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hram777 [196]
3 years ago
6

Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h

elp you. 1 AU 1.5 × 108 km 1 ly 63,000 AU 1 pc 3.262 ly distance to the star Betelgeuse: 640 ly distance to the star VY Canis Majoris: 3.09 × 108 AU distance to the galaxy Large Magellanic Cloud: 49976 pc distance to Neptune at the farthest: 4.7 billion km ↓ ↓ ↓
Geography
1 answer:
adelina 88 [10]3 years ago
8 0

Answer:

Distance to Neptune at the farthest < distance to the star Betelgeuse < distance to the star VY Canis Majoris < distance to the galaxy Large Magellanic Cloud

Explanation:

Firstly, we need to work with the same units in order to make the comparison among the given distances. So, let's work with kilometers (km):

distance to the star Betelgeuse: 640 ly

640 ly \frac{63000 AU}{1 ly} \frac{1.5(10)^{8} km}{1 AU}=6.048(10)^{15} km

distance to the star VY Canis Majoris: 3.09 × 108 AU

3.09(10)^{8} AU \frac{1.5(10)^{8} km}{1 AU}=4.635(10)^{16} km

distance to the galaxy Large Magellanic Cloud: 49976 pc

49976 pc \frac{3.262 ly}{1 pc} \frac{63000 AU}{1 ly} \frac{1.5(10)^{8} km}{1 AU}=1.54(10)^{18} km

distance to Neptune at the farthest: 4.7 billion km

4.7 billion km=4.7(10)^{12} km

Now that we have all the distances in kilometers we can compare and arrange them in order from least to greatest:

Distance to Neptune at the farthest: 4.7 billion km=4.7(10)^{12} km

Distance to the star Betelgeuse:  640 ly=6.048(10)^{15} km

Distance to the star VY Canis Majoris: 3.09(10)^{8} AU=4.635(10)^{16} km

Distance to the galaxy Large Magellanic Cloud: 49976 pc=1.54(10)^{18} km

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