Re: [Phys-L] Doppler

[Anthony Lapinski <alapinski@pds.org>]

Thanks for these details!

I think I (sort of) figured out the calculation using v/c = delta
lambda/rest lambda

The delta lambda would be (3/c) x 12.5 cm = 1.25 nm, which is the shift
from either edge of the planet.

So then a total shift (widening of spectral line) of 2 x 1.25 = 2.5 nm -
precisely what you wrote!

My students might have difficulty with this, especially unit conversions!

On Tue, May 27, 2025 at 1:41 PM Todd Pedlar <pedlto01@luther.edu> wrote:

> So I've not looked up the original papers, but the shift for a
> non-relativistic velocity from which radar waves are reflected is twice the
> ratio of the object speed to that of light, times the original frequency.
> So the frequency shift for 3 m/s on the limb of Mercury would result in
> about a 48 Hz shift in the original 2.4 GHz emission frequency.  I imagine
> that since Mercury is moving, and so all the light is shifted in an overall
> sense because of that much larger speed, they'd have to have compared the
> range of frequencies received and compare the directly reflected light
> (from the center) to that maximally shifted light (from the limbs) and
> subtract.
>
> The exercise of showing students how the shift of 2*f*v_obj/c would be a
> good exercise for them, though, starting from the shift of
> frequency received by the limb, f_limb = f_sent * sqrt((1-beta)/1+beta)),
> and then shifting again because the reflected waves are being
> reflected from a moving surface, f_received = f_limb *
> sqrt((1-beta)/(1+beta)) = f_s * (1-beta)/(1+beta).
>
> Then the overall shift would be f_sent - f_received = f_s * (1 -
> (1-beta)/(1+beta)) = f_s * 2 * beta / (1+beta), but since beta is 10^-8,
> this is about 2*f_s*beta.  With a wavelength of 12.5cm sent to Mercury,
> which corresponds to a frequency of 2.4 GHz, a frequency shift of 48 Hz
> would be expected if the rotation speed is 48 Hz.
>
> From that 48 Hz shift (relative shift of 20 x 10^-9) would mean a shift in
> wavelength of 2.5 nm.
>
> So... yeah, tiny :)
>
> T
>
> On Tue, May 27, 2025 at 9:58 AM Anthony Lapinski via Phys-l <
> phys-l@mail.phys-l.org> wrote:
>
> > Does anyone know how to calculate Mercury's rotation rate? I mean, Doppler
> > data was taken in the 1960s to determine its period to be around 59 days.
> > This corresponds to an equatorial speed of only 3 m/s. I think a radio
> > telescope was used, with a wavelength of 12.5 cm. What is the
> > corresponding
> > wavelength shift? I imagine very tiny! No calculations found online for
> > this, just the rotational period result. Wanted to make this a problem for
> > my HS astronomy students, but 3 m/s << speed of light in the Doppler
> > formula.
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>
> --
> Todd K. Pedlar
> Professor of Physics and Physics Department Head
> Luther College, Decorah, IA
> pedlto01@luther.edu
> (563) 387-1628
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