General > General Technical Chat
Acceleration of gravity from earth, on objects, traveling at near the SOL.
tszaboo:
--- Quote from: sourcecharge on December 19, 2020, 07:44:02 pm ---
--- Quote from: NANDBlog on December 19, 2020, 01:03:02 pm ---As I understand, if you accelerate an object, which is going at relativistic speeds, not all the energy is increasing the speed. Most of the energy is going to increase the mass of the object, not the speed, the faster it goes, the higher percentage of the energy goes into the mass. Slowly reaching 100%. So none of the answers.
--- End quote ---
It was a hypothetical question...how the space ship got to it's speed is improbable, this is true, but that part was supposed to be hypothetical.
--- End quote ---
But it is still an answer. Your answer(s) assume that the gravitational pull will increase the speed. What actually happens, that the gravitational pull will accelerate the object with some newtons. For most falling objects (in vacuum) this will increase the speed. 9.8 m/s2 is an oversimplification, because the mass of the object increases the force (newtons). For a relativistic speed object, the force (newtons) will be spent on increasing the mass, not the speed. So even 0.98m/s2 could be a reasonable answer. Probably not though.
sourcecharge:
--- Quote from: NANDBlog on December 19, 2020, 07:54:56 pm ---
--- Quote from: sourcecharge on December 19, 2020, 07:44:02 pm ---
--- Quote from: NANDBlog on December 19, 2020, 01:03:02 pm ---As I understand, if you accelerate an object, which is going at relativistic speeds, not all the energy is increasing the speed. Most of the energy is going to increase the mass of the object, not the speed, the faster it goes, the higher percentage of the energy goes into the mass. Slowly reaching 100%. So none of the answers.
--- End quote ---
It was a hypothetical question...how the space ship got to it's speed is improbable, this is true, but that part was supposed to be hypothetical.
--- End quote ---
But it is still an answer. Your answer(s) assume that the gravitational pull will increase the speed. What actually happens, that the gravitational pull will accelerate the object with some newtons. For most falling objects (in vacuum) this will increase the speed. 9.8 m/s2 is an oversimplification, because the mass of the object increases the force (newtons). For a relativistic speed object, the force (newtons) will be spent on increasing the mass, not the speed. So even 0.98m/s2 could be a reasonable answer. Probably not though.
--- End quote ---
I think it might be possible that the acceleration of gravity on the ship is lower, that is why I was interested in vad's answer and wanted more information.
Non-Abelian:
--- Quote from: sourcecharge on December 18, 2020, 12:57:38 pm ---The question is, if the space ship traveled past earth at near the speed of light, how much acceleration would gravity act upon the space ship?
--- End quote ---
Not enough to be noticed. In the relativistic limit, you can neglect the mass of the rocket, so E~pc, in which case you're asking how much will a light ray be bent by the earth's gravitational field.
--- Quote ---Would it be 9.8 m/sec^2, or would it be 9.8 m/(sec/10)^2?
--- End quote ---
None of the above. First of all, the acceleration due to gravity depends on how far from the earth you are. Second, for the relativistic case, gravity is involved, so you would need to use the schwarzschild metric to calculate something, which in this case isn't enough of an effect to calculate. But, the solution is that the trajectory will be deflected by approximately theta ~ 4m/r, where m is the mass of the earth in kilometers (the mass of the earth in these units is 0.8 cm = 0.000008 km) and the radius at the grazing angle of the earth is its radius (about 6400km).
Nominal Animal:
--- Quote from: sourcecharge on December 19, 2020, 07:40:13 pm ---
--- Quote from: Nominal Animal on December 19, 2020, 01:36:44 pm ---
--- Quote from: sourcecharge on December 19, 2020, 12:46:20 pm ---@ Nominal Animal & Ground_Loop
Relativistic mass does not create gravitational force like real mass.
--- End quote ---
Of course it does. Energy and momentum of whatever matter and radiation are present, affect the curvature of spacetime.
Even light, which has as close to zero rest mass as we can ascertain – photon rest mass is less than 10-18 eV/c2, or less than 1:100,000,000,000,000,000,000,000th of the rest mass of a single electron –, bends space; exactly because they have momentum. In Einstein's field equations, it is the energy-momentum tensor that determines the curvature of space.
In fact, mass is not directly related to spacetime curvature (and therefore gravitational acceleration) at all; only energy and momentum is. Mass is just one way to affect the energy and momentum present; and those determine the curvature of space.
--- End quote ---
This explains why relativistic mass does not have gravitational force.
Can you go fast enough to get enough mass to become a black hole?
https://wtamu.edu/~cbaird/sq/2013/06/18/can-you-go-fast-enough-to-get-enough-mass-to-become-a-black-hole/
--- End quote ---
Unfortunately, Dr. Baird is wrong. It is obvious general relativity is way outside his area of expertise. I'm sure that a PhD specializing in optics finds the fact that even light bends space frightening.
Like I said, in Einstein field equations energy and momentum cause the curvature of spacetime. Mass affects curvature only through energy and momentum. Dr. Bairds fundamental error is assuming that mass itself is what curves spacetime. That is simply not true.
Relativistic mass is actually just a notional measure. It is equal to the energy that causes the same spacetime curvature as the momentum of the point-like mass. It is only needed when you need something that has an analog in Newtonian physics. To correctly model spacetime, you need to use Einstein field equations.
vad:
--- Quote from: sourcecharge on December 19, 2020, 07:40:13 pm ---I don't think I've seen this before, do you have a reference link to this?
--- End quote ---
https://en.m.wikipedia.org/wiki/Acceleration_(special_relativity)
Equation (1c).
If we choose axis x to be the Earth’s radial along which spacecraft is moving, then:
a’x = a’, a’y = a’z = 0, ux = v, uy = uz = 0.
Therefore:
a = a’x = a / (gamma^3 * (1 - v^2/c^2)^3)
Where:
gamma = (1-v^2/c^2)^-0.5
Hence:
a’ = a’x = a / (gamma^3 * gamma^-6) = gamma^3 * a
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version