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A Journey through Space - The Solar System and its Components

Grade 6CBSE

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

🔑Concepts

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The Solar System consists of the Sun, eight planets, satellites, and other celestial bodies known as asteroids and meteoroids.

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The Sun is the center of the solar system and is a huge ball of extremely hot gases. It provides the pulling force that binds the solar system. The Sun is about 1.5×108 km1.5 \times 10^8 \text{ km} away from the Earth.

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There are eight planets in our solar system. In order of their distance from the sun, they are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. A common mnemonic is 'My Very Efficient Mother Just Served Us Nuts'.

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All the eight planets of the solar system move around the sun in fixed paths. These paths are elongated and are called orbits.

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The Earth is the third nearest planet to the sun. In size, it is the fifth largest planet. It is slightly flattened at the poles, a shape described as a Geoid.

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The Moon is Earth's only natural satellite. Its diameter is only 1/41/4 that of the Earth. It is about 384,400 km384,400 \text{ km} away from us.

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Asteroids are numerous tiny bodies which also move around the sun. They are found between the orbits of Mars and Jupiter.

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Meteoroids are small pieces of rocks which move around the sun. Sometimes these come near the earth and tend to drop upon it due to gravity.

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Light travels at the speed of about 3×105 km/s3 \times 10^5 \text{ km/s} (or 300,000 km/s300,000 \text{ km/s}). Even with this speed, the light of the sun takes about 88 minutes to reach the earth.

📐Formulae

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

Speed of Light (c)≈3×108 m/s\text{Speed of Light (c)} \approx 3 \times 10^8 \text{ m/s}

Time taken by Sunlight≈1.5×108 km3×105 km/s=500 seconds\text{Time taken by Sunlight} \approx \frac{1.5 \times 10^8 \text{ km}}{3 \times 10^5 \text{ km/s}} = 500 \text{ seconds}

💡Examples

Problem 1:

Calculate the time taken in minutes for light to reach Earth from the Sun if the distance is 150,000,000 km150,000,000 \text{ km} and the speed of light is 300,000 km/s300,000 \text{ km/s}.

Solution:

t=150,000,000 km300,000 km/s=500 secondst = \frac{150,000,000 \text{ km}}{300,000 \text{ km/s}} = 500 \text{ seconds} To convert to minutes: t=50060≈8.33 minutest = \frac{500}{60} \approx 8.33 \text{ minutes}

Explanation:

We use the basic physics formula t=dvt = \frac{d}{v} where dd is distance and vv is velocity (speed). Dividing the total distance by the speed of light gives the result in seconds, which is then converted to minutes.

Problem 2:

If a planet is 2 AU2 \text{ AU} (Astronomical Units) away from the Sun, and 1 AU≈1.5×108 km1 \text{ AU} \approx 1.5 \times 10^8 \text{ km}, what is its distance in kilometers?

Solution:

Distance=2×(1.5×108 km)=3.0×108 km\text{Distance} = 2 \times (1.5 \times 10^8 \text{ km}) = 3.0 \times 10^8 \text{ km}

Explanation:

An Astronomical Unit (AUAU) is the average distance between the Earth and the Sun. To find the distance of another body, we multiply its AUAU value by the base value of 1 AU1 \text{ AU}.

The Solar System and its Components - Revision Notes & Key Formulas | CBSE Class 6 Science