-1

I realise this is a (slightly) nihilistic question but lock down and all...

Is it possible to destroy the solar system by launching a proton at the sun at a speed approaching c (as seen in cern 0.999999991) and then using the gravitational pull of the sun to accelerate it to an even faster speed such that its mass increases to a ridiculous amount and the resulting collision either creates a black hole or otherwise generally wreaks havoc?

Just wondering and hoping the answer is no or that no one builds a cyclotron in orbit!

Qmechanic
  • 201,751
user106271
  • 25
  • 6
    @xXx_69_SWAG_69_xXx "Technically" is a really misleading term here. Also, even Einstein himself wasn't enamoured with the concept of 'relativistic mass'. Better relativistic momentum or kinetic energy. https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass_2 – Gert Mar 12 '21 at 16:54
  • 4
    "It is not good to introduce the concept of the mass ${\displaystyle M=m/{\sqrt {1-v^{2}/c^{2}}}}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ $m$. Instead of introducing $M$ it is better to mention the expression for the momentum and energy of a body in motion."

    — Albert Einstein in letter to Lincoln Barnett, 19 June 1948

     – Gert Mar 12 '21 at 16:55
  • 1
    Related: the OMG particle: https://en.wikipedia.org/wiki/Oh-My-God_particle – Gert Mar 12 '21 at 17:01
  • Happy to go with the momentum definition, according to a couple of fag packet calculations I did earlier launching the proton from a distance of 1AU at the speed above achieves rapidly approaching infinite momentum about half way to the sun, so what happens at that point (does the proton suck itself into its own black hole?) and assuming it doesn't and it collides with the sun is that super bad or have I missed something? – user106271 Mar 12 '21 at 17:05
  • 1
    Possible duplicates: https://physics.stackexchange.com/q/3436/2451 , https://physics.stackexchange.com/q/3584/2451 and links therein. – Qmechanic Mar 12 '21 at 17:24
  • Hint: Check the GZK limit. – Qmechanic Feb 02 '24 at 14:22
  • By conservation of energy, to pump up a proton Kinetic Energy to a value close to the Sun's rest mass energy, you'll need to consume a whole star! If you had an high energy laboratory in the solar system with that high rest mass energy, you'll already be perturbing significantly the whole solar system! – Cham Feb 02 '24 at 14:55

1 Answers1

0

TL;DR: Probably not.

The Sun has a lot of gravitational binding energy, roughly $3.8×10^{41}$ joules. Your proton needs to have kinetic energy in that neighbourhood, and it needs to deposit it deep inside the Sun to do much damage. From that Wikipedia article:

According to the virial theorem, the gravitational binding energy of a star is about two times its internal thermal energy in order for hydrostatic equilibrium to be maintained.

When a star forms, gravitational potential energy is converted to thermal energy. Half of that energy stays inside the star as its internal thermal energy, the other half is radiated away. This is known as the Kelvin-Helmholtz mechanism. Lord Kelvin thought that it was the process that powers the Sun. However, that mechanism would only produce enough energy for the Sun to shine with its present luminosity for around 8.9 million years. Of course, the Sun is actually powered by nuclear fusion. Still, 8.9 million years worth of total solar output is nothing to sneeze at. And that's roughly how much energy your proton accelerator needs to supply to your killer proton. I suspect you'll face extreme difficulties in getting that much energy, and in pumping it into a single proton. ;)

Your plan is to shoot the KP (killer proton) at the Sun from a distance of 1 au, and for it to pick up additional momentum due to the Sun's gravity. There are a couple of problems with that. Firstly, interplanetary space isn't a perfect vacuum. It contains numerous protons and other particles from the solar wind. True, the density of those particles is small, but when any of those particles collide with the KP they will take some of its momentum & KE, essentially acting as a form of friction. And the Sun's magnetic field will also deflect the KP's path and act as another form of friction.

Secondly, the KP won't actually pick up much momentum falling towards the Sun. At low speeds, a body falling from 1 au to the Sun gains about 40 km/s, but that's insignificant to your ultra-relativistic KP, which has kinetic energy many orders of magnitude greater than its rest mass. At high speeds, you can't just add speeds together, you need to use the relativistic formula for composition of velocities: $$w = \frac{u+v}{1+uv/c^2}$$

PM 2Ring
  • 11,873
  • Thank you! Your second point was the piece I was missing. It's good to know that some tech billionaire, say that's pissed off because his phallic shaped rocket keeps blowing up, can't construct a doomsday device... – user106271 Mar 12 '21 at 20:13