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Beyond Earth - Our Solar System

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, their satellites (moons), and other celestial bodies like asteroids and meteoroids.

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The Sun is the center of the solar system and provides the pulling force (gravitational force) that binds the entire system. It is approximately 150,000,000 km150,000,000 \text{ km} away from Earth.

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The eight planets in order of distance from the Sun are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. A mnemonic to remember this is 'My Very Efficient Mother Just Served Us Nuts'.

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Inner Planets (Mercury, Venus, Earth, Mars) are small and made of rocks, while Outer Planets (Jupiter, Saturn, Uranus, Neptune) are giant planets made of gases and liquids.

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The Earth is the 3rd3^{rd} nearest planet to the sun and the 5th5^{th} largest planet. Its shape is described as a 'Geoid', meaning an Earth-like shape (slightly flattened at the poles).

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The Moon is Earth's only natural satellite. It moves around the Earth in about 2727 days and takes exactly the same time to complete one spin on its axis.

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Light travels at a speed of about 300,000 km/s300,000 \text{ km/s}. Even at this high speed, it takes about 88 minutes for sunlight to reach the Earth.

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An Astronomical Unit (AUAU) is the average distance between the Earth and the Sun, which is roughly 1.5×108 km1.5 \times 10^{8} \text{ km}.

📐Formulae

Time(t)=Distance(d)Speed(v)\text{Time} (t) = \frac{\text{Distance} (d)}{\text{Speed} (v)}

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

1 AU≈1.5×108 km1 \text{ AU} \approx 1.5 \times 10^8 \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:

Calculate the time (in seconds) it takes for sunlight to reach Earth 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=150,000,000300,000=500 secondst = \frac{150,000,000}{300,000} = 500 \text{ seconds}

Explanation:

By using the formula t=dvt = \frac{d}{v}, we divide the total distance of Earth from the Sun by the speed at which light travels. 500 seconds500 \text{ seconds} is approximately 88 minutes and 2020 seconds.

Problem 2:

If the distance of Mercury from the Sun is 58,000,000 km58,000,000 \text{ km} and Earth is 150,000,000 km150,000,000 \text{ km} from the Sun, what is the difference in their distances?

Solution:

150000000−5800000092000000\begin{array}{r} 150000000 \\ - 58000000 \\ \hline 92000000 \end{array}

Explanation:

To find the difference in distance between the two planets, we subtract Mercury's distance from Earth's distance from the Sun. The result is 92,000,000 km92,000,000 \text{ km}.

Problem 3:

A planet is located at a distance of 3 AU3 \text{ AU} from the Sun. Convert this distance into kilometers.

Solution:

3×(1.5×108)=4.5×108 km3 \times (1.5 \times 10^8) = 4.5 \times 10^8 \text{ km}

Explanation:

Since 1 AU1 \text{ AU} is roughly 150,000,000 km150,000,000 \text{ km} or 1.5×108 km1.5 \times 10^8 \text{ km}, a distance of 3 AU3 \text{ AU} is calculated by multiplying 33 with the value of one AUAU.