It is recommended that you follow along with a pen and paper for this article, as although the equations and derivations here are really very simple, they can get confusing quickly if you are only reading the words.
Derivation is something I’m sure most maths and physics students have heard of, but for those who don’t know, derivation is essentially a method of getting equations for things. It is a very useful tool, and many scientific discoveries have been predicted essentially by messing around with equations until they say something new.
There are lots of ways to derive an equation. For example, you can use the definitions of those things to derive a very simple relationship such as distance = speed times time. Speed is defined as how much distance you cover in a given time, essentially distance divided by time, and therefore multiplying by time has to give you distance.
These tend to become more complicated as you find definitions that are less obviously related.
Angular Speed and Velocity:
Angular speed is defined as the amount of radians (an angle unit) that are moved through by the object per second.
In order to find the modular (not direction specific) velocity of this, you need to find the distance the object is moving through, which is equal to the radius times the number of radians. This is a basic equation for finding the length of a segment of a circle, and is usually first found at early A-level or even GCSE level.
You now must times the top and the bottom of this definition of angular speed by radius, which gives you:
Which is the same as:
Because L is the same as θr (as we saw in the previous equation I introduced). Since we know that speed (or modular velocity in this case) is equal to the amount of space moved through per unit time, the velocity of the object will be the length of the arc the object is moving through divided by time. This means we divide the top and bottom of the fraction by t, giving us:
Which is the same as:
Because the L in this situation is the distance, and speed is equal to distance over time. Multiplying both sides by r gives us:
So starting from the definitions of these quantities, we have managed to derive an equation for velocity in terms of angular speed.
This is, in my opinion, a more fun one to derive, as it involves a couple more definitions and isn’t as intuitive.
Escape velocity is defined as when an object’s gravitational potential energy (GPE) is equal to its kinetic energy (KE). GPE is how much energy the object stands to gain in KE by falling to somewhere it can’t fall anymore, eg. the surface of a planet, and KE is just “movement energy”. This means we can equate the equations for KE and GPE:
The first thing we can notice is that both sides have the mass of the object as part of the equations, so we can cancel both of them out by dividing both sides by m:
The second thing you can notice is that Δh is essentially the radius of the object who’s gravity you need to escape from, so we can replace Δh with r:
The only thing left that is going to cause a little trouble is that g, which stands for the acceleration of an object under the gravity of another object. It is the force per unit mass that is exerted by gravity, and on Earth it is taken to be about 9.81N/kg. We don’t want to have to know the acceleration under gravity for every single object however, and so we want to find what g is in terms of other things, like the mass and volume of the object. We can, again, use the definition of g to find this:
F=GMm/r² (This is Newton’s law of gravitation.)
g=(GMm/r²)/m (Substitute in Newton’s law of gravitation for the force, as the force experienced by the mass is the force that is being exerted on it by the mass it is trying to escape from, eg. a planet.)
Substituting this equation for g into our KE=GPE equation, we get:
Then, because we want to get v on its own on one side of the equation (because v is the escape velocity), we have to times both sides by two (because 0.5 x 2 is 1). Finally, we square root both sides to get:
This is the equation used to find the escape velocity of any object with a known radius and mass. The G in this equation stands for Newton’s gravitational constant, which is 6.67×10^-11. To find the escape velocity of Earth, you can substitute in the values for M and r as M=5.97×10^24kg and r=6.37×10^6m. This will give you a value of about 11000m/s, or 11km/s, answering the age old question from every ten-year-old of “if we drove really fast, could we shoot into space?” Yeah, you could, given a car capable of going about 28 times the speed of sound (see: “rocket”).
We can use our previous derivation for escape velocity to find the Schwarzschild radius of a black hole, which is the point at which the escape velocity becomes the speed or light. Essentially, it is the radius you would need to crush an object of a given mass down to in order to have a black hole, as a black hole is defined as an object for which the escape velocity is equal to or over the speed of light.
The first thing we need to do is take a step back in our escape velocity equation by squaring both sides:
The next thing we need to do is replace the v for escape velocity with c, as the escape velocity must be equal to the speed of light at this point.
We then re-arrange the equation to get r on its own on one side, which we can to by “swapping” the c² and r (in reality we are multiplying both sides of the equation by r and then dividing both sides by c²).
The value of c is about 3×10^8 m/s, which if you substitute in the previous values for the Earth’s mass, you can get the Schwarzschild radius of the Earth, which comes out at around 8.8mm.
The equation for Schwarzschild radius is not particularly accurate for actual black holes, unfortunately, because it doesn’t take into account the fact that most black holes rotate extremely quickly, which, due to relativistic effects, makes their event horizons smaller. The Schwarzschild radius you calculate here is only for non-rotating black holes.
So, your equation isn’t working. Here’s what could be wrong:
The units you use in an equation are really very important, and you need to remember this especially while deriving things. The units you are expected to use for all these equations are Système international (SI) units, which are m/s for speed, kg for mass, N for force etc. You can easily modify equations to be for whatever units you want though. For example, if you are measuring speed in km/h and trying to find the distance travelled using the time taken, then using time in hours will give you the distance travelled in km. If you’re measuring time in seconds though, I would recommend dividing your time in seconds by 3600 (the number of seconds in an hour) in order to get the distance travelled in km.
If, like I was until quite recently, you are a little unconfident with algebra and re-arranging equations, I would suggest going back to basics. Review “power rules” (rules pertaining to how you deal with powers of things) as well as just getting the very basics fresh in your mind. For example, I would recommend looking over the “keep, change, flip” rule we all learned in primary school to show us how to divide fractions by other fractions. You shouldn’t feel bad about being bad at algebra though, because it is quite impossible unless you really understand the basics.
- Sometimes things don’t be like that, you think they do, but they don’t, i.e. you have thought one quantity is the same as another one when it actually isn’t. I used to do this all the time with questions about gravity, and I would often assume two distances represented the same distance and end up cancelling them out, confusing myself.
- You’ve not used the right definition of a quantity. The best way to check for this is to just go back to your original equations and make sure they are all correct and refer to the right units.
- You’re using the wrong equation. This may sound like the same thing as the previous point, but it’s slightly different. Some equations in physics are for the same thing, but for different things at the same time. F=BIL and F=BQv are both equations for the force exerted on something by a magnetic field, but F=BIL is for a length (L) of wire, while F=BQv is for an individual charge in the field.
I am a strong believer in the usefulness of maths as a language, and I think anyone is capable of mastering it. You may wonder “well, when will this ever be useful to me in day-to-day life?” and the truth is, it probably won’t be, unless you pursue a STEM career. But, in the end, being a better slave to the capitalist machine shouldn’t be the only reason to pursue knowledge and understanding. We should learn about things and do things because we enjoy them, not because they make us more valuable employees. I personally find a lot of joy in seeing where equations come from and why they work. If you don’t care, that’s fine. Do what makes you happy.
Note: Hopefully articles will be more regular from now on, as all of the writers are finished with exams. My next article should be back on the Astrophysics track with Brown Dwarfs, one of my all-time favourite types of celestial objects. Until then, if you’re looking for something a little trickier than the stuff here, I would recommend looking up the derivation of Kepler’s Law of Planetary Motion.