<p>Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at only the "simple" versions of these equations.</p>

<p>You say that the you've heard that the source of the two is because of receiving the signal and reflecting it, which is entirely true. Let's consider the simpler case of a police officer using radar to track a car's speed (just so we don't have to drag in the more complicated concepts of orbital motions of satellites, dispersion relations, etc.) You have a police officer standing by the side of the road with a radar gun, shooting radar waves of some frequency $f_o$ at the car as it recedes from him (note I'm going to use frequency but it is completely analogous to wavelength, assuming a simple dispersion relation). That wave travels to the car and the first peak of the wave hits the car. The car then has time to move slightly forward by the time the next peak hits the car. From the car's perspective, because of his relative motion to the police officer, the frequency of the wave hitting the car has been Doppler shifted to a new frequency, $f_1$. This new frequency is given by the Doppler equation</p>

<p>$$\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)$$</p>

<p>Note that this equation is the same as what you presented, except that it uses frequency instead of wavlength. So now, our car has been hit by these frequency shifted waves. It will now reflect those waves back at the police officer. The process of reflection effectively makes the car an emitter of the frequency $f_1$ but since it's obviously moving with respect to the police office, the police officer receives a Doppler shifted frequency, which I'll call $f_2$. Since the speed between the car and the police officer is still $v$, we know that the Doppler equation for the frequency shift of the reflection is given by</p>

<p>$$\frac{f_2}{f_1} = \left(1+\frac{v}{c}\right)$$</p>

<p>Now the police officer receives back a frequency $f_2$, but he originally emitted $f_o$, so he wants to know the velocity which caused the shift from $f_o$ to $f_2$. Fortunately, this can be determined by combining the two shifts above.</p>

<p>$$\frac{f_2}{f_o} = \frac{f_2}{f_1}\frac{f_1}{f_o} = \left(1+\frac{v}{c}\right)\left(1+\frac{v}{c}\right)$$</p>

<p>$$\frac{f_2}{f_o} = \left(1+\frac{v}{c}\right)^2 = 1 + 2\frac{v}{c} + \frac{v^2}{c^2}$$</p>

<p>At this stage, we make the assumption that $v^2/c^2 \ll v/c$ and this second order term can be safely ignored. We then wind up with</p>

<p>$$\frac{f_2-f_o}{f_o} = \frac{\Delta f}{f_o} = \frac{2v}{c}$$</p>

<p>What the police officer measures by using the radar gun and applying the Doppler equation is $2v$, but of course we know this to be twice the speed of the car, so of course to get the actual speed of the car, the speed measured by the radar gun must be divided by two. As you can hopefully see, this 2 comes from the double receive and reflect of the radio waves by the car. The same exact principle holds whether we're talking about radar guns and cars or satellites and planets.</p>

<hr>

<p><sub>Note: As I said at the start, thing's are vastly more complicated than I've presented. For example, we assumed that $v^2/c^2 \ll v/c$. In reality, to get a more correct answer, you wouldn't want to drop this term and you'd find that $v_r \ne 2v_p$ as you've indicated in your answer, and there's actually more to it than that. This is all further exacerbated by the fact that the doppler equation I started with already had some assumptions built in. The fully correct equation is more complicated than that.</sub></p>

CC BY-SA 3.0
2017-03-29T15:05:35.033Z
2017-03-29T15:05:35.033Z
zephyr
<p>We know that we can use Doppler effect to determine equatorial velocity of a planet. We send a electromagnetic wave with wavelength $\lambda_0$ to an equatorial edge of planet and we get wavelength $\lambda$. So we can use Doppler Effect formula, (assuming $v \ll c$)</p>

<p>$$\frac{\Delta\lambda}{\lambda_0} = \frac{v_r}{c}$$</p>

<p>such that $v_r$ is radial velocity. But $v_r$ = $2*v_p$. And $v_p$ is equatorial velocity of planet.</p>

<p>I want to know where does the $``2"$ comes from. Someone said it is because of receiving the signal and then reflecting it. I want to know the math behind this not just to say receiving and reflecting. Why do we need to multiply the velocity by 2 to get the radial velocity?</p>

<velocity><wavelength><doppler-effect>
Question about doppler effect
titansarus
# Question about doppler effect

(Asked by: username_0 on 2017-03-29)

We know that we can use Doppler effect to determine equatorial velocity of a planet. We send a electromagnetic wave with wavelength $\lambda\_0$ to an equatorial edge of planet and we get wavelength $\lambda$. So we can use Doppler Effect formula, (assuming $v \ll c$)

$$\frac{\Delta\lambda}{\lambda\_0} = \frac{v\_r}{c}$$

such that $v\_r$ is radial velocity. But $v\_r$ = $2\*v\_p$. And $v\_p$ is equatorial velocity of planet.

I want to know where does the $``2"$ comes from. Someone said it is because of receiving the signal and then reflecting it. I want to know the math behind this not just to say receiving and reflecting. Why do we need to multiply the velocity by 2 to get the radial velocity?

**Tags:** velocity, wavelength, doppler-effect


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## 1 answers

### Answer by username_1 on 2017-03-29. Score: 4 | Accepted answer

Before I start, let me just say that this topic is vastly more complicated than you've presented and what I will be showing. The trouble here is that ultimately, everything you've done and I will do (to a lesser extent) uses approximations and assumptions. Because of this, it can sometimes be hard to understand the true underlying physics when looking at only the "simple" versions of these equations.

You say that the you've heard that the source of the two is because of receiving the signal and reflecting it, which is entirely true. Let's consider the simpler case of a police officer using radar to track a car's speed (just so we don't have to drag in the more complicated concepts of orbital motions of satellites, dispersion relations, etc.) You have a police officer standing by the side of the road with a radar gun, shooting radar waves of some frequency $f\_o$ at the car as it recedes from him (note I'm going to use frequency but it is completely analogous to wavelength, assuming a simple dispersion relation). That wave travels to the car and the first peak of the wave hits the car. The car then has time to move slightly forward by the time the next peak hits the car. From the car's perspective, because of his relative motion to the police officer, the frequency of the wave hitting the car has been Doppler shifted to a new frequency, $f\_1$. This new frequency is given by the Doppler equation

$$\frac{f\_1}{f\_o} = \left(1+\frac{v}{c}\right)$$

Note that this equation is the same as what you presented, except that it uses frequency instead of wavlength. So now, our car has been hit by these frequency shifted waves. It will now reflect those waves back at the police officer. The process of reflection effectively makes the car an emitter of the frequency $f\_1$ but since it's obviously moving with respect to the police office, the police officer receives a Doppler shifted frequency, which I'll call $f\_2$. Since the speed between the car and the police officer is still $v$, we know that the Doppler equation for the frequency shift of the reflection is given by

$$\frac{f\_2}{f\_1} = \left(1+\frac{v}{c}\right)$$

Now the police officer receives back a frequency $f\_2$, but he originally emitted $f\_o$, so he wants to know the velocity which caused the shift from $f\_o$ to $f\_2$. Fortunately, this can be determined by combining the two shifts above.

$$\frac{f\_2}{f\_o} = \frac{f\_2}{f\_1}\frac{f\_1}{f\_o} = \left(1+\frac{v}{c}\right)\left(1+\frac{v}{c}\right)$$

$$\frac{f\_2}{f\_o} = \left(1+\frac{v}{c}\right)^2 = 1 + 2\frac{v}{c} + \frac{v^2}{c^2}$$

At this stage, we make the assumption that $v^2/c^2 \ll v/c$ and this second order term can be safely ignored. We then wind up with

$$\frac{f\_2-f\_o}{f\_o} = \frac{\Delta f}{f\_o} = \frac{2v}{c}$$

What the police officer measures by using the radar gun and applying the Doppler equation is $2v$, but of course we know this to be twice the speed of the car, so of course to get the actual speed of the car, the speed measured by the radar gun must be divided by two. As you can hopefully see, this 2 comes from the double receive and reflect of the radio waves by the car. The same exact principle holds whether we're talking about radar guns and cars or satellites and planets.

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Note: As I said at the start, thing's are vastly more complicated than I've presented. For example, we assumed that $v^2/c^2 \ll v/c$. In reality, to get a more correct answer, you wouldn't want to drop this term and you'd find that $v\_r \ne 2v\_p$ as you've indicated in your answer, and there's actually more to it than that. This is all further exacerbated by the fact that the doppler equation I started with already had some assumptions built in. The fully correct equation is more complicated than that.


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