If you were to record the spectrum of Saturn as well as the A ring, you would find light from one edge of the rings redshifted and light from the other edge blueshifted. If you observed a spectral line at a wavelength of 480 nm, what difference in wavelength (in nm) should you expect between the opposite edges of the rings? (Hints: The radius of the edge of the A ring is 136,500 km. Use the formula for orbital velocity, Vc =  GM r
If you were to record the spectrum of Saturn as well as the A ring, you would find light from one edge of the rings redshifted and light from the other edge blueshifted. If you observed a spectral line at a wavelength of 480 nm, what difference in wavelength (in nm) should you expect between the opposite edges of the rings? (Hints: The radius of the edge of the A ring is 136,500 km. Use the formula for orbital velocity,
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The correct answer and explanation is:
To solve the problem, we need to calculate the Doppler shift difference in the observed wavelength caused by the orbital motion of the A ring. Here are the steps:
Step 1: Orbital velocity of the A ring (Vc)
The formula for orbital velocity is: Vc=GMrV_c = \sqrt{\frac{GM}{r}}
- GG (gravitational constant): 6.674×10−11 m3 kg−1 s−26.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}
- MM (mass of Saturn): 5.683×1026 kg5.683 \times 10^{26} \, \text{kg}
- rr (radius of the A ring): 136,500 km=1.365×108 m136,500 \, \text{km} = 1.365 \times 10^8 \, \text{m}
Vc=(6.674×10−11)(5.683×1026)1.365×108≈19,700 m/sV_c = \sqrt{\frac{(6.674 \times 10^{-11})(5.683 \times 10^{26})}{1.365 \times 10^8}} \approx 19,700 \, \text{m/s}
Step 2: Doppler shift formula
The Doppler shift formula relates the velocity of the source to the observed wavelength change: Δλλ0=Vrc\frac{\Delta \lambda}{\lambda_0} = \frac{V_r}{c}
- λ0\lambda_0: original wavelength (480 nm480 \, \text{nm})
- Δλ\Delta \lambda: wavelength shift
- VrV_r: radial velocity (orbital velocity VcV_c)
- cc: speed of light (3.0×108 m/s3.0 \times 10^8 \, \text{m/s})
Substitute values for one edge of the ring: Δλ=λ0Vcc=480×19,7003.0×108≈0.0315 nm\Delta \lambda = \lambda_0 \frac{V_c}{c} = 480 \times \frac{19,700}{3.0 \times 10^8} \approx 0.0315 \, \text{nm}
Step 3: Total wavelength difference
The total difference between the redshifted and blueshifted edges is twice this value: Δλtotal=2×0.0315≈0.063 nm\Delta \lambda_\text{total} = 2 \times 0.0315 \approx 0.063 \, \text{nm}
Final Answer:
The difference in wavelength between the opposite edges of Saturn’s A ring is 0.063 nm.
Explanation:
This phenomenon occurs because one side of the ring moves toward the observer (blueshift), while the other moves away (redshift). The orbital velocity VcV_c determines the magnitude of this effect. By calculating the velocity using Saturn’s mass and the radius of the ring, we find how much the wavelength changes. The total Doppler shift combines the contributions from both edges, giving the final wavelength difference of 0.063 nm. This is a direct result of the relative motion of the ring particles and demonstrates the application of both orbital mechanics and the Doppler effect in astrophysical observations.