In a scale Universe the sun is the size of a fine sand grain

blar

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ok these questions are sort of linked and i really hope this question belongs in this board since it does deal with astronomy

Question 1 In a scale Universe with the sun the size of a spherical grain of fine sand (diameter= 0.2 millimeters), how big a box (in cubic meters) would you need to hold all the stars in the Milky Way Galaxy. Packing efficiency of sand is a very complex problem, but assume that each grain of sand occupies 0.023 cubic millimeters.

i believe i got this one right and that the 0.2mm isn't used. here is what i got

200 billion grains of sand
Density of a grain of sand = 0.0023 mm3

200 billion x 0.0023mm3 = 460000000mm3

460000000mm3 = 0.46m3 (google.com)

The box would have to be the size of 0.46m3


ok but the next problem is tricky and I think i am really doing it wrong

Question 1b How big, in meters, would the Milky Way galaxy be in this scale universe? If the center of the Milky way was centered on Toronto, approximately where would the edge of the galaxy be? (eg, near Kitchner? Near Jupiter?).

I was told to scale the sun to a grain of sand so i have to use the 0.2mm from Question 1 to help figure this out


i really dont know if I am doing this right here is what i got

Milky Way = 100,000 light years in diameter = 9.4605284 × 1020 m
Sun’s diameter = 1,400,000 km (Taken from 100,000 light years = 946,091,000,000,000,000km
(100,000 x 9.46091E+12)
Sun’s diameter = 1,400,000 km
0.2mm


946,091,000,000,000,000km divided by 1,400,000km x 0.2=135155857142.857142

i really don't know what i did wrong here it seems that what i am doing is correct but this numer seems wrong so i am wondering if any one can help me with this question

soory again if this question does not belong in this board and in another one here
 
blar said:
i really hope this question belongs in this board since it does deal with astronomy
But this board deals with arithmetic, not astronomy...?

blar said:
200 billion grains of sand
I will guess that somewhere (unstated) you were told to assume a total of 200 billion stars for the galaxy.

blar said:
Density of a grain of sand = 0.0023 mm3

200 billion x 0.0023mm3 = 460000000mm3

460000000mm3 = 0.46m3 (google.com)
I'm not sure where Google comes into play here...? In any case, since there are, by definition, 1000 mm to each meter, then there are (1000)<sup>3</sup> mm<sup>3</sup> = 1000000000 mm<sup>3</sup> in one cubic meter. Doing the division gives the result of 0.46 m<sup>3</sup>.

Note: It would probably be more standard, when doing volumes, to give results in cc's or liters. But, in cases of doubt, it's almost always best to follow the lead of your textbook.

blar said:
946,091,000,000,000,000km divided by 1,400,000km x 0.2=135155857142.857142
I'm not quite sure I follow the reasoning...?

Here's a more explicit outline:

. . . . .Sun : model = 1 400 000 km : 0.2 mm

In equivalent terms, this is:

. . . . .Sun : model = 1 400 000 km : 0.000 000 2 km

To go from the Sun's number to the scale-model's number, you'd have to multiply by a very small number:

. . . . .(0.000 000 2) / (1 400 000) = 1.42857... × 10<sup>-12</sup>

Given the diameter of the galaxy, we then get:

. . . . .(100 000 l.y.) (946 091 × 10<sup>12</sup> km / l.y.) = 9.46091 × 10<sup>22</sup> km

. . . . .[galaxy dimension] [really small number] = [model dimension]

. . . . .[9.46091 × 10<sup>22</sup>] [ (0.000 000 2) / (1 400 000) ] = 13 515 585 714.3 km

So I arrived at the same value you did, and now we can see how. :wink:

blar said:
i really don't know what i did wrong here
On what basis do you conclude that you did something wrong? If the input numbers are correct (I haven't checked), then the methodology is going to provide the correct value.

Most of space is empty. "Astronomical" has come to mean "really, really big", because astronomical distances are staggering. That's probably the point of this exercise.

Eliz.
 
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