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Beyond Earth - Night Sky Watching

Grade 6CBSE

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

🔑Concepts

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Celestial Bodies: All objects existing in the sky, such as stars, planets, satellites, comets, and asteroids, are called celestial bodies.

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Stars: Huge balls of hot gases that emit their own light. The Sun is the nearest star to Earth, located at approximately 1.5×1081.5 \times 10^{8} km away.

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Constellations: Groups of stars that form recognizable patterns in the night sky. Examples include Ursa Major (Great Bear or Saptarshi), Orion (The Hunter), and Cassiopeia.

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Pole Star (Polaris): A star located in the North direction that appears stationary from Earth because it lies on the axis of rotation of the Earth.

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The Moon: Earth's only natural satellite. It does not have its own light but reflects the light of the Sun. The changing shapes of the bright part of the moon as seen from Earth are called 'Phases of the Moon'.

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Light Year: Since distances in the universe are very vast, they are measured in a unit called a light year. It is the distance traveled by light in one year at a speed of approximately 3×108 m/s3 \times 10^{8} \text{ m/s}.

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Planets: Celestial bodies that revolve around a star in fixed paths called orbits. Unlike stars, they do not twinkle.

📐Formulae

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

1 Light Year=Speed of Light×Time in one year\text{1 Light Year} = \text{Speed of Light} \times \text{Time in one year}

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

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

💡Examples

Problem 1:

If the distance of a star from Earth is 88 light years, calculate its distance in kilometers.

Solution:

Distance in km =8×9.46×1012 km=75.68×1012 km= 8 \times 9.46 \times 10^{12} \text{ km} = 75.68 \times 10^{12} \text{ km} or 7.568×1013 km7.568 \times 10^{13} \text{ km}.

Explanation:

We multiply the number of light years by the value of one light year in kilometers (9.46×1012 km9.46 \times 10^{12} \text{ km}) to find the total distance.

Problem 2:

Light from the Sun takes approximately 500500 seconds to reach Earth. If the speed of light is 300,000 km/s300,000 \text{ km/s}, find the distance between the Sun and the Earth.

Solution:

300,000×500150,000,000\begin{array}{r} 300,000 \\ \times 500 \\ \hline 150,000,000 \end{array} Distance =150,000,000 km= 150,000,000 \text{ km} or 1.5×108 km1.5 \times 10^{8} \text{ km}.

Explanation:

Using the formula Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}, we multiply the speed of light by the time taken.

Problem 3:

How can one locate the Pole Star using the Ursa Major (Saptarshi) constellation?

Solution:

Look at the two stars at the end of the 'bowl' of Ursa Major, known as the 'pointers'. Imagine a straight line passing through these two stars and extend it towards the North. This line will lead to a star that is not very bright; this is the Pole Star.

Explanation:

The pointers in Ursa Major always align with the Pole Star, which remains fixed in the northern sky.

Night Sky Watching Class 6 Notes & Examples | CBSE Science