The planet Earth orbits around the Sun and also spins around its own axis

The planet Earth orbits around the Sun and also spins around its own axis. Part (a) Calculate the angular momentum of the Earth in its orbit around the Sun in kg m^2/s. L_orb = 2.76 × 10^40 kg m^2/s Part (b) Calculate the angular momentum of the Earth spinning on its axis in kg m^2/s. Part (c) How many times larger is the angular momentum of the Earth in its orbit than the angular momentum of the Earth around its axis? Grade Summary: L_orb Deductions: 000

The Correct Answer and Explanation is:

Correct Answer:

Part (a)
Given:
Lorb=2.76×1040 kg⋅m2/sL_{\text{orb}} = 2.76 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}Lorb​=2.76×1040kg⋅m2/s

Part (b): Angular momentum of the Earth spinning on its axis

To calculate the Earth’s rotational angular momentum:Lspin=I⋅ωL_{\text{spin}} = I \cdot \omegaLspin​=I⋅ω

Where:

• III is the moment of inertia of Earth, approximated as a solid sphere: I=25MR2I = \frac{2}{5} M R^2I=52​MR2
• M=5.97×1024 kgM = 5.97 \times 10^{24} \, \text{kg}M=5.97×1024kg (mass of Earth)
• R=6.37×106 mR = 6.37 \times 10^6 \, \text{m}R=6.37×106m (radius of Earth)
• ω=2πT\omega = \frac{2\pi}{T}ω=T2π​ is the angular velocity
• T=86400 sT = 86400 \, \text{s}T=86400s (rotation period of Earth)

Step 1: Calculate moment of inertiaI=25⋅(5.97×1024)⋅(6.37×106)2=25⋅5.97×1024⋅4.06×1013=9.69×1037 kg⋅m2I = \frac{2}{5} \cdot (5.97 \times 10^{24}) \cdot (6.37 \times 10^6)^2 = \frac{2}{5} \cdot 5.97 \times 10^{24} \cdot 4.06 \times 10^{13} = 9.69 \times 10^{37} \, \text{kg} \cdot \text{m}^2I=52​⋅(5.97×1024)⋅(6.37×106)2=52​⋅5.97×1024⋅4.06×1013=9.69×1037kg⋅m2

Step 2: Calculate angular velocityω=2π86400≈7.27×10−5 rad/s\omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \, \text{rad/s}ω=864002π​≈7.27×10−5rad/s

Step 3: Calculate rotational angular momentumLspin=9.69×1037⋅7.27×10−5≈7.05×1033 kg⋅m2/sL_{\text{spin}} = 9.69 \times 10^{37} \cdot 7.27 \times 10^{-5} \approx 7.05 \times 10^{33} \, \text{kg} \cdot \text{m}^2/\text{s}Lspin​=9.69×1037⋅7.27×10−5≈7.05×1033kg⋅m2/s

Part (c): Compare the two angular momentaLorbLspin=2.76×10407.05×1033≈3.91×106\frac{L_{\text{orb}}}{L_{\text{spin}}} = \frac{2.76 \times 10^{40}}{7.05 \times 10^{33}} \approx 3.91 \times 10^6Lspin​Lorb​​=7.05×10332.76×1040​≈3.91×106

Final Answers:

• (a) Lorb=2.76×1040 kg⋅m2/sL_{\text{orb}} = 2.76 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}Lorb​=2.76×1040kg⋅m2/s
• (b) Lspin≈7.05×1033 kg⋅m2/sL_{\text{spin}} \approx 7.05 \times 10^{33} \, \text{kg} \cdot \text{m}^2/\text{s}Lspin​≈7.05×1033kg⋅m2/s
• (c) Orbital angular momentum is about 3.91 million times larger than rotational angular momentum

Explanation

Angular momentum is a physical quantity that describes how much rotation an object has. It depends on both how mass is distributed and how fast it is spinning. For Earth, we can study angular momentum in two major ways: from its motion around the Sun (orbital angular momentum) and from its rotation about its own axis (spin angular momentum).

The angular momentum of Earth in its orbit around the Sun is already given as 2.76×1040 kg⋅m2/s2.76 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}2.76×1040kg⋅m2/s. This is extremely large because the Earth travels a vast distance each year and its orbital radius is about 150 million kilometers.

To calculate Earth’s spin angular momentum, we model it as a solid sphere. Its moment of inertia depends on its mass and radius. The angular velocity tells us how fast Earth spins each day, and multiplying these values gives the rotational angular momentum, which we found to be about 7.05×1033 kg⋅m2/s7.05 \times 10^{33} \, \text{kg} \cdot \text{m}^2/\text{s}7.05×1033kg⋅m2/s.

By dividing the orbital angular momentum by the rotational one, we find that the Earth’s orbital angular momentum is about 3.91 million times greater. This huge difference is mainly due to the Earth’s large distance from the Sun and the speed of its orbital motion. While the Earth spins quickly, the sheer scale of its orbit leads to much greater angular momentum. This comparison helps highlight the significance of mass distribution and radius in rotational motion.