se from your new vantage point. (Hir itudes. Note: 1 pc = 2.1 x 105 AU).
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I am struggling to answer this question. There are two parts to this question (as shown in the picture) I have already tried the answers 6.29x10^-13 for part 1 and 65.53 for part 2 and they are both incorrect. Any help is greatly appreciated, thank you!!
Transcribed Image Text:**Question:**
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 6 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 10^5 AU).*
**Incorrect Attempt:**
6.29 × 10^−13 times fainter
**Question:**
What would its apparent magnitude be?
**Incorrect Attempt:**
65.53
**Explanation:**
The text includes questions related to the apparent magnitude of Venus from different distances, applying astronomical concepts like the inverse square law. The attempts show incorrect answers that require reevaluation using the correct formula for magnitude and distance calculations.
Expert Solution
Step 1
Apparent magnitude and absolute magnitude of a celestial object are related by the equation
And flux or brightness of an object is related to its luminosity as
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