is 6.96 times 6.96 10 to 8 meters on DH. We need to find the power for unit area off the sun at the surface of the sun. And so, in order to do this, we need to recall that the power the total power is given by the powerful unit area well supplied by the area of the sun tell it to the total power radiated by the sun is equal to the power unit area at the surface of the sun multiplied by the surface area of surface area of the sun is equal tio full pie times the radius of the sun squared So this is a total power that is radiated by the sun, and this total power is can be reasonably assumed to be a constant when it reaches the other. So assuming that there is no power, you know, being alone from the from the from the, you know, on the territory between the under the trajectory between the sun India, then the total power of the sun when it reaches the the fear that the earth is rotating on will be the same as the total Paulo at the surface of the sun. And so this is equal to P at the position of the off. So on the other hand, p, the total power at the position of the earth is equal to H E, multiplied by the surface area off this whole fear. Right here. So this is intern is equal to a full pie Time's philistines between the Earth and the Sun Square, and it is equal to P s on. This is equal Tio. Hmm. Multiply last full pie R square. Until from here we can rearrange the equation to find that the power the total power per year area at the surface of the sun is equal to the power per unit area at the position of the Earth multiplied by a distant between your square divided by the radius with son quick and then and so all we have to do now is sew quality in these numbers, so this's equal to 1.5 kilowatt per meter square. Multiply by 1.5 times 10 to the 11 divided by 6.9 times 10 to 8. Swing. This is me. You can't go out. So if you actually do the calculation, you'll get 69,671 points by kilowatt per meter squared. So that is a chest that is part ay off the problem. Pardon me, It's where we assume that the sun is the perfect black body, and we need to find ah, we need to find the temperature, the surface temperature of the sun. So in order to do this, we recall Stefan's Bozeman's Law, which say that the power per you'd area is equal. Teo signal multiplied by excellent not supplied by T to the fourth, where Sigma is Stefan Bolton and Constant and Signal is estimated to be 5.67 times 10 to the negative And after Lan for perfect back black Bori is equal to one and open here. Since we already know the power the power valued area at the proficient son from part a re simply just plugging numbers tea would be equal Tio hs divided by signal times as alone to the 14 and this is equal tio 69,671 0.5 times. Send to the third what our meter squared divided by, um divided by 95.67 climb 10 to the negative to the wonderful And this turns out to be 5920 0.63 Captain. So that's what stuff it temperature office and something will. This problem is that we can estimate the surface temperature of the sun just by measuring power continued area at this position of yours, which is something that we are able to do, so that is no problem

Let's do not see as the speed of light. Now this means that the distance that is given Toby 1.5 times 10 to the part 11 send the middle. This is equal to speed off the light, Einstein. So the time can be calculated by dividing their distance. Let's see, not substituting the values. So the morning ran you off the speed of light. So that's what we're gonna use. That is three times tend to the bar eight. You talk for a second to convert the distance to meter on, then substitute off the values. We want the units to be consistent so the length of dimension off land should be seen. Now the unit off length should the same. I'm sorry. Now, substituting the values he find the time to be 500 seconds. Now the time for light across the diameter off parts of it will be twice this number. So why you Soft is 1000 seconds now the astronomical only Rome are found at the time. Let's do not Nash. So he calculated Oh, he didn't mind the time to be toting 100 seconds on what you found the time to be Hunt 1000 seconds. So this is a little bit less than the astronomic determination, but he got reasonably close, so yeah, on the tie intact you found over here. 500 seconds. This is nearly 1/4 of a 5000.3 minutes. The 1000 seconds is merely called to 16 points experience. Thank you.

So the distance, which is the distance from art to son, is equal to speed off light times the time taken by light to cover the distance to travel from the sun. To our this means that he will be equal to be overseas. My day is going to be 1.5 times 10 to the part 11 bitter and see that it's a speed off life, the modern value of speed off life. So that is three times 10 to the party ate meat up a second. Storing the consolation, you find the time to be 500 seconds. Therefore, the time for light to cross the diameter off its orbit is twice this number. So be cross this choice off the Stein and therefore this is equal to do 1000 seconds. Now, if you compare this time So the one that Ole Roan was measured. So that is 300 seconds, so d no, I'm more so the astronomical determine the time to be talking under seconds. Therefore, your calculation is less than his value. So you can say that although it's less but a letter Mars got reasonably close to the actual time. Thank you.


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