Doppler effect

awepratization

2010-12-06 6:05 am

The planet Mercury may be assumed to be at a constant distance from the Earth and rotates about an axis which is perpendicular to the line joining the centres of the Earth and Mercury.
A short pulse monochromatic radio waves travelled from the Earth to Mercury and back in 1260s.
(i) What is the distance for the planet from the Earth?
(ii) When the reflected pulse was received back on the Earth, its duration was found to have been extended by 0.0163s and its spectrum had been broadened in a band of frequencies of width df given by the expression
df / f = 4.00 x 10^-8
where f is the original frequency.
Explain these two observations and determine the radius of the planet Mercury and its period of rotation.

更新1:

df = (delta) f

回答 (1)

天同

2010-12-07 1:28 am

✔ 最佳答案

(i) Distance = 1260 x (3x10^8)/2 m = 1.89 x 10^11 m

(ii) The extension of pulse duration is because of waves reflected on different part of the spherical surface of Mercuy. The frequency band is due to echoesfrom the two sides of the planet and which undergone a Doppler shift in frequency.

Let R be the radius of Mercury,
2R = (3x10^8) x 0.0163
i.e. R = 2.4 x 10^6 m

The shift in frequency for echoes from the side of the planet moving away from earth = -(2v/c).f
where v is the speed of rotation of Mercury
c is the speed of light
f is the frequency of the radio wave

The shift in frequency for echoes from the side of the planet moving towards earth = (2v/c).f

Hence, total shift in frequency = [(2v/c) - (-2v/c)].f
= (4v/c).f
thus, 4v/c = 4 x 10^-8
v = (4x10^-8) x (3x10^8)/4 m/s = 3 m/s

But period of rotation T = 2.pi.R/v = 2 x (3.14159) x (2.4x10^6)/3 s
T = 5 x 10^6 s = 57.87 days

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