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Prepare your scantron:

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Presentation on theme: "Prepare your scantron:"— Presentation transcript:

1 Prepare your scantron:
Use a pencil, not a pen! Fill in your name and fill the bubbles under your name. LAST NAME FIRST, First name second Put your 4-digit code instead of “IDENTIFICATION NUMBER”. --- (The last 4 digits of your OleMiss ID.) Test 1: Wednesday during class!! (Prepare; have scantron & pencil) Question # 1: answer A Question # 2: answer A Question # 3: answer E Setup: Note extra credit deadlines: Signup: Sept > send , (no extension) Read the first 13 pages of < Image Proc. Lab at 9:00 Tuesday, Sept. 11; turned in: Sept. 17 Recall reading assignment: Chapter 2, pp. 25 – 52 Please take a moment to mute your cell phone!

2 We skip the review questions today because we must finish all before Wednesday’s test

3 Apparent and absolute brightness
Recall: A parsec (pc) is a unit of distance, 3.26 light years. Why this unit? Because the orbit of Earth looks 1 arc sec radius from the distance of 1 pc. How bright it looks (m) How bright it really is (M) Solar neighborhood (~100 pc) A star’s absolute magnitude is how bright it would look from 10 pc (32 light years) away. The Sun: M = 5mg (an average star)

4 Distance makes stars dimmer
The rule: 2.5 times as far away, looks 2mg dimmer Distance makes stars dimmer Star A and star B have the same absolute magnitude M = 5mg Star B is 10 pc away: apparent magnitude m = 5mg Star A is 25 pc away: apparent magnitude m = 7mg Reality: Looks like: B 10 pc B 25 pc A A

5 Distance modulus The distance modulus m - M = 5  lg d - 5
Relation: M (abs. magn.), m (appt. magn.), d (distance in pc) Meaning of relation: the farther a star, the fainter it looks m - M = 5  lg d - 5 The name of “m - M” is: the “distance modulus” How to use this? Centauri m=1.3mg, distance=1.3 pc => absolute M=5.7mg : a faint star looking bright; distance modulus = mg (close to us) Crucis m=1.6mg, distance=27 pc => absolute M=-0.6mg distance modulus = mg (not that close) Centauri m=2.6mg, distance=120 pc => absolute M=-2.8mg: bright star looking faint; distance modulus = mg (far from us - still in the solar neighborhood) Centauri m=3.5mg, distance=5200 pc => absolute M=-17.1mg: whole star cluster looking faint; distance modulus = mg (very far from us, halfway to the center of the Galaxy)

6 Questions coming …

7 sec 10 9 11 12 15 14 13 8 7 1 2 3 5 4 6 16 26 17 27 28 30 29 24 25 23 19 18 20 21 22 Question 4 How bright would the Sun look in the sky from a distance of 10 parsec? A -12mg (as bright as the full Moon). B 1mg (as bright as the brightest stars in the sky). C 5mg (barely visible to the naked eye). D 15mg (very faint). E Invisible: we cannot see that far. Next question coming …

8 sec 10 9 11 12 15 14 13 8 7 1 2 3 5 4 6 16 26 17 27 28 30 29 24 25 23 19 18 20 21 22 Question 5 Which of the following data of Sirius cannot be directly measured from Earth, only calculated? A Its apparent brightness, which is m =  1.5mg. B Its absolute brightness, which is M = + 1.5mg. C Its speed of motion in the sky. D The color of its light. Next question coming …

9 sec 10 9 11 13 14 8 12 7 2 1 3 4 6 5 15 17 26 25 27 28 30 29 16 24 19 18 23 20 22 21 Question 6 Polaris, the North Star, is m = 2mg, not particularly bright. Its absolute magnitude is M = - 4mg. If you imagine Polaris placed where the Sun is now, would it look brighter, or fainter than the Sun? A Ten thousand times brighter. B Somewhat brighter. C About the same. D Somewhat fainter. E Ten thousand times fainter.

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