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Beyond Earth - The Milky Way Galaxy

Grade 6CBSE

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

🔑Concepts

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A galaxy is a massive system of billions of stars, clouds of gas, and dust, all held together by the force of gravity.

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The Milky Way is the galaxy that contains our Solar System. It is a spiral-shaped galaxy.

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In ancient India, the Milky Way was called Akash Ganga, meaning a 'river of light' flowing in the sky.

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Distances in space are so vast that they are measured in a unit called a Light Year. A light year is the distance light travels in one year.

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The speed of light is approximately 3,000,00,000 m/s3,000,00,000 \text{ m/s} (or 3×108 m/s3 \times 10^8 \text{ m/s}).

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The Milky Way contains between 100100 to 400400 billion stars, and its diameter is approximately 100,000100,000 light years.

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The Andromeda Galaxy is the closest large spiral galaxy to our Milky Way, located about 2.52.5 million light years away.

📐Formulae

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

1 Light Year≈9.46×1012 km1 \text{ Light Year} \approx 9.46 \times 10^{12} \text{ km}

Speed of Light (c)≈3×105 km/s\text{Speed of Light (c)} \approx 3 \times 10^5 \text{ km/s}

💡Examples

Problem 1:

If a star is 88 light years away from Earth, how long does it take for the light from that star to reach us?

Solution:

88 years

Explanation:

By definition, a light year is the distance light travels in one year. Therefore, if a star is nn light years away, its light takes nn years to reach Earth.

Problem 2:

The diameter of the Milky Way is approximately 100,000100,000 light years. If a smaller nebula is 24,50024,500 light years wide, calculate the difference in their sizes using vertical subtraction.

Solution:

100000−2450075500\begin{array}{r} 100000 \\ - 24500 \\ \hline 75500 \end{array}

Explanation:

To find the difference, we subtract the width of the nebula from the diameter of the galaxy: 100,000−24,500=75,500100,000 - 24,500 = 75,500 light years.

Problem 3:

Calculate the distance of 11 light year in kilometers if there are 31,536,00031,536,000 seconds in a year and light travels at 300,000 km/s300,000 \text{ km/s}.

Solution:

300,000 km/s×31,536,000 s≈9.46×1012 km300,000 \text{ km/s} \times 31,536,000 \text{ s} \approx 9.46 \times 10^{12} \text{ km}

Explanation:

Distance is calculated by multiplying the speed of light by the total number of seconds in a year.