Calculate the angular momentum of the Earth in its orbit around the Sun. Compare this angular momentum with the angular momentum of Earth on its axis.
The Correct Answer and Explanation is:
To calculate and compare the angular momentum of Earth in two scenarios — orbiting the Sun and rotating on its axis — we use the following physics principles:
1. Angular Momentum of Earth in Its Orbit Around the Sun
This motion can be treated as circular:
Formula:
Lorbit=mvrL_{\text{orbit}} = mvrLorbit=mvr
Where:
- mmm = mass of Earth = 5.97×1024 kg5.97 \times 10^{24} \, \text{kg}5.97×1024kg
- vvv = orbital speed of Earth ≈ 2.98×104 m/s2.98 \times 10^4 \, \text{m/s}2.98×104m/s
- rrr = average distance from Sun = 1.496×1011 m1.496 \times 10^{11} \, \text{m}1.496×1011m
Calculation:Lorbit=(5.97×1024)(2.98×104)(1.496×1011)Lorbit≈2.66×1040 kg⋅m2/sL_{\text{orbit}} = (5.97 \times 10^{24})(2.98 \times 10^4)(1.496 \times 10^{11}) \\ L_{\text{orbit}} ≈ 2.66 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}Lorbit=(5.97×1024)(2.98×104)(1.496×1011)Lorbit≈2.66×1040kg⋅m2/s
2. Angular Momentum of Earth Spinning on Its Axis
We treat Earth as a rotating solid sphere:
Formula:
Lspin=IωL_{\text{spin}} = I \omegaLspin=Iω
Where:
- I=25mr2I = \frac{2}{5}mr^2I=52mr2
- ω=2πT\omega = \frac{2\pi}{T}ω=T2π, with T=86400 sT = 86400 \, \text{s}T=86400s (1 day)
Using Earth’s radius:
r=6.371×106 mr = 6.371 \times 10^6 \, \text{m}r=6.371×106mI=25(5.97×1024)(6.371×106)2≈9.72×1037 kg⋅m2I = \frac{2}{5}(5.97 \times 10^{24})(6.371 \times 10^6)^2 ≈ 9.72 \times 10^{37} \, \text{kg} \cdot \text{m}^2I=52(5.97×1024)(6.371×106)2≈9.72×1037kg⋅m2ω=2π86400≈7.27×10−5 rad/s\omega = \frac{2\pi}{86400} ≈ 7.27 \times 10^{-5} \, \text{rad/s}ω=864002π≈7.27×10−5rad/sLspin=(9.72×1037)(7.27×10−5)≈7.07×1033 kg⋅m2/sL_{\text{spin}} = (9.72 \times 10^{37})(7.27 \times 10^{-5}) ≈ 7.07 \times 10^{33} \, \text{kg} \cdot \text{m}^2/\text{s}Lspin=(9.72×1037)(7.27×10−5)≈7.07×1033kg⋅m2/s
3. Comparison
- Angular momentum in orbit: ≈2.66×1040 kg⋅m2/s\approx 2.66 \times 10^{40} \, \text{kg} \cdot \text{m}^2/\text{s}≈2.66×1040kg⋅m2/s
- Angular momentum on axis: ≈7.07×1033 kg⋅m2/s\approx 7.07 \times 10^{33} \, \text{kg} \cdot \text{m}^2/\text{s}≈7.07×1033kg⋅m2/s
Conclusion:
Earth’s angular momentum due to orbiting the Sun is about a million times greater than its rotational angular momentum. This shows that the bulk of Earth’s angular momentum is tied to its motion through space rather than its spinning motion. These values help scientists understand planetary dynamics, gravitational interactions, and conservation of momentum in celestial systems.
