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Our Home: Earth, a Unique Life Sustaining Planet - What Makes the Earth Suitable for Life to Exist?

Grade 8CBSE

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

🔑Concepts

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The Goldilocks Zone: Earth is located at an ideal distance (approx. 1.5×1081.5 \times 10^8 km) from the Sun, which allows water to exist in liquid form.

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Atmospheric Composition: Earth's atmosphere contains a precise balance of gases: 78%78\% Nitrogen (N2N_2), 21%21\% Oxygen (O2O_2), and 0.03%0.03\% Carbon Dioxide (CO2CO_2), which supports life and photosynthesis.

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The Greenhouse Effect: Greenhouse gases like CO2CO_2 and CH4CH_4 trap heat, maintaining an average surface temperature of about 15∘C15^\circ C. Without this, the Earth would be frozen at −18∘C-18^\circ C.

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The Ozone Layer: A layer of Ozone (O3O_3) in the stratosphere absorbs nearly 97%−99%97\% - 99\% of the Sun's harmful Ultraviolet (UV) radiation.

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Magnetic Field: Earth's core generates a magnetic field that deflects solar winds and protects the atmosphere from being stripped away.

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The Hydrosphere: Water covers about 71%71\% of the Earth's surface, acting as a universal solvent and a thermal regulator due to its high specific heat capacity.

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Biosphere Interaction: Life exists in the narrow zone where the Lithosphere (land), Hydrosphere (water), and Atmosphere (air) interact.

📐Formulae

6CO2+6H2O+light energy→C6H12O6+6O26CO_2 + 6H_2O + \text{light energy} \rightarrow C_6H_{12}O_6 + 6O_2

C6H12O6+6O2→6CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}

3O2→UV rays2O33O_2 \xrightarrow{\text{UV rays}} 2O_3

g=GMR2≈9.8 m/s2g = \frac{GM}{R^2} \approx 9.8 \text{ m/s}^2

💡Examples

Problem 1:

The total surface area of Earth is approximately 510510 million km2\text{km}^2. If 71%71\% of the Earth is covered by water (Hydrosphere), calculate the approximate area of the land (Lithosphere) in millions of km2\text{km}^2.

Solution:

Land Percentage=100%−71%=29%\text{Land Percentage} = 100\% - 71\% = 29\% Land Area=29100×510≈147.9 million km2\text{Land Area} = \frac{29}{100} \times 510 \approx 147.9 \text{ million km}^2 510.0−362.1147.9\begin{array}{r} 510.0 \\ - 362.1 \\ \hline 147.9 \end{array}

Explanation:

To find the land area, we first subtract the water percentage from the total to get the land percentage (29%29\%). Then, we multiply the total surface area by this percentage. The vertical subtraction shows the difference between the total area and the estimated water area (362.1362.1 million km2\text{km}^2).

Problem 2:

A sample of 500500 liters of pure air is collected. Calculate the volume of Oxygen (O2O_2) present in this sample based on its standard atmospheric percentage.

Solution:

Volume of O2=21% of 500 L\text{Volume of } O_2 = 21\% \text{ of } 500 \text{ L} Volume=21100×500=21×5=105 L\text{Volume} = \frac{21}{100} \times 500 = 21 \times 5 = 105 \text{ L}

Explanation:

Since Oxygen makes up approximately 21%21\% of the Earth's atmosphere, we apply this percentage to the total volume of the air sample to determine the specific quantity of life-sustaining Oxygen.