this post was submitted on 19 Jan 2024
18 points (100.0% liked)

Physics

1332 readers
14 users here now

founded 2 years ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
[โ€“] [email protected] 3 points 10 months ago* (last edited 10 months ago) (1 children)

EDIT: Ah! After sending this answer I saw jorge's answer!! So this answer is redundant, but at least you can see that two people arrived at virtually the same conclusion ๐Ÿ˜€


We need to define a threshold of energy that we consider "ionizing radiation", and we also need to a more precise definition of "starlight".

I will arbitrarily select the ionizing radiation threshold to be at 10 eV (124 nm). As for "starlight", let's just say that we want to push the 750 nm red light all the way until the point where it becomes ionizing. One thing to consider is that in this situation you will also push infra-red light from the stars towards the visible, so if a star emits a lot infrared this IR light will become "starlight". So the answer can be muddled up by all of these definitions as well as the emission properties of the star.

To keep it simple... Let's shift 750 nm red light to 124 nm ionizing radiation. You can rearrange the Doppler expression from this website to solve for the "v" to get the velocity needed to transform 750 nm to 124 nm. The solution I get is -284,035,329 m/s, with the "-" sign indicating movement of the receiver towards the source.

You can double-check by inputting 750 nm as the wavelength from the light emitted by the source, -284,035,329 m/s as the velocity, and the speed of light as "c":

Then, if you agree with the assumptions, definitions, and the analysis, the receiver needs to move at about 94.68% the speed of light to shift the redder starlight into the ionizing radiation range.

[โ€“] [email protected] 2 points 10 months ago

Much thanks! Was coming at it from a sci fi perspective, so definitions are laymans definitions, and you've made good assumptions. It also lines up nicely with the other answers :)

In conclusion, it'll become a massive problem for a probe at 0.95c ish.