Supernova energy reached by detector disk 1560 light years away

[skibum143]

Homework Statement

A super nova releases 1.9E45 J of energy over a 10 day period. It is 1560 ly from earth. A detector, facing the star, is a disk of radius 7cm. How much energy reaches the detector? (The detector is in orbit, so it can face the star for the entire 10-day period)

Homework Equations

Intensity = Power / Area
Surface area of disk = pi r ^2
light year = 10E16 m

The Attempt at a Solution

First, I found the initial power of the supernova. 1.9E45Joules / 864,000 seconds (10 days) = 2.199E39 W

Then, I found the initial intensity: this is where I was confused - does the supernova radiate energy as a sphere or a disk? I used a sphere (4 pi r^2)
Intensity = 2.199E39 W / (4 pi (1560E16 m) ^2) = .719 W/m^2
Then I found the power that the disk receives:
Power = Intensity * area = (.719 W/m^2)(pi (.07)m^2) = .011 W
Then .011W * 864,000 seconds = 9562.9 Joules

Not sure what I'm doing wrong?

 

[ideasrule]

That's a correct, though quite roundabout, way of doing it. Your answer is off by a little bit because a light year is 9.46E15 m, not 10E15 m.

An easier way of doing it is to notice that the entire 1.9E45 J of energy is "pasted" onto the surface of a sphere with area A=4*pi*r^2. Dividing 1.9E45 by 4*pi*r^2 gives you the energy/square meter deposited, and multiplying that by the detector's area gives you the answer.

If you're thinking that 10 000 J is too much, that's because supernovae are extremely bright and 1560 ly is extremely nearby . In 1054 AD, Chinese and Arab astronomers recorded a supernova that was visible in daytime for 23 days and easily outshone everything in the sky except the Sun and Moon. That supernova was 6500 light years away. No historically documented supernova has been 1560 ly away or less, so such a supernova would look quite spectacular to say the least!

 

[skibum143]

ah, i see. that makes it simpler, i often make the problems harder than they need to be. thanks for your help!