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# Nuclear physics question

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1. Imagine the Earth to be part of a sphere of radius 1.35×10^11m, with the sun at the centre. The sun converts about 4×10^6 tonnes of matter/sec into energy. Calculate the intensity of energy reaching the Earth from the sun. (intensity is power per square metre).

Don't really understand the question/where to start (ignore the image, it's not the right orientation).
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2. (Original post by theguywhosaidhi)
Imagine the Earth to be part of a sphere of radius 1.35×10^11m, with the sun at the centre. The sun converts about 4×10^6 tonnes of matter/sec into energy. Calculate the intensity of energy reaching the Earth from the sun. (intensity is power per square metre).

Don't really understand the question/where to start (ignore the image, it's not the right orientation).
Intensity = Power/Area.

It already gives you the power as 4x10^6 joule tonnes since power is defined as the amount of energy being transferred per second. Convert joule tonnes per second to Joules per second which is 4x10^12 Watts.(1Joule tonne/second = 1,000,000 watts). Then do I = P/A where A is the are of the sun and you can calculate the Surface area of the sun using A = 4Pir^2.

Should arrive at answer of 1.74x10^-11 Wm-2. I think this or right, anyone tell me if I'm wrong.

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3. (Original post by RossB1702)
Intensity = Power/Area.

It already gives you the power as 4x10^6 joule tonnes since power is defined as the amount of energy being transferred per second. Convert joule tonnes per second to Joules per second which is 4x10^12 Watts.(1Joule tonne/second = 1,000,000 watts). Then do I = P/A where A is the are of the sun and you can calculate the Surface area of the sun using A = 4Pir^2.

Should arrive at answer of 1.74x10^-11 Wm-2. I think this or right, anyone tell me if I'm wrong.

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not quite what i thought after a little while, what do you think of this?

1 tonne = 1000kg btw, so it's 10^9

so we know that its converting that much mass/sec to energy, so i use E per sec = M per sec C^2. Then you get Energy per second = (4 * 10^9)kg * (speed of light squared) = 3.6 * 10^26 Js^-1

Js^-1 = Watts = power

then i did what you did with the I = P/A, but with the different power value above to get I = 1565 Wm^-2.

let me know what you think
4. (Original post by theguywhosaidhi)
not quite what i thought after a little while, what do you think of this?

1 tonne = 1000kg btw, so it's 10^9

so we know that its converting that much mass/sec to energy, so i use E per sec = M per sec C^2. Then you get Energy per second = (4 * 10^9)kg * (speed of light squared) = 3.6 * 10^26 Js^-1

Js^-1 = Watts = power

then i did what you did with the I = P/A, but with the different power value above to get I = 1565 Wm^-2.

let me know what you think
Sorry but 1 joule tonne does not equal 1 billion joules, it's 1 million. Besides that you're working looks good though.

1 j to 1kj = 10^3
1kj to 1 joule tonne(mega joules) = 10^3

Therefore 1j to 1 joule tonne is 10^6.
So 4x10^12 joules/second. If it was 4x10^9 it would Kj/second but you want it in joules.

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5. (Original post by theguywhosaidhi)
not quite what i thought after a little while, what do you think of this?

1 tonne = 1000kg btw, so it's 10^9

so we know that its converting that much mass/sec to energy, so i use E per sec = M per sec C^2. Then you get Energy per second = (4 * 10^9)kg * (speed of light squared) = 3.6 * 10^26 Js^-1

Js^-1 = Watts = power

then i did what you did with the I = P/A, but with the different power value above to get I = 1565 Wm^-2.

let me know what you think
That's definitely in the right ballpark - the book value is about 1400 Wm-2 at the top of the earths atmosphere iirc
6. (Original post by Joinedup)
That's definitely in the right ballpark - the book value is about 1400 Wm-2 at the top of the earths atmosphere iirc
Hey I think I've done it wrong but do you mind looking at the way I did it and see if that's a viable way of doing it. I tried to do I = P/A where A was the surface area of the sun. But I'm quite confused I haven't done this type of question in a while. Thanks.

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7. (Original post by RossB1702)
Sorry but 1 joule tonne does not equal 1 billion joules, it's 1 million. Besides that you're working looks good though.

1 j to 1kj = 10^3
1kj to 1 joule tonne(mega joules) = 10^3

Therefore 1j to 1 joule tonne is 10^6.
So 4x10^12 joules/second. If it was 4x10^9 it would Kj/second but you want it in joules.

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ah, you see I think it's talking about tonnes as in mass. My working is turning that mass (tonnes) to kilograms is only 10^3 difference, hope that makes sense. I don't think it's talking about Joule tonnes, but mass instead.
8. (Original post by theguywhosaidhi)
ah, you see I think it's talking about tonnes as in mass. My working is turning that mass (tonnes) to kilograms is only 10^3 difference, hope that makes sense. I don't think it's talking about Joule tonnes, but mass instead.
I thought you were turning it to just joules. Yeah tonnes to kilojoules would just be 10^3. So it's not a mega joule ?

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9. (Original post by theguywhosaidhi)
ah, you see I think it's talking about tonnes as in mass. My working is turning that mass (tonnes) to kilograms is only 10^3 difference, hope that makes sense. I don't think it's talking about Joule tonnes, but mass instead.
What level are you studying at ? I'm doing Scottish advanced higher physics.

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10. (Original post by RossB1702)
Hey I think I've done it wrong but do you mind looking at the way I did it and see if that's a viable way of doing it. I tried to do I = P/A where A was the surface area of the sun. But I'm quite confused I haven't done this type of question in a while. Thanks.

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well 1.35×10^11m is about 1AU, not the radius of the sun... so you might have got the right answer for the wrong reason... cos you don't need to worry about the surface area of the sun.

assuming the sun radiates the energy away equally in all directions finding the power per m2 of the sphere around the sun with radius 1AU is the required calculation.

there's a calculus way of expressing it formally but hopefully you can see that it makes sense.
11. a level (a2) physics, the question only tells us the mass per second and not anything about the value of energy - i think its something you need to work out . i convert the mass tonnes (10^6) into Kg (10^3) because its the standard unit for mass and gives an answer in Joules for E = MC^2
12. (Original post by theguywhosaidhi)
a level (a2) physics, the question only tells us the mass per second and not anything about the value of energy - i think its something you need to work out . i convert the mass tonnes (10^6) into Kg (10^3) because its the standard unit for mass and gives an answer in Joules for E = MC^2
Yeah thanks. Good luck on your A levels.

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13. (Original post by RossB1702)
Yeah thanks. Good luck on your A levels.

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you too with your Scottish highers!

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