Archive for the ‘acceleration’ Category
an interminable conversation 10: more basic physics – integrals
that’s for sure
Jacinta: So I watched episode 3 of the crash course physics videos, on integrals, and found it so overwhelming I had to immediately go and take a nap. I’m surely too old and thick for this stuff, but I must soldier on.
Canto: We can do battle together – so this episode is about integrals, the inverse of derivatives. I’m not sure we gleaned much from the episode about derivatives, but if we combine this with exercises from Brilliant and some other practical application-type videos and websites we might make some more progress before we die.
Jacinta: Okay so equations, at least some of them, can be plotted on graphs with x-y axes, and the integral of the equation is the area between the curve and the horizontal x axis. Dr Somara is going to teach us some shortcuts for calculating these integrals, which sounds ominous.
Canto: I didn’t really understand the stuff about derivatives, but I’ll keep going, hoping for a light-bulb moment. So integrals help us to understand how things move, she says, which in itself sounds weird. And then she mentions the displacement curve, which I’ve forgotten.
Jacinta: Looking elsewhere, I’ve found a simple video showing displacement-time graphs. Displacement, which is simply movement from one position to another, is shown on the y (vertical) axis, and time on the x axis. A graph showing a straight horizontal line would mean no displacement, therefore zero velocity. A graph showing a vertical straight line would mean displacement in zero time, which would indicate something impossible – infinite velocity? Anyway, a straight line between the horizontal and the vertical would indicate a fixed velocity – neither accelerating nor decelerating. A curve would indicate positive or negative shifts in velocity. I think. Sorry – the terms used are constant velocity and variable velocity. That’s much neater. Oh and there’s also negative velocity, but that’s a weird one.
Canto: Thanks, that’s useful. Need to point out though that ‘curve’ is just the term for representing the data on a graph – in that sense the curve could be a straight line, or whatever. So Dr Somara starts with a gravity problem. You want to know the distance between your bedroom window and the ground, in a multi-storey building. You have a ball, a stop-watch and a knowledge of gravity. The ball will fall at g, 9.81 m/sec². What is the distance? According to the Doctor, discussion so far about motion has involved three aspects, position, velocity and acceleration, and has focussed on velocity as the derivative of position, and acceleration as the derivative of velocity. The connection has to be reversed to work out the distance problem. So velocity is the integral of acceleration.
Jacinta: And of course position is the integral of velocity. And this is important – velocity is the area under the acceleration curve, and position the area under the velocity curve. Which might be difficult to calculate. Areas within polygons, without too many sides, is easy enough, sort of, but under complex curvy stuff, not so much. And when we talk about ‘under’ here, it’s the area, often, to the axis, which represents some sort of zero condition, I think. Anyway, one method of calculating this area is to treat it as a series of rectangles, growing more or less infinitely smaller. Imagine dividing a circle into squares to determine its area – a big one extending from four equidistant points on the circumference, which would account for most of the area, and then progressively smaller squares in the interstices. You get the idea?
Canto: Yes, with that method you’d get infinitely closer to the precise area… Anyway, Dr Somara shows us a curve which is apparently the graphic representation of a formula, x⁴- 3x² + 1, and shows us how to find the integral, or at least shows us how we can divide the curve and its connection to the x axis by dividing it into rectangles. But what’s a more practical way of doing it? Well I’ll follow her precisely here. ‘If you know that your velocity is equal to twice time (v = 2t), then you know this is the derivative of position. So to find the equation for position, you have to look for an equation whose derivative is 2t, as for example, x = t². So x = 2t is the integral of v = t²
Jacinta: Yeah I can barely follow that. But the good doctor assures us that integral calculation is a bit messy. But apparently we can use the ‘power rule’ which we used for derivatives, and reverse it, in some instances. To quote: ‘Basically, you add one to the exponent, then divide the variable by that number’. Here’s an example: with v = 2t, x = 2/2t¹+¹, so x = 2/2t², so x = t². x = t² is the integral of v = 2t. She shows another more complex example, but I can’t do the notation for it with my limited keyboard skills. It involves some division. Anyway, with these mathematical methods we can look at trigonometric derivatives and do them backwards, e.g. the integral of cos(x) is sin(x).
Canto: We need to look at a variety of explanations of all this to bed it down methinks. I can only say I know a little more than I did, and that’s progress. And next we get onto constants. So what’s a constant (c)? It’s a number, and can literally be any number, positive, negative, fractional, whatever. It can be a placeholder, as for gravity, g (here on Earth), or presumably the speed of light, or ye olde cosmological constant, which is apparently still alive and well. Anyway, the derivative of a constant is always 0. That’s because a derivative’s a rate of change, and constants don’t change, by definition.
Jacinta: The derivative of t² is 2t, which presumably works by the power rule. Add any number and the derivative will always be 2t. That’s to say, the derivative of t² +/- (any number) is 2t. So, ‘if you’re looking for the integral of an equation like x = 2t, you have infinite choices, all of which are equally correct’. It could be t² or perhaps t² -18 or t² + 0.456. But I’m not clear on what this has to do with constants.
Canto: We’re flying blind, but it’s not too dangerous. The idea seems to be that the integral of x = 2t is t² + c. And here, if not back there, is where it gets tricky. With a bit of practice, we might know what the graph of the integral would look like, but not so much where it will lie vis-a-vis the vertical axis. For that we need to know more about the constant, ‘in order to know where to start drawing its integral. Whatever the constant is equal to, that’s where the curve will intersect with the vertical axis’. If it’s just t², it will intersect at zero, if it’s t² – 10, it’ll intersect at minus 10, etc. To avoid this infinity of integrals, the practice is to add c at the end of the integral, to stand for all the possible constants. So, saying that the integral of x = 2t is t² + c covers all the infinite options for c.
Jacinta: Well, Dr Somara next talks about the ‘initial value’, which you can apparently use to work out ‘where your integral is supposed to be on the y-axis’ without knowing the value for c – I think. For a graph of position, the initial value would be your starting position – where it intersects the vertical axis. This is the c value.
Canto: So returning to the bedroom window and the ball, the ball is dropped from the window sill at the same time the stopwatch starts. It hits the ground at 1.7 secs. So we know the time and the acceleration, 9.81m/sec². We need an equation for the ball’s position. We do this by finding its velocity, working out the integral of its acceleration. If you have a graph with the y-axis representing acceleration, which in this case is constant, and the x-axis representing time, the uniformly accelerating ball would be represented as a flat line, making the area under the ‘curve’ – between it and the x-axis – fairly easy to calculate. The area would be rectangular, and would be calculated by base x height. The base is t, the amount of time the ball took from release to hitting the ground, and the height is a, the acceleration, so it’s just a matter of a x t. The integral is at plus c, the constant. We need this constant, according to Dr Somara, because we can see that the velocity graph will be diagonal, a line ‘slanted in such a way that, every second, it rises by an amount equal to the acceleration’.
Jacinta: But, where to put it the line on the vertical axis? We’re looking for the integral of the acceleration, so we may use the power rule, which I still don’t get. So I’ll quote the doctor, for safety: ‘The acceleration, a, is a constant, but we could also say that it’s a x tº (t to the power zero)’. And anything ‘to the power zero’ is always 1. So, according to this mysterious power rule, the integral of acceleration (i.e the velocity) would be equal to the acceleration multiplied by the time – plus c. The c is added because we didn’t know where to place it on the y axis when time, on the x axis, is zero.
Canto: Yeah right. Let’s continue to quote the doctor – ‘Now here’s where the initial value [??] comes in. The velocity graph tells you what the velocity is for each moment in time. But we had to add the c, because we didn’t know…’ the initial value, being the velocity at time zero. ‘So the integral of the acceleration could have just been a x t, or at’. Or at plus or minus whatever. The c in the integral represents these options. But if we can work out the velocity at time zero (v0) we won’t need c. So, according to our Doctor, ‘if we write our equation with that v0 in it, as a placeholder for the velocity when time equals zero, we end up with the full equation for velocity, v = at + v0′. This is the kinematic equation, the definition of velocity. So the equation tells us that the final velocity of out tennis ball is at, that’s to say 9.8m/sec x 1.7 secs, that’s to say, 16.7m/sec (down, towards the Earth’s centre of gravity).
Jacinta: All of which seems to complicate something not quite so complicated. Anyhow, the pain isn’t over yet – we need to link acceleration with position, and this requires further integration, apparently. So, according to the power rule, which we should have learned, the integral of at is half at squared, and to get the integral of v0 you multiply it by t. Get it?
Canto: No. Let me quote from a highlighted comment: ‘i cant imagine how the avg viewer with no prior knowledge of calculus would actually understand calculus just by watching this video’.
Jacinta: Hmmm. Maybe we’ll try Khan Academy next. Anyway, if you put these integrals together you’ll get something that looks like the kinematic equation, the displacement curve. So for our example, we can work out the height from which the ball was dropped, or the distance the ball has travelled, using the initial position and time (zero and zero) plus half at squared. a was 9.81 m/sec², and t was 1.7 secs. t² is 2.89. Multiplying them makes 28.35, which, halved, is 14.175. metres. At least I got the calculation right, but as to the why….
References
https://sites.google.com/a/vistausd.org/physicsgraphicalanalysis/displacement-position-vs-time-graph
an interminable conversation 9: some basic physics
yer most basic sine graph
Jacinta: So it’s time for us oldies to go to school, and get into physics from scratch, including the maths.
Canto: Yes, we’re not going to go all historical this time, much as we love all the nerdy characters to be encountered, instead we’re going to go with the concepts, from simple to complicated. I’ve found a collection of videos, called ‘crash course physics’, and we’re going to follow the ineffable logic of the presenter, Dr Shini Somara, to reach the pinnacle of sagesse en physique. Starting with basic motion in a straight line.
Jacinta: Exciting. I’ve done that. But in this first episode she deals with cars and acceleration, inter alia, including its maths. Equations! Time, position, velocity and acceleration will be explained/analysed in simple terms for this starter.
Canto: Kinematic equations – we’re going to the Kinema! So, motion in one direction on a straight line. You’re stopped at a red light, and then put your foot on the accelerator when it goes green. Seven seconds later (precisely), there’s a siren behind you – a police car is asking you to stop. They give you a ticket for speeding in a 100kph zone.
Jacinta: So, in 7 seconds you’re up to more than 100kph? I know nothing about cars but that’s – unusual? Is it?
Canto: I’m sure car nerds can tell us, but so can google. There are plenty of cars that can get to 100 in less than 4 secs, even less than 3. Supposedly. Anyway, you’re doubtful about the police claim, but you can’t be sure, your speedometer is stuffed. How can you challenge the police claim, using maths?
Jacinta: You can’t, and anyway in Australia you’d be defected for a stuffed speedo.
Canto: But this is the USA, the land of shitty libertarian laws. So you’re travelling in one direction, one-dimensionally, so to speak. So the key variables here are the afore-mentioned time, position, velocity and acceleration. We also have to bear in mind change in position, aka displacement, which could have a positive or negative value – in this example, clearly positive. Now, velocity is about how that displacement occurs over time. It also can have a positive or negative value. Acceleration is about changes in velocity over time. You can feel that change – positive or negative – when you’re ‘thrown’ forward or backward on acceleration or braking.
Jacinta: So Dr Somara presents graphs that are fairly easy to read for a stationary vehicle, and one moving at a constant velocity. The vertical x-axis measures position or displacement in metres from an initial position, the y axis measures time. A stationary vehicle will show a straight horizontal line from the moment it stopped until it starts to move again. Constant velocity will show a straight line moving diagonally along both axes. An accelerating vehicle will of course show a curving line, curving up to the vertical, while a decelerating one will be curving to the horizontal.
Canto: So that’s a simple position v time graph, now to look at velocity and acceleration slightly differently, with velocity in metres/second on the vertical axis and time on the horizontal, and with acceleration in metres per second per second, that is, metres per second squared, on the vertical axis, and time, in seconds, on the horizontal. So this relates all our variables, time, position, velocity and acceleration. Average velocity is the change in position over time, and acceleration is the change in velocity over time. To get average velocity you divide change in position by change in time.
Jacinta: But as Dr Somara says, subtraction is also a feature – to find out ‘change’ you subtract initial value from final value, which sounds right but somehow seems to contradict the previous….
Canto: One’s talking about a change, the other about an average. They’re quite different. So the change in a particular value, or variable, is symbolised or abbreviated as delta, ∆. So, v = ∆x/∆t, average velocity (the v should have a bar above it, but I haven’t learned how to do that – will I need an extra keyboard?) equals change in position over change in time. For Dr Somari’s example, the car moved from 4 metres to 13 metres (the change in position), i.e. a value of 9 metres for ∆x. This occurred over 3 seconds, apparently, which divides as 3m/sec for average velocity over that period. But of course the car was accelerating during that period. The equation for acceleration is a = ∆v/∆t, for average acceleration.
Jacinta: Okay, and we can, apparently handily, rearrange the equation to get v(average) = v(at time zero) + at. This equation is called the Definition of Acceleration. Tadaaa! Constant acceleration is equal to the change in velocity divided by change in time. This is the first of the two main kinematic equations, which links velocity acceleration and time.
Canto: Okay now our physicist turns to gravity (g), which here on Earth is a force causing acceleration at 9.81 m/sec squared. But then she talks about the second kinematic equation, the Displacement Curve, which involves acceleration, starting velocity and time in order to calculate displacement:
x(position) – x(at time zero, initial position) = v(initial velocity)t +1/2a(acceleration)t(squared).
All of which looks very messy because I haven’t learned how to do the proper notation. Anyway this links acceleration as change in velocity to velocity as change in position. Right?
Jacinta: Uhhh, yeah. And the other kinematic equations, we’re assured, are just rearrangements of this dynamic duo. So apparently this takes us back to our speeding issue at the start. The initial velocity was 0, the time was 7 seconds. The displacement curve²equation/formula can be used to work it all out, or at least the acceleration. Our physicist tells that x – x (initial position) is 122 metres, which equals initial velocity (zero) multiplied by 7 seconds (which must surely be zero?) plus I/2a (which is to be found) multiplied by 7s squared, which is 49 seconds. So 122m = 0 + 49 (49) multiplied by half the acceleration, which by calculation I discovered to be close to 2.5, so the acceleration was approximately 5 metres per second squared.
Canto: It works out! And, following our expert, we can use the Definition of Acceleration formula to arrive at final velocity. It’s basically V + at, or 0 + 5 X 7, so a speed of 35 m/sec, which in km terms is about 126 km/h. Amazing! We got the maths. There is hope!
Jacinta: Well they’re diving into the deep end with crash course physics, as the next video is all about calculus and derivatives. About which I have no idea.
Canto: Yes, maths are the basis of physics, and we lost contact with complex maths decades, though I’m quite good at multiplication. But calculus, duh. Though our teacher tells us that it’s all just about accurately describing change.
Jacinta: Important – she goes on to explain things called derivatives, but I note in the inset:
Not all equations have derivatives! When we say ‘equations’ here, we really mean a function – an equation with only one output for each input. More specifically, we’re talking about functions that have derivatives.
I’m looking forward to clarification of all that.
Canto: So calculus explains the why’s and wherefores of change through derivatives. She also mentions integrals early on, as ways of calculating area under a curve – which we actually mentioned in those terms in a previous post.
Jacinta: We sound smart sometimes. So, derivatives. Dr Somara returns to the car and speeding example. The car drives off after the police incident, accelerating of course. But we don’t have a direct measure of the acceleration, but we know positional change over time. This is apparently equal to amount of time driving, squared, X = t². After 20 secs of driving, some kind of roadside ‘detector’ reveals the car’s speed. The driver takes time to register that she’s going even faster than 126 mph.
Canto: Dumb blond? Maybe not, maybe the detector is dodgy. How to find the velocity at the moment she passes it? Which, according to Dr Somara is the derivative of her change in position. And this is also about limits. These are key ideas:
Limits are based on the idea that if you have an equation on a graph, you can often predict what it’s going to look like at one point, just by knowing what it looks like at the surrounding points.
Jacinta: So our teacher gives the example of graphing x = t² when t approaches the limit of 0. So remember we have our time on the horizontal, and distance covered (or displacement, or positional change – it seems ‘distance’ is a no-no in this maths) on the vertical axis. So, moving back to zero from t=1 and x=1 she finds that when t=0.5, x=0.25, and when t reaches 0.1, x=0.001, so both values approach zero. This apparently shows what happens when you make intervals smaller. Another definition:
An interval is just a range on a graph. It’s the space between two points on the horizontal axis.
Of course, because that’s the time axis, generally. This is great parroting, but then when parrots copy their trainer perfectly they’re regarded as brilliant.
Canto: So we’re calculating the average velocity over a particular interval – from 15 to 20 secs. We use the equation v = ∆x/∆t (∆x is change in position, ∆t is change in time). The change in position, after subtraction, was 175 metres, the change in time 5 secs. So the average velocity works out as 35 m/sec. But this is only an average, and doesn’t take into account acceleration. But using limits gets us closer to the number we want. You can calculate your average over increasingly small intervals to arrive at an increasingly accurate figure.
Jacinta: So, sticking with our teacher, velocity is an equation that describes change in position, acceleration describes change in velocity. Velocity is thus the derivative of position and acceleration is the derivative of velocity. This is expressed in writing, using, for example, the power rule, expressed using variables and their numbered exponents. So x = t² is an equation that works here. To calculate the derivative, you take the exponent, 2, and put it in front of the variable, and subtract 1 from the exponent, and that’s the derivative, 2t. In full, the derivative of x = t² is 2t.
Canto: That’s a trick, as Dr Somara said, but it’s not really explained. She says ‘no matter how [you’re accelerating], your velocity will be 2t – double the number of seconds’. So I think it depends on those seconds. After 5 seconds, say, you’re travelling at 5m/sec, but after 20 secs, your speed is 40m/sec. So dx/dt = 2t ‘which is just a way of saying, mathematically, we’re taking the derivative of x with respect to t’. But it’s also written differently sometimes: if f(t) = t², then f'(t) = 2t. And I’m guessing that f stands for function, but I don’t quite know what a function is.
Jacinta: A function is:
in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are ubiquitous in mathematics and are essential for formulating physical relationships in the sciences.
That’s from Britannica online. So to continue, if f(t) = t², then f'(t) = 2t. That’s to say, f prime (t) = 2t, according to our teacher, who doesn’t explain ‘prime’. Do we have to do a maths course before we do this physics course? Does it have to do with prime numbers?
Canto: Apparently not. The symbol can serve a number of purposes in maths. Let’s just leave it for now. Using the power rule we can find other derivatives, e.g. x = 7t to-the-power-6. This equation has the variable t, and its exponent 6. We take the exponent and put it in front of the 7t variable, multiplying the number and subtracting 1 from the exponent, 42t to-the power 5. That’s to say dx/dt = 42(t to the power of 5). But maybe that shouldn’t be bracketed. And when the exponent is a fraction or decimal, the derivative of, say t to the power of one half is 1/2t to the negative one half. You always minus one, I don’t know why.
Jacinta: Ours is clearly not to reason why, at least not yet. This derivative trick works for negatives too. In the case of x = t to-minus-2, the derivative (dx/dt) = -2t to minus 3. Not very comprehensible, and then she mentions the dread word, trigonometry, used for calculating triangles, their angles and sides. Apparently physics uses right-angled triangles a lot. We shall see.
Canto: Indeed, let’s get into it. The derivatives of sine x and cosine x. If you have a right-angled triangle with an adjacent angle x, sin(x) = the length of the opposite side divided by the hypotenuse. For cosine, cos(x) it’s the length of the adjacent side divided by the hypotenuse. So, sin(x) = o/h, cos(x) = a/h. ‘So the graphs will tell you what those ratios will be, depending on the angle’.
Jacinta: I’m not sure if I really understand this, but let’s move on into further weird territory, in which sin(x) is plotted on a graph going from -360° to 360° on the x (horizontal) axis (that’s the ‘phase’, in degrees) and -1 to 1 on the y axis. At x = -90° and x = 90° the curve turns – that’s at every 180°. At those points the equations aren’t changing and the derivative is zero. Between the points the derivative oscillates from positive to negative. That derivative is in fact cos(x). I’m not sure why, but the derivative of cos(x) is -sin(x), the derivative of -sin(x) is -cos(x) and the derivative of -cos(x) is sin(x), for future reference. I’m hoping it’ll all become clear some day. Graphing all these will provide the proofs, evidently.
Canto: Yes, so Dr Somara finishes off this vid with another derivative that’s important in calculus, e×, the derivative of which is also e×, always. e, like π, is an irrational number which is quite vital to calculus, apparently. And even finance. Can’t wait to find out. So with the preceding we can, supposedly, take any equation for position and calculate the derivative, and so, velocity. And for velocity, your acceleration. Using integrals, which we’ll soon learn about, we can go backwards from acceleration to velocity, and from velocity to position. Presumably that will be next time.
Jacinta: So easy…
References
towards James Clerk Maxwell 6: Newton’s universal law of gravitation and G
Newton’s law of gravity goes like this:
where F is the force of gravitational attraction, G is the constant of proportionality or gravitational constant, m is an entity, particle or object with a particular mass, and r is the distance between the centres of mass of the two entities, particles or objects.
What’s the relation between all this and Maxwell’s electromagnetic work? Good question – to me, it’s about putting physics on a mathematical footing. Newton set us on this path more than anyone. The task I’ve set myself is to understand all this from the beginning, with little or no mathematical expertise.
The law of gravity, in its un-mathematical form, says that every object of mass attracts every other massive object with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
It seems often to be put about that Newton was revolutionary because he was the first to wonder why objects fell to the ground. This is unlikely, and Newton wasn’t the first to infer an inverse square law in relation to such falling. Two Italian experimenters, Francesco Grimaldi and Giovanni Riccioli, investigated the free fall (where no force acts besides gravity) of objects between 1640 and 1650, and noted that the distance of the fall was proportional to the time taken. Galileo had previously conducted free fall experiments and found that objects fell with uniform acceleration – an acceleration that is proportional to the square of the elapsed time. Nor was he the first to find a time-squared relationship. The point of all this is that science doesn’t proceed via revolutions proceeding from one brilliant person (which shouldn’t diminish Galileo or Newton’s genius). The more you find out about it, the more incremental and fascinatingly collaborative and confirmative over time it is.
Galileo used the geometry of his time to present his time squared law, but algebraic notation, invented principally by Descartes, superseded this approach in the seventeenth century.
What about the gravitational constant? This appears to be a long and complicated mathematical story. I think it tries to answer the question – why do objects fall to Earth at such and such a rate of acceleration? But I’m not sure. The rate of acceleration would have been easy enough to measure – it’s approximately 9.8 m/sec2. This rate would appear to be caused by the mass of the Earth. The Moon has a fraction of Earth’s mass, and I believe the gravitational force it exerts is approximately one sixth that of Earth. It has been measured as 1.62 m/sec² (for Mars it’s 3.71).
It’s frustratingly difficult to get an explanation online of what the gravitational constant (G) is or really means – without very quickly getting into complex (for me) mathematics. Tantalisingly, Wikipedia tells us that the aforementioned Grimaldi and Riccioli ‘made [an attempted] calculation of the gravitational constant by recording the oscillations of a pendulum’, which means nothing to me. Clearly though, there must be some relationship between G and the mass of the Earth, though how this can be ascertained via pendulums is beyond me. Anyway, on with the struggle.
We do have a number for G, or ‘Big G’, as it’s called (explanation to come), and it’s a very very small number, indicating that, considering that the multiplied masses divided by the square of the distance between them then get multiplied by G, gravitation is mostly a very small force, and only comes into play when we’re talking about Big Stuff, like stars and planets, and presumably whole galaxies. Anyway here’s the actual number:
G = 0.0000000000667408, or 6.67408 × 10-11
I got the number from this useful video, though of course it’s easily available on the net. Now, my guess is that this ‘Big G’ is specific to the mass of the Earth, whereas small g is variable depending on which mass you’re referring to. In other words, G is one of the set of numbers in g. We’ll see if that’s true.
Now, looking again at the original equation, F stands for force, measured in newtons, m for mass, measured in kilograms, and and r for distance in metres (these are the SI units for mass and distance). The above-mentioned video ‘explains’ that the newtons on one side of the equation are not equivalent to the metres and kilograms squared on the other side, and G is introduced to somehow get newtons onto both sides of the equation. This has thrown me into confusion again. The video goes on to explain how G was used by Einstein in relativity and by Max Planck to calculate the Planck length (the smallest possible measure of length). Eek, I’m hoping I’m just experiencing the storm before the calm of comprehension.
So, to persist. This G value above isn’t, and apparently cannot be, precise. That number is ‘the average of the upper and lower limit’, so it has an uncertainty of plus or minus 0.00031 x 10-11, which is apparently a seriously high level of uncertainty for physicists. The reason for this uncertainty, apparently, is that gravitational attraction is everywhere, existing between every particle of mass, so there’s a signal/noise problem in trying to isolate any two particles from all the others. It also can’t be calculated precisely through indirect relation to the other forces (electromagnetism, the strong nuclear force and the weak nuclear force), because no relationship, or compatibility, has been found between gravity and those other three forces.
The video ends frustratingly, but providing me with a touch of enlightenment. G is described as a ‘fundamental value’, which means we don’t know why it has the value it does. It is just a value ‘found experimentally’. This at least tells me it has nothing to do with the mass of the Earth, and I was quite wrong about Big G and small g – it’s the other way round, which makes sense, Big G being the universal gravitational constant, small g pertaining to the Earth’s gravitational force-field.
Newton himself didn’t try to measure G, but this quote from Wikipedia is sort of informative:
In the Principia, Newton considered the possibility of measuring gravity’s strength by measuring the deflection of a pendulum in the vicinity of a large hill, but thought that the effect would be too small to be measurable. Nevertheless, he estimated the order of magnitude of the constant when he surmised that “the mean density of the earth might be five or six times as great as the density of water”
Pendulums again. I don’t quite get it, but the reference to the density of the Earth, which of course relates to its mass, means that the mass of the Earth comes back into question when considering this constant. The struggle continues.
I’ll finish by considering a famous experiment conducted in 1798 by arguably the most eccentric scientist in history, the brilliant Henry Cavendish (hugely admired, by the way, by Maxwell). I’m hoping it will further enlighten me. For Cavendish’s eccentricities, go to any online biography, but I’ll just focus here on the experiment. First, here’s a simplification of Newton’s law: F = GMm/R2, in which M is the larger mass (e.g. the Earth), and m the smaller mass, e.g a person. What Cavendish was trying to ascertain was nothing less than the mass and density of the Earth. In doing so, he came very close – within 1% – of the value for G. Essentially, all that has followed are minor adjustments to that value.
The essential item in Cavendish’s experiment was a torsion balance, a wooden bar suspended horizontally at its centre by a wire or length of fibre. The experimental design was that of a colleague, John Michell, who died before carrying out the experiment. Two small lead balls were suspended, one from each end of the bar. Two larger lead balls were suspended separately at a specific distance – about 23cms – from the smaller balls. The idea was to measure the faint gravitational attraction between the smaller balls and the larger ones.
Wikipedia does a far better job than I could in explaining the process:
The two large balls were positioned on alternate sides of the horizontal wooden arm of the balance. Their mutual attraction to the small balls caused the arm to rotate, twisting the wire supporting the arm. The arm stopped rotating when it reached an angle where the twisting force of the wire balanced the combined gravitational force of attraction between the large and small lead spheres. By measuring the angle of the rod and knowing the twisting force (torque) of the wire for a given angle, Cavendish was able to determine the force between the pairs of masses. Since the gravitational force of the Earth on the small ball could be measured directly by weighing it, the ratio of the two forces allowed the density of the Earth to be calculated, using Newton’s law of gravitation.
To fully understand this, I’d have to understand more about torque, and how it’s measured. Clearly this weak interaction is too small to be measured directly – the key is in the torque. Unfortunately I’m still a way from fully comprehending this experiment, and so much else, but I will persist.
References
https://en.wikipedia.org/wiki/Newton’s_law_of_universal_gravitation
https://en.wikipedia.org/wiki/Gravitational_constant
https://energyeducation.ca/encyclopedia/Gravitational_constant
https://en.wikipedia.org/wiki/Cavendish_experiment
https://www.school-for-champions.com/science/gravitation_cavendish_experiment.htm#.XSrCrS3L1QI
Go to youtube for a number of useful videos on the gravitational constant
towards James Clerk Maxwell 4: a detour into dimensional analysis and Newton’s laws
Canto: Getting back to J C Maxwell, I’m trying to learn some basic physics, which may or may not be relevant to electromagnetism, but which may help me to get in the zone, so to speak.
Jacinta: Yes, we’re both trying to brush up on physics terms and calculations. For example, acceleration is change in velocity over time, which is hard to put in notation form in a blog post, but I can steal it from elsewhere
in which the triangle represents ‘change in’. Now velocity is a vector quantity, therefore so is acceleration – it’s a particular magnitude in a particular direction. So imagine a car that goes from stationary to, say 50 kms/hour in 5 seconds, what’s the acceleration? According to the formula, it’s 50 – 0 kph/5 seconds, or 10kph/sec, which we can write out as a change of velocity of ten kilometres per hour per second.
Canto: So every second, the velocity of the car is increasing by 10 kilometres per hour. I’m trying to picture that. It’s quite hard.
Jacinta: Okay while you’re doing that, let’s introduce dimensional analysis, so that we reduce everything to the same dimension, sort of. I mean, we have hours and seconds here, so let’s take it all to seconds. I won’t be able to do this properly without an equation-writing plug-in, which I can’t work out how to get without paying. Anyhow..
10 kms/hour.second.1/3600 hour/second. Cancelling out the hours, you get 10 kms/3600 seconds squared, or 1/360 km/s2
Canto: I wonder if there’s a way of hand-writing equations in the blog, that’d be more fun and easy. So can you briefly explain dimensional analysis?
Jacinta: Well physical quantities are often measured in different units – for example, quantities of time – time is called the base quantity – are measured in seconds, hours, days etc. So, it’s just a matter of getting such measurements to be commensurate, so that an equation can be simplified – all in seconds, or all in metres, when they can be. Though actually it’s more complicated than that, and I’ve probably got it wrong.
Canto: So talking of brushing up on stuff, or actually knowing about stuff for the first time, I thought it might be good to go back to Newton, his three laws of motion, in written and mathematical form.
Jacinta: Go ahead.
Canto: Well, the first law, which really comes from Galileo, is often called the law of inertia. Newton formulated it this way, in the Principia (translated from Latin):
Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.
And as Sal Khan and others point out, Newton is talking about an unbalanced force, one that isn’t matched by an equal and opposite force (which would be a balanced force – see Newton’s third law). This law doesn’t come with a mathematical formula.
The second law, which I filched from The Physics Classroom, can be stated thus:
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
It’s famous formula is this:
Fnet = m • a
It can be written different ways, for example simply F =m.a, or with the vector sign (an arrow) above force (F) and acceleration (a), showing the same direction, but it’s certainly important to explain net force here. It’s essentially the sum of all the forces acting on the mass, in vector or directional terms. It’s this net force that produces the acceleration.
So to the third law, and this is how Newton presented it, again translated from Latin:
To every action there is always an equal and opposite reaction: or the forces of two bodies are always equal and are directed in opposite directions.
It’s often stated in this ‘wise proverb’ sort of way: ‘for every action there’s an equal and opposite reaction’.
Jacinta: What goes around comes around.
Canto: That’s more of a wise-guy thing. Anyway, the best formula for the third law is:
FA = −FB
where force A is the action and force B the reaction. This law is sort of counter-intuitive and also sort of obvious at the same time! I think it’s the most brilliant law. Sal Khan gives a nice extra-terrestrial example of how you might utilise it. Imagine you’re in outer space and you’ve been cut off from your spaceship and are accelerating away from it. To save yourself, take something massive, if you can, something on your suit or a tool you’re carrying, and push it hard away from you in the opposite direction to the ship, and this should send you accelerating back to the ship. But make sure your aim is true!
Jacinta: Okay, so this seems to have taken us absolutely no closer to Maxwell’s equations.
Canto: Well, yes and no. It makes us think of forces and energy, albeit of a different kind, and it makes us think in a logical, semi-mathematical way. but we’ve certainly got a long way to go…
References
https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
https://www.physicsclassroom.com/class/newtlaws/Lesson-1/Newton-s-First-Law
https://www.livescience.com/46558-laws-of-motion.html
https://www.physicsclassroom.com/class/newtlaws/Lesson-3/Newton-s-Second-Law