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3 years ago

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Astronomers observe a spectral analysis of a distant star where a particular element has a spectral line with a wavelength of 66

3 nm. In the laboratory, the same element has a spectral line with a wavelength of 645 nm.
a. How fast is the star moving, and in what direction is it moving?

b. The same element on Earth’s surface is observed from a space shuttle orbiting at 7800 m/s,
350 km above Earth. Is the same shift in the spectral line seen from the space shuttle? Why or why not?

1 answer:

tekilochka [14]3 years ago

7 0

Answer:

(a). The velocity of star is $8.1\times10^{6}\ m/s$ and the direction of star toward the earth.

(b). The shift is 0.0168 nm.

Explanation:

Given that,

Wavelength of spectral line = 663 nm

Wavelength of spectral line in lab = 645 nm

(a). We need to calculate the velocity

Using doppler's effect

$\Delta \lambda=\dfrac{v}{c}\lambda$

Where, $\Delta\lambda$ = change in wavelength

v = velocity

c = speed of light

Put the value into the formula

$663-645=\dfrac{v}{3\times10^{8}}\times663$

$v=\dfrac{3\times10^{8}(663-645)}{663}$

$v=8144796.38\ m/s$

$v=8.1\times10^{6}\ m/s$

The direction of star toward the earth.

(b). Speed = 7800 m/s

We need to calculate the shift

Using formula of shift

$\Delta \lambda=\lambda(\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}-1)$

Put the value into the formula

$\Delta \lambda=645\times(\sqrt{\dfrac{1+\dfrac{7800}{3\times10^{8}}}{1-\dfrac{7800}{3\times10^{8}}}}-1)$

$\Delta\lambda=0.0168\ nm$

This shift is small compare to the the movement of Earth around the sun.

Hence, (a). The velocity of star is $8.1\times10^{6}\ m/s$ and the direction of star toward the earth.

(b). The shift is 0.0168 nm.

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" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

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Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

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$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

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These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

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Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

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