21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
6th Edition
ISBN: 9780393874921
Author: PALEN
Publisher: Norton, W. W. & Company, Inc.
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Chapter 10, Problem 26QP
To determine
The explanation for Great Red Spot and its changes over time.
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Venus can be as bright as apparent magnitude -4.7 when at a distance of about 1 AU. How many times fainter would Venus look from a distance of 5 pc? Assume Venus has the same illumination
phase from your new vantage point. (Hints: Recall the inverse square law; also, review the definition of apparent visual magnitudes. ote: 1 pc = 2.1 x 10° AU).
times fainter
What would its apparent magnitude be?
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Calculate the rate of spread in the hypothetical hot spot islands below. Your answer should be in cm/year. Round to the nearest tenth.
arded
7 mya
6.3 mya
4.5 mya
3 mya
1500 kms
1.2 mya
0.8 mya
Chapter 10 Solutions
21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
Ch. 10.1 - Prob. 10.1CYUCh. 10.2 - Prob. 10.2CYUCh. 10.3 - Prob. 10.3CYUCh. 10.4 - Prob. 10.4CYUCh. 10.5 - Prob. 10.5CYUCh. 10 - Prob. 1QPCh. 10 - Prob. 2QPCh. 10 - Prob. 3QPCh. 10 - Prob. 4QPCh. 10 - Prob. 5QP
Ch. 10 - Prob. 6QPCh. 10 - Prob. 7QPCh. 10 - Prob. 8QPCh. 10 - Prob. 9QPCh. 10 - Prob. 10QPCh. 10 - Prob. 11QPCh. 10 - Prob. 12QPCh. 10 - Prob. 13QPCh. 10 - Prob. 14QPCh. 10 - Prob. 15QPCh. 10 - Prob. 16QPCh. 10 - Prob. 17QPCh. 10 - Prob. 18QPCh. 10 - Prob. 19QPCh. 10 - Prob. 20QPCh. 10 - Prob. 21QPCh. 10 - Prob. 22QPCh. 10 - Prob. 23QPCh. 10 - Prob. 24QPCh. 10 - Prob. 25QPCh. 10 - Prob. 26QPCh. 10 - Prob. 27QPCh. 10 - Prob. 28QPCh. 10 - Prob. 29QPCh. 10 - Prob. 30QPCh. 10 - Prob. 31QPCh. 10 - Prob. 32QPCh. 10 - Prob. 33QPCh. 10 - Prob. 34QPCh. 10 - Prob. 35QPCh. 10 - Prob. 36QPCh. 10 - Prob. 37QPCh. 10 - Prob. 38QPCh. 10 - Prob. 39QPCh. 10 - Prob. 40QPCh. 10 - Prob. 41QPCh. 10 - Prob. 42QPCh. 10 - Prob. 43QPCh. 10 - Prob. 44QPCh. 10 - Prob. 45QP
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- 1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forward1. These images were taken six months apart, first when Earth was as far to one side of Alpha Centauri as it can get and again when Earth was as far to the other side of Alpha Centauri as it can get. Consequently, the baseline between the two observing positions is how many AU across? Answer: 1.7 arcsec USE 1.7 arcsec NOT 2.946 2. First, convert this to kilometers using your measurement of how many kilometers are in an AU. 3. Now convert the baseline to kilometers using the true value for the number of kilometers in an AU. 4. Calculate the distance to Alpha Centauri using parallax and the true baseline in kilometers. 5. Google and record the true value. 6. Calculate your percent error 7. Discuss significant sources of errorarrow_forwardQ1arrow_forward
- Did hydrogen gas condense from the nebula as the nebula cooled? What about helium gas? How do you know?arrow_forwardVenus can be as bright as apparent magnitude −4.7 when at a distance of about 1 AU. How many times fainter would Venus look from a distance of 1 pc? What would its apparent magnitude be? Assume Venus has the same illumination phase from your new vantage point. (Hints: Light follows an inverse square law as does gravity, review Section 5-1c; also, review the definition of apparent visual magnitudes, Chapter 2.) (Note: 1 pc = 2.1 × 105 AU.)arrow_forwardVenus can be as bright as apparent magnitude -4.7 when at a distance of about 1 AU. How many times fainter would Venus look from a distance of 5 pc? Assume Venus has the same illumination phase from your new vantage point. (Hints: Recall the inverse square law; also, review the definition of apparent visual magnitudes. Note: 1 pc = 2.1 x 105 AU). times fainter What would its apparent magnitude be?arrow_forward
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