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Physics: Space and Astrophysics - Scale of the Universe

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Astronomical Unit (AU) is defined as the average distance between the Earth and the Sun, which is approximately 1.5×108 km1.5 \times 10^{8} \text{ km} or 1.5×1011 m1.5 \times 10^{11} \text{ m}. It is primarily used to measure distances within our Solar System.

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A Light-year (ly) is the distance light travels in a vacuum in one Earth year (365.25365.25 days). It is a unit of distance, not time, and is approximately 9.46×1012 km9.46 \times 10^{12} \text{ km} or 9.46×1015 m9.46 \times 10^{15} \text{ m}.

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The Speed of Light (cc) is a fundamental constant, approximately 3×108 m/s3 \times 10^{8} \text{ m/s} in a vacuum. This means light can travel around the Earth about 7.57.5 times in a single second.

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The hierarchy of the universe follows a scale: Planets →\rightarrow Planetary Systems (Solar System) →\rightarrow Star Clusters →\rightarrow Galaxies (Milky Way) →\rightarrow Galaxy Groups/Clusters (Local Group) →\rightarrow Superclusters (Laniakea Supercluster) →\rightarrow Observable Universe.

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Distances in space are so large that we use Scientific Notation (a×10na \times 10^{n}) to represent them. For example, the distance to the nearest star, Proxima Centauri, is 4.24.2 ly or approximately 4.0×1013 km4.0 \times 10^{13} \text{ km}.

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The Observable Universe has a radius of approximately 4.6×10104.6 \times 10^{10} light-years, reflecting the expansion of space since the Big Bang.

📐Formulae

d=v×td = v \times t

1 AU≈1.5×1011 m1 \text{ AU} \approx 1.5 \times 10^{11} \text{ m}

1 ly=c×(1 year in seconds)≈9.46×1015 m1 \text{ ly} = c \times (1 \text{ year in seconds}) \approx 9.46 \times 10^{15} \text{ m}

t=dct = \frac{d}{c}

💡Examples

Problem 1:

Light from the Sun takes approximately 88 minutes and 2020 seconds to reach the Earth. Given the speed of light is 3×108 m/s3 \times 10^{8} \text{ m/s}, calculate the distance from the Sun to the Earth in meters.

Solution:

t=(8×60)+20=500 st = (8 \times 60) + 20 = 500 \text{ s} d=v×t=(3×108 m/s)×500 sd = v \times t = (3 \times 10^{8} \text{ m/s}) \times 500 \text{ s} d=1.5×1011 md = 1.5 \times 10^{11} \text{ m}

Explanation:

First, convert the time into seconds. Then, use the distance formula d=v×td = v \times t by multiplying the speed of light by the total seconds.

Problem 2:

A distant galaxy is 2.5×1062.5 \times 10^{6} light-years away from Earth. How many years does it take for light from this galaxy to reach us?

Solution:

t=dct = \frac{d}{c} Since d is in light-years, time t=2.5×106 years\text{Since } d \text{ is in light-years, time } t = 2.5 \times 10^{6} \text{ years}

Explanation:

By definition, a light-year is the distance light travels in one year. Therefore, if an object is XX light-years away, its light takes XX years to reach the observer.

Problem 3:

The distance between Jupiter and the Sun is approximately 7.78×108 km7.78 \times 10^{8} \text{ km}. Convert this distance into Astronomical Units (AU), using 1 AU=1.5×108 km1 \text{ AU} = 1.5 \times 10^{8} \text{ km}.

Solution:

Distance in AU=7.78×108 km1.5×108 km\text{Distance in AU} = \frac{7.78 \times 10^{8} \text{ km}}{1.5 \times 10^{8} \text{ km}} Distance in AU≈5.19 AU\text{Distance in AU} \approx 5.19 \text{ AU}

Explanation:

To convert a distance into AU, divide the given distance by the value of 1 AU1 \text{ AU} in the same units.