How much energy was Superman exposed to?

[RosutoTakeshi]

In Action Comics #867, Superman survived the explosion of a sun

What I would like to know is, how much energy was he exposed to?
- I know there's the Inverse square law to consider
- And I know because of that, he did not get hit by the entire energy of that explosion because his surface area is very small

I just want to know how much energy actually hit him
For analysis purposes
- Total energy of exploding sun is 1e44 joules
- Supermans distance from explosion is 35.98 million milesquare
- His surface area is 2.35 m^2

Thanks in advance

 

[Orodruin]

So what is the problem? You have correctly identified the inverse square law. Essentially you need to multiply the total energy release by the quotient between Superman's area and the area of a sphere with the distance to the explosion as its radius. This should be fairly straight forward.

- His surface area is 2.35 m^2

That is a quite fat Superman. The area exposed would also depend on Superman's orientation.

 

[RosutoTakeshi]

So what is the problem? You have correctly identified the inverse square law. Essentially you need to multiply the total energy release by the quotient between Superman's area and the area of a sphere with the distance to the explosion as its radius. This should be fairly straight forward.That is a quite fat Superman. The area exposed would also depend on Superman's orientation.

I do not know the formula to properly calculate it. The necessary steps

And I apologize, his "exposed" body surface area is 1.175m^2

 

[Janus]

I do not know the formula to properly calculate it. The necessary steps

And I apologize, his "exposed" body surface area is 1.175m^2

Surface area of sphere:
[tex]A = 4 \pi r^2[/tex]

 

[RosutoTakeshi]

Surface area of sphere:
[tex]A = 4 \pi r^2[/tex]

Yea, the surface area for the sun is a bit above 6.08e18m^2

But I still do not know what to do next with that information. What is the next step?

 

[Orodruin]

But I still do not know what to do next with that information. What is the next step?

you need to multiply the total energy release by the quotient between Superman's area and the area of a sphere with the distance to the explosion as its radius.

 

[RosutoTakeshi]

That's what I don't know how to do, that's what I'm saying

 

[RosutoTakeshi]

I don't know how to do that. I want to learn

 

[Orodruin]

That's what I don't know how to do, that's what I'm saying

Exactly what do you have a problem with? Multiplication and division?

 

[RosutoTakeshi]

Exactly what do you have a problem with? Multiplication and division?

No no, lol, I mean which one do I do first? You said multiply total energy release (1e44 joules) between Superman's area (1.175m^2) and area of the sphere (6.08e18m^2) with distance to explosion as its radius (432,288 miles)

So... is it..

... yeah I'm confused

 

[RosutoTakeshi]

Exactly what do you have a problem with? Multiplication and division?

I got four different numbers here and no idea what to do with them. Do I divide the surface area of the sun by supes surface area before multiplying it by the total energy release? This is where I'm confused

 

[Janus]

That's what I don't know how to do, that's what I'm saying

Okay. First, you know how much energy is released by the sun's explosion, 1E44 joules.

As that energy travels outward from the sun, the sphere that contains that energy expands. Now while the total energy does not decrease, it is being spread thinner and thinner over the surface of the sphere. Since the surface area increases by the square of the radius, the energy per square meter decrease by the square of the radius (this is where the inverse square law come in. The total energy released by the star does not decrease with distance, just the intensity per square meter.
Thus if you take the total energy given off by the sun, and multiply this by Superman's exposed surface area divided by the surface area of a sphere with the radius of Superman's distance from the sun, you get the total energy striking Superman.

 

[RosutoTakeshi]

Thus if you take the total energy given off by the sun, and multiply this by Superman's exposed surface area divided by the surface area of a sphere with the radius of Superman's distance from the sun, you get the total energy striking Superman.

Forgive my lack of intellect, but "that" is what is confusing me. How does "that" look? I know I'm thinking too hard about it and I'm sure it's very simple, but I don't know how to do "that". How does that look in an equation? Would you mind showing me an example of how it looks? I hate feeling stupid

 

[RosutoTakeshi]

Does it look like this?

1e44 × (1.175m^2/6.08E18)

 

[Janus]

Forgive my lack of intellect, but "that" is what is confusing me. How does "that" look? I know I'm thinking too hard about it and I'm sure it's very simple, but I don't know how to do "that". How does that look in an equation? Would you mind showing me an example of how it looks? I hate feeling stupid

E= total energy released by the star
R = the distance of Superman from the sun
a = Superman's exposed surface

then
if A equals the surface area of a sphere with a radius equal to Superman's distance from the sun, then
A = 4*pi*R

And e, the amount of energy striking Superman is

e = E * a/A

 

[Janus]

Does it look like this?

1e44 × (1.175m^2/6.08E18)

close, but you are using the surface area that you gave for the Sun earlier. You don't even need to know that to solve your equation. You need the surface area of a sphere with radius of 35.98 million miles . (which is what I assumed you meant even though you wrote "milesquare")
Be sure to convert miles to meters before making the calculation.

 

[Noel]

Also consider that the totality of energy does not arrive in a zero thickness expanding shell. Thus Superman's rate of energy exposure will depend on the actual temporal thickness of the energy shell, and his distance from the nova will presumably increase during that time period, thus reducing his actual energy impact.

Furthermore, it is likely that his orientation will change like a flag (cape?) in the wind, thus reducing his areal exposure.

 

[RosutoTakeshi]

E= total energy released by the star
R = the distance of Superman from the sun
a = Superman's exposed surface

then
if A equals the surface area of a sphere with a radius equal to Superman's distance from the sun, then
A = 4*pi*R

And e, the amount of energy striking Superman is

e = E * a/A

Ok... ok I'm going to attempt this

E = 1e44 joules
R = 35,980,000 miles (57,904,197,120 meters)
a = 1.175m^2

A = 4 * 3.14 * 57,904,197,120
A = 4.21E22m^2

e = 1e44 * (1.175m^2/4.21E22m^2)
e = 2.79e21 joules/m^2

 

[RosutoTakeshi]

Also consider that the totality of energy does not arrive in a zero thickness expanding shell. Thus Superman's rate of energy exposure will depend on the actual temporal thickness of the energy shell, and his distance from the nova will presumably increase during that time period, thus reducing his actual energy impact.

Furthermore, it is likely that his orientation will change like a flag (cape?) in the wind, thus reducing his areal exposure.

Yes, but for simplicity sake, I'm just going with (his distance from the explosion did not increase upon impact)

 

[RosutoTakeshi]

close, but you are using the surface area that you gave for the Sun earlier. You don't even need to know that to solve your equation. You need the surface area of a sphere with radius of 35.98 million miles . (which is what I assumed you meant even though you wrote "milesquare")
Be sure to convert miles to meters before making the calculation.

Did I get it right?

 

[Janus]

Ok... ok I'm going to attempt this

E = 1e44 joules
R = 35,980,000 miles (57,904,197,120 meters)
a = 1.175m^2

A = 4 * 3.14 * 57,904,197,120
A = 4.21E22m^2

e = 1e44 * (1.175m^2/4.21E22m^2)
e = 2.79e21 joules/m^2

Looks good.

 

[RosutoTakeshi]

Looks good

Thanks a lot for the assistance, I appreciate it