My attempt at physics...
It'd probably be more interesting if the ball was rotating. But, this is what I got so far for this physics problem.
MLB ball mass is 5.25 ounces or
The ball’s rest energy is E=mc^2
v=c * .9
v=299792458m/s * .9
E-rest=0.148834996kg * (269813212.2m/s)^2 <- mistake using .9c instead of c, see Update below
E-rest=0.148834996kg * 7.27991694776822E16m^2/s^2
4.184E15 J = 0.4184E16 J = is 1 megaton, so
E-rest=1.08350640980142E16 J =2.58964247084469 megaton TNT rest energy
The impact energy that created Meteor Crater, Arizona was thought to be from a 10 megaton impact.http://en.wikipedia.org/wiki/Meteor_Crater
For a velocity ~ c, the formula for relativistic kinetic energy is to subtract the rest mass energy from the total energy:http://en.wikipedia.org/wiki/Kinetic_energy#cite_note-6http://www.wolframalpha.com/entities/calculators/relativistic_kinetic_energy/http://www.wolframalpha.com/input/?i=Lorentz+factor+formula
K = (γ-1)mc^2
K = (γ-1)*1.08350640980142E16J
γ = 1/sqrt(1-((v/c)^2))
γ = 1/sqrt(1-((.9/1)^2)
γ = 1/sqrt(1-.81)
γ = 1/0.435889894354067
γ = 2.2941573387056 Lorentz factor
K = (2.2941573387056-1) * 1.08350640980142E16 J
K = 1.40222777177906E16 J
Or 3.35140480826735 metaton TNT kinetic energy
Total energy would be: K + E-rest = 2.48573418158048E16 J total?
or 5.94104727911205 megaton TNT ?
MLB ball circumference is: 9 1/4 or 9.25 inches
Diameter: 2.9443664472000637 inches
Radius: 1.47218322360003 inches
Surface area: 27.235389636601 cubic inches
Volume: 13.365161237071 cubic inches = 2.19015752562E-4 cubic meter
Flat Circle Area: 6.808847409150148 inches^2 or
Distance from pitcher to home plate is 18.4m
The density of air at sea level at around 80 degrees F or 30 C is about 1.1644kg/m^3
Volume of air cylinder from pitcher to home: 0.0043927959944873m^2 * 18.4m =
Total air mass opposing the ball would be 0.09411547847005kg ?
This air mass would have a rest energy of:
E-air=0.09411547847005kg * (299792458m/s)^2
E-air=8458677367425090 J (much less than the .9c baseball)
If 0.0043927959944873 cubic meter baseball cylinder sized volume of air is:
0.00511497165598 kg mass then that is a rest mass of:
Then how many meters would equal the KE of the .9c baseball? ?m of air?
= K / rest mass energy for 1 cubic meter of baseball cylinder sized air volume
= 1.40222777177906E16 J / 459710726490406 J
= 30.5023940268734 m?
So, 0.0043927959944873m^2 * 30.5023940268734m is:
A 0.13399079430352 cubic meter volume with:
0.15601888088702 kg mass, which is a bit more than the rest baseball at 0.148834996 kg, thus this 30.5m air volume equals the extra mass gained by the .9c baseball through relativistic kinetic energy:
E-airbrake = 1.40222777177932E16 J = K-.9cBball ?
If you then assume that the baseball is fusing with most of this rest mass energy, it's gaining more mass, wouldn't that take away from it's relativistic kinetic energy, and slow it down?
Or rather, it might need to interact with double the rest mass energy to stop since the air mass can be considered to be relatively at rest compared to the baseball. Newtonian kinetic energy = (1/2)mv^2. So maybe 60m of airbraking?
I'm first assuming a rigid body that maintains it's integrity to first understand the farthest it might go based on the conservation of energy. One could also assume a single test particle or rigid disc which it hits dead center. I'm also ignoring the Newtonian kinetic energy of the air molecules, they're basically frozen in time but they have some momentum. So, I'm assuming the baseball is a cylinder for now, hitting edge on with the flat edge face/disc so that there would be no odd angles involved between the ball and the air mass due to the curved profile of the real spherical baseball. And the air molecules would be going in random directions anyway with some PSI at 80F/30C sea level.
I stopped here because I'm probably wrong about something so far. And, I don’t understand the fusion calculations using the cross section or anything. I also think there are probably more relativistic corrections that would need to be made for the interactions, including thermal Bremsstrahlung.
An interesting question might be how much airbraking of a given average atmospheric density might be needed to slow the .9c baseball down enough to be captured by Earth's gravity in a stationary orbit (assuming the baseball had a theoretical heatshield)? And what angle, if any?
Someone also posted this link, appears I wasn't too far off.http://physics.stackexchange.com/questions/31740/relativistic-baseball
See comments on Yonatan's post.Update:
It looks like I made a few mistakes, which I kinda figured I would make. Instead of using c, I mistakenly used .9c to determine rest mass when calculating the relativistic kinetic energy. Using the Lorentz factor and checking the total energy helped me see the mistake. But I wasn't far off because .9 is close to 1.
Instead of 30m-60m it's more like 37.657m-75.315 m of maximum travel before I think it would have to stop were it a rigid body.
I've determined that traveling 37.6572765763794m through a volume of air cut out by the sphere with a 0.0373934539m radius (0.0043927959944873 meter squared area), means that the ball will cut through a volume of 0.16542073370802 cubic meter of air (at sea level, 80F, density of 1.1644kg per cubic meter) which has a mass of roughly of 0.19261590232962kg. This is the same amount of mass that the ball gains at .9c, so it shouldn't be able to travel through not more than 1-2x this much rest mass if it was rigid in a worse case scenario while conserving energy.
If the ball has a rest mass of 0.148834996kg and infinitely accelerates to .9c, the Lorentz factor γ will be 2.2941573387056, giving the ball a total relativistic mass of 0.34145089832962kg, a difference of 0.19261590232962kg more mass.
The ball has a rest mass energy of 1.33766223432274E16 J. At .9c, it should have a total energy of 3.06880763158084E16 J. The difference in energy is an additional 1.7311453972581E16 J of energy which equals the mass difference.
If you use Einstein's relativistic kinetic energy formula K = (γ-1)mc^2 using γ = 2.2941573387056, and a rest mass energy of 1.33766223432274E16 J, you get:
K = (2.2941573387056-1)*1.33766223432274E16 J
K = 1.7311453972581E16 J
Or 4.13753680033007 metaton TNT from K alone.
If you add the rest mass energy, you get 7.3346 megaton TNT. This total energy is nearly the energy released by the Krakatoa eruption. Or, the impact energy that created Meteor Crater, Arizona (10 megaton).
To oppose the relativistic kinetic energy only, worst case scenario with rigid bodies, I think you'll need double the rest mass energy since the ball is moving and hitting what is basically stationary matter with stationary inertia. But in any event, you will need between 1 and 2 times the relativistic kinetic energy in the form of stationary rest mass energy.
opposing rest mass energy om = (K/c^2)2
om = 0.19261590232962 kg * 2
Now the question becomes, how long does an air mass volume in the shape of cylinder need to be to interact with the ball with between 0.1926kg and 0.3852kg air mass? It can't be any worse than this in terms of air volume length with a given density (ignoring PSI and Newtonian kinetic energy, and aerodynamics).
mass = V * density
V = mass/density
V = 0.19261590232962kg/(1.1644kg/m^3)
V = 0.16542073370802 cubic meter
Since we know the area being cut out:
L = V/area
L = 0.16542073370802 m^3 / 0.0043927959944873 m^2
L = 37.6572765763794 m
At most, between 37.657m-75.315 m of travel through air would cause the baseball to interact with rest mass equal to it's relativistic kinetic mass-energy of 1.33766223432274E16 J or 0.19261590232962 kg.
Since it's not rigid, now I'm wondering how much conversion of the rest mass to photon energy, gamma rays, plasma, electron cloud or maybe even neutrinos through nuclear disintegration would propel some of it's mass-energy farther than 75m? Obviously, some photons would go quite far. But everything would still be buffeted by the opposing air mass density and have to adhere to the conservation of energy between the baseball and the volume of air mass in it's way, worst case scenario without accounting for the spread which makes it stop even shorter (exponentially with time and volume). The leading half of the mass density would probably be largely converted to byproducts before it had time to thermally fuse with the back half of itself within at least the first 18.4m.
I'm also thinking that the mass conversion and subsequent release of energy of both the air and the baseball would mean that it would slow down pretty fast, but I'm not yet brave enough to do the calculus to try to find out. My intuition says it's going to be at least partially a bomb within the first 18.4m and no longer have as much of that kinetic relativistic energy, although it should have some when it hits the bat because there's not enough air density in 18.4m to fully vaporize it? The baseball is not as highly dense and massive as meteorite. So, while the air molecules are going to vaporize it super fast within between 75m-37m, it will releases a vast amount of energy from rest mass and kinetic energy via fusion like nuclear bomb.
But I still don't quite know how much, if any, remaining mass & kinetic energy of the original will make it to home plate. This is a fascinating thought experiment, even if I might be getting some of the physics wrong or my math sucks.
At .9c does it even have enough Planck time over 18.4m between the stationary air and the relativistic ball, assuming it's rigid, to thermalize/vaporize that much of itself from the fusion? Time is also time dilated between the .9c ball and the air molecules. It seems like it really needs to slow a lot first to have local time to let this work out and develop outside the ball's reference frame, which is a faster one.