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Supernovae

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Type Ia supernovae all emit the same amount of energy. Some appear brighter to us than others because they are at different distances from Earth. Astronomers can use these facts to determine distances to these supernovae, and so can you!

Astronomers find that if all Type Ia supernovae were at a distance of 100,000 parsecs, they would have a peak magnitude of +1 (about as bright as the star Aldebaran). They treat this distance and magnitude as a "standard". The actual distance to a particular supernova is found by first comparing its peak magnitude with this standard. Now remember that the magnitude scale is set up so that each difference of 5 magnitudes (e.g. between magnitude +1 and +6) represents a difference in brightness of 100 times. A difference of 10 magnitudes represents a difference in brightness of 100 x 100 (= 1002), or 10,000 times, and a difference of 15 magnitudes is a difference in brightness of 1003 or 1,000,000 times.

The inverse square law compares the brightness of objects with their distances. An object twice as far away from you appears 22 or 4 times fainter, and one 3 times farther appears 32 or 9 times fainter. Likewise, the inverse square law tells us that we can find the difference in distance by taking the square root of the difference in brightness. So by comparing the brightness of two objects, we can determine their relative distances. Then if we know the actual distance to one of the objects, we can determine the distance to the other. With our standard magnitude of +1 at a distance of 100,000 parsecs, we can thus determine distances to Type Ia supernovae.

inverse square law
An illustration of the Inverse Square Law.
The Intensity (or brightness) of a source of light falls off as the inverse square of the distance from the source. This is because the light's intensity can be thought of as being spread out uniformly over the surface of a sphere. At greater distances, the sphere is larger, so the same amount of light is spread over a larger area. This makes the light source appear dimmer as the distance from it increases.

Example: A type Ia supernova has a peak magnitude of +11. How far away is it ?

Solution: This supernova is 10 magnitudes fainter than our standard supernova of magnitude +1. Since each difference of 5 magnitudes gives a factor of 100 times fainter in brightness, our supernova is (10/5) = 2 of these factors fainter than our standard. It is thus 100 x 100 (= 1002), or 10,000 times fainter than our standard. We next use the inverse square law to determine how much farther away it is than our standard. We can find the difference in distance by taking the square root of the difference in brightness. We find our supernova is

sqrt(10,000) = 100 times
farther away than our standard. Recall that the standard magnitude for supernovae is for supernovae at a distance of 100,000 parsecs. So our supernova is at a distance of
 100 x (100,000 parsecs) = 10,000,000 parsecs
Because the distances to supernovae are usually large, astronomers often express them in Megaparsecs (= 1,000,000 parsecs), which is abbreviated Mpc. So our supernova is 10 Mpc away!

Now try some yourself:
For each of these observed magnitudes of type Ia supernova, determine their distances:

(Note: Don't use commas or spaces in your answers.)

Observed Magnitude Distance (Mpc)
*
*Try this as a bonus question. You'll need a calculator.


This activity is based on an idea by Cheryl Carter, Bladensburg High School, Bladensburg, MD.

If words seem to be missing from the articles, please read this.

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This page last updated: Tuesday, 25-Sep-2007 14:35:17 EDT