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Beyond Earth - The Universe

Grade 6CBSE

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

🔑Concepts

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The Universe is a vast space that contains all matter and energy, including galaxies, stars, planets, and moons. Our solar system is a tiny part of the Milky Way galaxy, also known as 'Akash Ganga'.

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Celestial Bodies are objects like the Sun, Moon, and stars that shine in the night sky. Stars are huge celestial bodies made of hot gases that emit their own heat and light; the Sun is the nearest star to Earth.

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The Solar System consists of the Sun, eight planets, satellites (moons), and other celestial bodies like asteroids and meteoroids. The planets in order of distance from the Sun are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.

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Earth is the third planet from the Sun and the fifth-largest planet. It is called a 'Blue Planet' because two-thirds of its surface is covered by water. Its unique shape is described as a 'Geoid', which means an earth-like shape (flattened at the poles).

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Light Year is the unit used to measure distances in space. It is the distance light travels in one year. Light travels at a speed of approximately 3,00,000 km/s3,00,000 \text{ km/s}.

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Asteroids are tiny rocky bodies that move around the Sun, mostly found between the orbits of Mars and Jupiter. Meteoroids are small pieces of rocks which move around the Sun and occasionally enter Earth's atmosphere.

📐Formulae

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

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}

Time taken by sunlight to reach Earth≈8 minutes and 20 seconds=500 seconds\text{Time taken by sunlight to reach Earth} \approx 8 \text{ minutes} \text{ and } 20 \text{ seconds} = 500 \text{ seconds}

💡Examples

Problem 1:

Calculate the approximate distance of the Sun from the Earth if light takes 500 seconds500 \text{ seconds} to reach the Earth at a speed of 3,00,000 km/s3,00,000 \text{ km/s}.

Solution:

Distance=3,00,000 km/s×500 s=15,00,00,000 km\text{Distance} = 3,00,000 \text{ km/s} \times 500 \text{ s} = 15,00,00,000 \text{ km}

Explanation:

By multiplying the speed of light (3×105 km/s3 \times 10^5 \text{ km/s}) by the time taken (500 seconds500 \text{ seconds}), we find that the Sun is approximately 150 million kilometers150 \text{ million kilometers} away.

Problem 2:

If the average distance of Earth from the Sun is 150,000,000 km150,000,000 \text{ km} and a certain comet is at 149,600,000 km149,600,000 \text{ km}, find the difference in distance using vertical subtraction.

Solution:

150000000−149600000400000\begin{array}{r} 150000000 \\ - 149600000 \\ \hline 400000 \end{array}

Explanation:

The difference between the two distances is 4,00,000 km4,00,000 \text{ km}.