To calculate the Earth's angular momentum about the Sun, we use the formula for angular momentum:

$L = m \cdot v \cdot r$

Where:

• $m$ is the mass of the Earth,
• $v$ is the orbital velocity,
• $r$ is the orbital radius.

Given Data:

• Earth's mass ($m$) = $5.97 \times 10^{24} \, \text{kg}$,
• Orbital velocity ($v$) = $29.9 \, \text{km/s} = 29.9 \times 10^3 \, \text{m/s}$,
• Orbital radius ($r$) = $150 \times 10^9 \, \text{m}$.

Calculation:

$L = (5.97 \times 10^{24}) \cdot (29.9 \times 10^3) \cdot (150 \times 10^9)$

1. Multiply the constants: $L = 5.97 \cdot 29.9 \cdot 150 \cdot 10^{24+3+9}$
2. Simplify the exponents: $L = (5.97 \cdot 29.9 \cdot 150) \cdot 10^{36}$
3. Calculate the numerical part: $5.97 \cdot 29.9 = 178.803$ $178.803 \cdot 150 = 26,820.45$ So: $L = 2.682 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}$

Final Answer:

The Earth's angular momentum about the Sun is: $\boxed{2.68 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}}$

Would you like a detailed explanation of the process? Here are related questions to deepen your understanding:

1. Why do we assume the Earth’s orbital motion to calculate angular momentum?
2. How does angular velocity ($\omega$) relate to angular momentum?
3. What would happen to the angular momentum if the orbital radius were halved?
4. How is angular momentum conserved in orbital mechanics?
5. Can angular momentum change if the Earth’s mass changes? How?

Tip: Always ensure units are consistent (e.g., meters, kilograms, seconds) when calculating angular momentum.