Re: [Phys-L] Doppler

[John Denker <jsd@av8n.com>]

On 5/27/25 07:57, Anthony Lapinski via Phys-l 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.

1) To answer the general question of how to "measure" it: The
MESSENGER spacecraft orbited Mercury from 2011 to 2015 and mapped the
surface plus or minus half a gnat's eyelash. So nowadays that's our
best data. The rotation rate is non-uniform, due to spin-orbit
interactions; this is called libration.
  https://science.nasa.gov/mission/messenger/
  https://science.nasa.gov/gallery/mercury/
 
https://www.astronomy.com/science/mercurys-movements-give-scientists-peek-inside-the-planet/

2) Even in the pre-MESSENGER days, NRAO was able to image the surface,
using the VLA plus Goldstone. The resolution is about 1 arc-second as
seen from earth. That doesn't give you a lot of pixels on the disk of
Mercury, but more than enough to let you watch surface features going
around and around, in real space (not just Doppler frequency space).
  https://public.nrao.edu/gallery/radar-image-of-mercury/
 
https://science.nrao.edu/facilities/vla/docs/manuals/oss/performance/resolution

3) To return to the topic indicated by the Subject line: The VLA is a
phased array, so you can image both limbs at once, in parallel. (Image
the center of the disk, too, while you're at it.) Then you get the
Doppler frequency by subtracting one limb-frequency from the other.

The subtraction means that other contributions such as the orbital
velocities of Earth and Mercury drop out. This could be done using
little more than an analog mixer circuit: Multiplying two sine waves
gives you the sum and difference of the frequencies. At some level,
that's just high-school trigonometry; OTOH I wouldn't expect HS
students to know whether it's hard or easy to multiply a pair of GHz
signals on the fly. Turns out it's easy. The radio in every cell phone
you've ever seen does it.

For Mercury, I reckon the Doppler shift is a few parts per billion. We
agree that's "small" ... but it's not crazy small.

To put that into perspective, note that the measured value of the
hydrogen hyperfine splitting is 1,420,405,751.768(2) Hz. That's a lot
of precision. The point being, if you need a microwave signal with
high spectral purity, you can get it.

The Doppler radar on your average sailboat can't resolve half a meter
per second, but NRAO can.

The Doppler formula is derived here:
https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/05%3A__Relativity/5.08%3A_Doppler_Effect_for_Light


On 5/28/25 14:24, John Sohl via Phys-l wrote:
>> John Denker noted that you have to take the orbital motion into 
account as
>> well. Yep.

I don't recall saying that.
I hope I didn't say that.

You have to worry about it for a few minutes, long enough to realize
that you ought to make a differential measurement, but then it drops
out. See item (3) above.