Question

If the Milky Way were the size of a nickel (about 2 centimeters).

a. How big would the Local Group be?
b. How big would the Local Supercluster be?
c. How big would the visible Universe be?

Answer 1

Answer on Question#55027 - Physics - Astronomy - Astrophysics

If the Milky Way were the size of a nickel (about $l_{n} = 2$ centimeters).

a. How big would the Local Group be?

b. How big would the Local Supercluster be?

c. How big would the visible Universe be?

Solution:

The size of the Milky Way is $L_{MW} = 10^{5}\mathrm{ly}$.

a. Since the size of the Local Group is $L_{LG} = 10^{7}\mathrm{ly}$, then the its new size $L_{LG}^{N}$ will be

$$L_{LG}^{N} = \frac{L_{LG}}{L_{MW}} l_{n} = \frac{10^{7}\mathrm{ly}}{10^{5}\mathrm{ly}} 2\mathrm{cm} = 200\mathrm{cm} = 2\mathrm{m}$$

b. Since the size of the Local Supercluster is $L_{LS} = 52 \times 10^{7}$ ly, then the its new size $L_{LS}^{N}$ will be

$$L_{LS}^{N} = \frac{L_{LS}}{L_{MW}} l_{n} = \frac{52 \times 10^{7}\mathrm{ly}}{10^{5}\mathrm{ly}} 2\mathrm{cm} = 104\mathrm{m}$$

c. Since the size of the visible Universe is $L_{U} = 91 \times 10^{9}$ ly, then the its new size $L_{U}^{N}$ will be

$$L_{U}^{N} = \frac{L_{U}}{L_{MW}} l_{n} = \frac{91 \times 10^{9}\mathrm{ly}}{10^{5}\mathrm{ly}} 2\mathrm{cm} = 18.2\mathrm{km}$$

Answer:

a. $2\mathrm{m}$

b. $104\mathrm{m}$

c. $18.2\mathrm{km}$

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