## THE QUESTION

*What would happen if you fired a bullet straight up at escape velocity?*

First let’s talk about what escape velocity means. Escape velocity is the speed at which an object fired from a planet will escape to an infinite distance.

The escape velocity for the moon is about 2.4 kilometers per second. Contrary to what a lot of people believe, a rocket doesn’t need to maintain this speed all the time in order to escape moon’s gravity. Rather, it means that at the moment the rocket is launched, if its speed is 2.4Km/s and it turns off its engines after that, it will reach an infinite distance. In fact, you could escape the moon’s gravitational field even if you maintained a speed of 0.00001 m/s if you had enough fuel. Escape velocity is actually more meaningful for projectiles like stones rather than rockets because stones don’t propel themselves. If you threw a rock at 2.4 Km/s from the moon, it would never come back!

The interesting thing about escape velocity is that it’s something of a misnomer because it’s actually a speed, not a velocity. People who made it through middle school will remember that speed is a scalar and velocity is a vector,meaning speed only requires a magnitude but a velocity requires both a direction and magnitude.

## THE PROBLEM(S)

Notice how I used the Moon rather than the Earth to explain escape velocity. That wasn’t me trying to be alternative and quirky. That explanation only holds for celestial bodies that aren’t surrounded by a blanket of gas because Escape velocity is derived in a way that ignores atmospheric forces on the projectile. Congress supporters will know that Jupiter’s escape velocity is 60 km/s and JEE aspirants will know that Earth’s is 11.2 Km/s. But if you threw a stone at that speed, it would be slowed down considerably by the effects of the atmosphere and would never reach space. You’d have to throw it at a much higher speed than Earth’s escape velocity in order to escape the Earth’s influence. So let’s talk about all the problems our bullet will face on its way up.

## THE GUN

11.2 Km/s is not a small speed. A 747 only goes 250 m/s. The concorde maxed out at about 600 m/s and even the SR-71 couldn’t cross 1 km/s. If you were travelling at 11 Km/s, you’d circle the earth in an hour. If you set out from east to west, the sun would appear to move in reverse (having said that, I should add that you could do that in just about any plane that can sustain speeds greater than the speed of the Earth’s rotation at that latitude).

Obviously,firing a bullet at 11.2 Km/s is not a feat that can be accomplished by a pistol. That would only get you to about mach 1. Even a tank will fire shells only at about 1 Km/s. The reason conventional guns can’t fire at much higher speeds has to do with the way guns work. The bullet is propelled out of the barrel by gases expanding behind it or rather the pressure difference between its back and front. This pressure difference forces the bullet to move towards the side of lower pressure: forward. Now this doesn’t pose a problem if your aim is to shoot your enemy 50 yards away, but it does if you want to fire a bullet into space. Pressure waves are limited by the speed of sound in the medium in which they travel. So you’d have no problem getting a bullet to a few hundred m/s but beyond that,you’re out of luck.

One way to fix the problem (partially) is to raise the speed of sound of the propellant gases.

The speed of sound in a medium can be calculated from the relation

Where R is the gas constant, T is the Temperature, M is the Average molar mass of the propellant gases and γ is a constant associated with each gas depending on its atomicity which we need not be concerned about at the moment.

By decreasing the molar mass of the propellant gas, we see that the speed of sound in it at a constant temperature should increase. This is why your voice becomes squeaky after breathing in helium. Helium has a molar mass of 4 grams/mole, while that of air is about 28 grams/mole. This principle is used in the light gas gun. Not the squeaky voice thing, although scientists with squeaky voices would be fun to watch.

Mechanism of the light gas gun.

-wikipedia

The light gas gun consists of 2 sections. The first wider chamber contains a gas with low molar mass like Hydrogen or Helium. Helium is safer, but hydrogen, which has a lower molar mass, will result in a higher velocity. At the back of this chamber is a propellant charge and a piston. When the charge explodes, the piston is pushed forward, which in turn pushes the light gas towards the end of the chamber which is blocked off by a disc that breaks after a certain threshold pressure is crossed. A conical section leads to a thinner second chamber which contains the projectile. This section is made thinner because to maintain a uniform flow rate, the product of cross sectional area x fluid velocity must remain constant, leading to an increase in velocity with a decrease in area, thus increasing the speed of the projectile. NASA has used light gas guns to accelerate projectiles to 7 km/s to study reentry effects and high speed meteorite impacts.

7 km/s is fast, but it’s not enough. The only way to eliminate the problems arising from the limits of gas propulsion is to eliminate our dependence on gases altogether. We’ll need an entirely new system that doesn’t use pressure to accelerate a projectile. How about magnets?

More specifically, electromagnets. Plans for a coil gun or Gauss gun have been around for a long time now. The idea is that a ferromagnetic projectile placed at one end of a solenoid will be accelerated when current starts to flow in it because a current carrying coil acts as an electromagnet. The current is switched off when the projectile reaches its center since this is the time when it’s moving fastest. In theory, a series of such arrangements could accelerate a projectile to enormous speeds.

However, theory isn’t all that matters. Coil guns are horribly impractical. Here are just a few reasons why nobody considers it a serious way to launch projectiles. In reality, coil guns offer more limitations than gas propelled systems!

● Efficiency is one of the biggest problems with coil guns. No matter what you do, there is no way to link 100% of the magnetic flux to the projectile. This means that the magnetic field generated by the coil will store energy when current flows and when it’s switched off, a reverse current will flow in the circuit. Since it is common to use large capacitors to supply bursts of current in the coils, this reverse current will produce LC oscillations which produce heat. At high speeds, the oscillations will happen fast and generate tremendous amounts of heat.

● Changing magnetic fields in a metal cause eddy currents to flow through it. This will heat up the projectile and melt it when the process happens fast enough. This is the basis of induction heating. Here’s a gif to demonstrate what happens when changing magnetic fields interact with iron

● They require huge amounts of power to accelerate projectiles to sufficiently high speeds. However, hobbyists make relatively slow coil guns all the time.

● One of the obvious problems is that at high speeds, you’d need to switch the coil on and off very quickly and there is a limit to how fast a switch can be switched.

To learn more about high speed projectiles and space guns, watch Scott Manley’s video: https://www.youtube.com/watch?v=Moo5nuLWtHs

## WHAT IF

There is no gun that can fire a bullet at the speed we need. And don’t bring up the operation plumbbob manhole cover. At best, it was vaporised but you will never convince me that it left the atmosphere. But what if we did have a hand held gun powerful enough to fire a 5 gram bullet at 11.2 Km/s? Well, first you’d feel the recoil. A quick calculation shows that the momentum transferred to the bullet and therefore to you is only about 56 Kgm/s which is the same as a 56 Kg boy walking at 3.6 km/h or a bus at 0.015 km/h. Yeah…

But recoil is a force, which is defined as change in momentum per unit time,so the force you experience depends greatly on how fast the momentum is transferred from the gun to the bullet. Let’s say it takes 1 millisecond, and I have no idea if that’s reasonable because I know almost nothing about guns. The force you’d experience is the same as being stood on by a male african bush elephant (courtesy: google). Your arm would probably be blown right off.

As soon as the bullet leaves the barrel of your impossible gun, you’d hear a sonic boom that would shatter your eardrums.

an F/A-18 travelling faster than the speed of sound- aka pace booklet cover

For more about this picture: http://www.truthorfiction.com/rumors/s/soundbarrier.htm#.U1I_cd_Ft20

Now let’s talk about the kinematics of the bullet.The only forces acting on the bullet are drag and gravity. The force of gravity is given by

Where m= mass of the bullet and g= acceleration due to gravity

Drag is given by

Where ρ=density of air, v= velocity of the bullet, C_{d}=drag coefficient and A is the area of the orthographic projection of the bullet. since we are assuming our bullet to be a lead sphere, A=πR^{2}

Since the density of lead is 11.2 g/cm^{3} and the mass of the bullet is 5 grams, a simple calculation tells us that the radius of the bullet will be 0.0047 m and A=6.94 x 10^{-5} m^{2}.

As for the drag coefficient, this paper leads me to believe that at the velocities we are working with, the drag coefficient of a sphere is 0.92. Of course, the drag coefficient will change with speed and viscosity of the air, but we will see why that becomes unnecessary soon.

For more on drag coefficients:

http://www.thermopedia.com/content/707/?tid=104&sn=1423

http://www.grc.nasa.gov/WWW/k-12/airplane/dragsphere.html

According to wikipedia, the density of air within the troposphere can be estimated as:

Where ρ= density of air, M=average molar mass of air = 0.028 Kg/mol, p_{o}= pressure at sea level= 101325 pa, L= temperature lapse rate= 0.0065 K/m, R= gas constant= 8.314 J/mol K , T_{o}= temperature at sea level= 288.15K, h= height above sea level and T= temperature at height h= T_{o}-Lh.

Now, to find density data for the stratosphere, I had to turn to research from NASA that used the X-15 to gather temperature, pressure and density information in the stratosphere and mesosphere. Extrapolating from this graph,

I determined the density relation to be

ρ=1.33 x (10^{-x/50000})kg/m^{3}

For calculations beyond the stratosphere, I’m assuming a perfect vacuum, which is inaccurate of course, but the air density beyond the stratosphere is so low that it would barely change the result. On reading further, you will see why this never became a problem either.

Now that I’m done explaining all the forces on the bullet, let’s talk about the changes that those forces will cause in the bullet’s motion.

The work energy theorem tells us that the change in kinetic energy of the bullet will be equal to the work done on it by the forces acting on it. That is

We also know that work done on a body is given by

where F is the force and S is the displacement of the object over the interval of time that the force acts. Since the displacement of the bullet is upwards and the forces both act downwards, the work done by both of them will be negative.

Over an infinitely small displacement ‘dh’ , the work done by both the forces is given by

And the change in kinetic energy will be

Where v= velocity at the start of the displacement dh and v’ = velocity after crossing the displacement dh

equating these two, we get

Obviously, v’ will be less than v and therefore the quantity v’^{2}-v^{2} will be negative, just like the total work done. Now we know the left hand side at least has the same sign as the right: we must be headed in the right direction.

In order to find the height at which the bullet stops moving, we need to sum the work done over n infinitely small displacements dh, where n→**∞ **and dh→0

This amounts to the integral

The first integral is straightforward, but the second one is complicated because v is dependant on h and we don’t know how they are related. In fact, the relation is what we are trying to find. Understand that if we knew the relation between v and h we could easily find out at what height the velocity becomes 0.

The only way I could think of to solve the problem was to use a computer to estimate the integral by summing over multiple small displacements which were not infinitely small, but were small and finite.

We understood earlier that we could equate the work done on the bullet to the change in its kinetic energy by this relation

A little bit of rearranging leads to the following relation:

This relation tells us that if the bullet was travelling at velocity v at the beginning of the track of length dh, it is possible to calculate how fast it will be going after it has crossed that displacement. Knowing the velocity of the bullet at the end of the displacement dh, we can go on to find its velocity after it has crossed the next displacement as well! Now all we have to do is write a program that will find the velocity of the bullet after some number of tracks with length dh, which is not that difficult. However , this will only work under the assumption that for every small displacement dh, the bullet will travel with a constant velocity and only slow down to velocity v’ at the end of that length. Also, we will work under the assumption that the length dh= 1cm ie this is only an approximation of an integral. For it to be accurate, dh would have to be as close to 0 as possible.

*Here’s the python program:*

http://labs.codecademy.com/CLgb/16#:workspace**

(*I was going to add in a piece in the program to facilitate entry into the stratosphere, but after running the program , I found that the bullet doesn’t even even reach a kilometer up so adding that would have been redundant. Earlier, I’ve noted the density-height relation for the stratosphere, so readers are welcome to add the pieces to the program and while you’re at it, you could play around with the values of dh, launch velocity and mass of the bullet to see what will yield the greatest range. I’ve added comments for every variable so you can understand what you’re doing.)

(** Yeah I use codecademy. You can stop laughing now.)

When you run the program, it will print a huge list of values of h and the corresponding value of v. The program terminates (actually, it shows an error but “terminate” sounds cool) when v’^{2} becomes less than zero and the height printed at that time will be the maximum height attained by the bullet ie 866m. I really wish I had the know how to plot a graph of h vs v but sadly , I don’t. If one of the readers knows how to do this, please leave it in the comments.

But all this assumes that our bullet CAN get that high. As it turns out, it can’t. A quick calculation using this data shows that you only need about 5620 Joules of heat to completely vaporise 5 grams of lead. As the bullet climbs, it loses kinetic energy and in an ideal situation, all of this will be stored as potential energy, but this is far from an ideal situation. Some of the energy will be released as sound energy, some will be used to heat the air passing by the bullet and some will be used to heat the bullet itself. All of this energy is being lost because of drag. Let’s assume half of it is released as sound, a quarter is used to heat the air and the remaining quarter is used to heat the bullet.

The energy loss can be given by

If you looked at the python code posted earlier, you will see that an easy tweak can be made to estimate the value of this integral.

Here’s the tweaked code:

http://labs.codecademy.com/CLgb/13#:workspace*

(*I’ve reduced the final height to 20m to decrease processing time because it turns out the bullet doesn’t go far before burning up)

On running this code, you will find that Heat becomes equal to 5618 Joules around h=5! That is unbelievably fast. Our bullet has changed from a solid to a gas in 0.4 milliseconds. If that doesn’t blow your mind, I’m afraid you’re an arts student. lol jkjk if you were an arts student, you wouldn’t have gotten this far. (here’s a version of the code that prints the time along with the other data http://labs.codecademy.com/CLgb/15#:workspace)

But the mechanism of this vaporisation isn’t your everyday water-boiling type. Our bullet will undergo a process called ablation. The surface of the bullet gets heated up so fast that the atoms on the surface don’t have enough time to transfer their energy to the bulk of the material, so instead they take the energy and evaporate without heating the center of the bullet. This is similar to spacecraft reentry. Reentry pods usually have a shield that is blown off by heating due to the atmosphere, keeping the astronauts from getting fried. Read more about it on wikipedia.

In conclusion, I’d like to say that all of this is pure speculation and I’m not a scientist. I didn’t even know drag coefficients depended on velocity before I started writing this article. I apologise for any inaccuracies and the horribly inefficient programming. If you’re an aeronauticist or a programmer, I can only imagine your faces as you read this article. If you have thoughts or corrections, leave it in the comments.

## ACKNOWLEDGEMENTS

I’d like to thank Aniruddha Sinha of the IITB aerospace department for not laughing at me when I called him to ask what happens when you fire a bullet at mach 30 and for giving me the idea of using a computer to iterate a function to find velocity as a function of height. A thank you as well to DP Mishra of the IITK aerospace department for explaining to me that at high mach numbers, there are plenty of effects that cause drag to increase. Sadly, even on research, I failed to understand any of it and ended up not considering it in my equations cuz im stoopid. D’oh!

-Vaibhav Nayel

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