Question

If the Sun’s 1,395,000 km diameter is shrunken down to 12 inches, and the next closest star is 4.20 × 1013 km from the Sun, how far apart should the two be placed in a properly scaled model of the Sun and its nearest neighbor, Proxima Centauri? (For your answer, convert the scale model distance into miles, since the separation in inches will be huge)

30,179,400 miles
340 miles
362,152,200 miles
2.295×1013 miles
5720 miles
30,107,500 miles
68,590 miles
Human beings have not evolved to the point where inches can be converted into miles

Answer 1

Answer:

5720miles

Explanation:

Given,

diameter of Sun= 1,395,000km

=(1395000*0.621371) miles

=8.67*105miles

distance between Sun and Proxima Centauri= 4.20*1013 km

=(4.20*1013 *0.621371) miles

=2.60*1013 miles

Scaled down diameter of sun = 12inch

=(12*1.58*10-5)miles

=1.89*10-4 miles

, therefore, the scale factor =

=2.18*10-10

therefore the scaled-down distance between Sun and Alpha Centauri = scale factor* actual distance

=2.18*10-10 * 2.60*1013 miles

=5668miles

Therefore the answer would be 5720 miles (error due to rounding off)

Answer 2

Answer:

5702 miles

Explanation:

First, let us find the rate at which the shrinking down is occurring.

If 1,395,000 km has been shrunk down to 12 inches, that means that 1 km will be:

1 km = $\frac{12}{1395000}$ = $8.602 * 10^{-6} inches$

Therefore, the  distance between the sun and the next closest star, in inches, will be:

$4.20 * 10^{13} km = 4.20 * 10^{13} * 8.602 * 10^{-6} = 361284000$ inches

Finally, we have to convert this distance to miles.

1 inch = $1.57828 * 10^{-5}$ miles

Therefore, 361284000 inches will be:

$361284000 * 1.57828 * 10^{-5}$

= $5702$ miles

The nearest neighbor of the sun is 5702 miles away from the sun.