Question

a hydrogen line in a stars spectrum has a frequency of $4.57 \times 10^{14}$hz when stationary. in altairs spectrum, it is shifted upward by $3.98 \times 10^{10}$hz. what is altairs velocity toward us? ? m/s

Explanation:

Step1: Recall Doppler shift formula for light

The Doppler shift formula for the frequency of light is given by $\Delta f = \frac{v}{c}f_0$, where $\Delta f$ is the change in frequency, $v$ is the velocity of the source (star) relative to the observer, $c$ is the speed of light ($c = 3\times10^{8}\ m/s$), and $f_0$ is the original (stationary) frequency. We need to solve for $v$, so we can rearrange the formula to $v=\frac{\Delta f\times c}{f_0}$.

Step2: Identify given values

We are given $\Delta f = 3.98\times 10^{10}\ Hz$, $f_0 = 4.57\times 10^{14}\ Hz$, and $c = 3\times 10^{8}\ m/s$.

Step3: Substitute values into the formula

Substitute the values into $v=\frac{\Delta f\times c}{f_0}$:
\[
v=\frac{(3.98\times 10^{10}\ Hz)\times(3\times 10^{8}\ m/s)}{4.57\times 10^{14}\ Hz}
\]

Step4: Calculate the numerator

First, calculate the numerator: $(3.98\times 10^{10})\times(3\times 10^{8})=3.98\times3\times10^{10 + 8}=11.94\times 10^{18}=1.194\times 10^{19}$

Step5: Divide by the denominator

Now divide by the denominator $4.57\times 10^{14}$:
\[
v=\frac{1.194\times 10^{19}}{4.57\times 10^{14}}=\frac{1.194}{4.57}\times10^{19 - 14}
\]
Calculate $\frac{1.194}{4.57}\approx0.2613$, and $10^{19-14}=10^{5}$. So $v\approx0.2613\times 10^{5}=2.613\times 10^{4}\ m/s$ (we can also do the calculation more precisely: $\frac{3.98\times3\times 10^{18}}{4.57\times 10^{14}}=\frac{11.94\times 10^{18}}{4.57\times 10^{14}}=\frac{11.94}{4.57}\times 10^{4}\approx2.613\times 10^{4}\ m/s$)

Answer:

$\approx 2.61\times 10^{4}\ m/s$ (or more precisely, after calculation: $\frac{3.98\times3\times 10^{8}}{4.57\times 10^{14}}\times10^{10}$? Wait, no, wait. Wait, let's re - check the formula. The Doppler shift for light when the source is moving towards the observer: the observed frequency $f = f_0(1+\frac{v}{c})$ (when $v\ll c$), so $\Delta f=f - f_0=\frac{v}{c}f_0$. So plugging in:

$\Delta f = 3.98\times 10^{10}$, $f_0 = 4.57\times 10^{14}$, $c = 3\times 10^{8}$

$v=\frac{\Delta f\times c}{f_0}=\frac{3.98\times 10^{10}\times3\times 10^{8}}{4.57\times 10^{14}}=\frac{11.94\times 10^{18}}{4.57\times 10^{14}}=\frac{11.94}{4.57}\times 10^{4}\approx2.61\times 10^{4}\ m/s$ (exact value: $11.94\div4.57\approx2.6127$, so $2.6127\times 10^{4}\approx2.61\times 10^{4}\ m/s$ or $26100\ m/s$)

So the final answer is approximately $\boxed{2.61\times 10^{4}}$ (or $26100$)