Question

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

Answer 1

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: $49976 pc=1.54(10)^{18} km$