- #1

- 112

- 0

## 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?