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Our Home: Earth, a Unique Life Sustaining Planet - Position of the Earth

Grade 8CBSE

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

🔑Concepts

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Earth is the third planet from the Sun, positioned at an average distance of approximately 1.5×1081.5 \times 10^8 km (1 Astronomical Unit).

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The Goldilocks Zone or Habitable Zone refers to the orbital region around a star where the temperature is moderate enough for liquid water to exist, which is essential for life.

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Earth's axis is tilted at an angle of 23.5∘23.5^\circ relative to its orbital plane. This axial tilt, combined with its revolution around the Sun, results in the change of seasons.

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The Earth's atmosphere acts as a protective shield, containing 78%78\% Nitrogen, 21%21\% Oxygen, and 0.04%0.04\% Carbon Dioxide (CO2CO_2), which helps maintain a stable surface temperature via the greenhouse effect.

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The Earth's magnetic field, generated by its iron-rich core, protects the planet from harmful solar winds and high-energy particles from space.

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The average surface temperature of Earth is approximately 15∘C15^\circ C, compared to a calculated temperature of −18∘C-18^\circ C if the greenhouse effect were absent.

📐Formulae

1 AU (Astronomical Unit)≈1.496×108 km1 \text{ AU (Astronomical Unit)} \approx 1.496 \times 10^8 \text{ km}

Time(t)=Distance(d)Speed of Light(c)\text{Time} (t) = \frac{\text{Distance} (d)}{\text{Speed of Light} (c)}

c≈3×108 m/s=3×105 km/sc \approx 3 \times 10^8 \text{ m/s} = 3 \times 10^5 \text{ km/s}

Axial Tilt=23.5∘\text{Axial Tilt} = 23.5^\circ

💡Examples

Problem 1:

If the distance between the Sun and Earth is 150,000,000 km150,000,000 \text{ km}, calculate how long it takes for sunlight to reach Earth in seconds.

Solution:

t=150,000,000 km300,000 km/st = \frac{150,000,000 \text{ km}}{300,000 \text{ km/s}} t=500 secondst = 500 \text{ seconds}

Explanation:

By dividing the average distance of Earth from the Sun by the speed of light, we find it takes approximately 88 minutes and 2020 seconds for light to travel the distance.

Problem 2:

Calculate the approximate distance of Earth from the Sun in meters using the scientific notation 1.5×108 km1.5 \times 10^8 \text{ km}.

Solution:

1.5×108 km×103 m/km1.5 \times 10^8 \text{ km} \times 10^3 \text{ m/km} =1.5×1011 meters= 1.5 \times 10^{11} \text{ meters}

Explanation:

To convert kilometers to meters, we multiply by 1,0001,000 (or 10310^3). Adding the exponents (8+3=118 + 3 = 11) gives the distance in meters.

Problem 3:

Determine the total days in a leap year based on Earth's orbital period of 365.25365.25 days.

Solution:

365.25×41461.00\begin{array}{r} 365.25 \\ \times 4 \\ \hline 1461.00 \end{array} Average per year=14614=365.25 days\text{Average per year} = \frac{1461}{4} = 365.25 \text{ days}

Explanation:

Because Earth takes 365365 days and 66 hours (0.250.25 days) to complete one revolution, the 0.250.25 day is accumulated over 4 years to add one full day (2424 hours) to the calendar in February.