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Are the 'Laws of Physics' Wrong?
A Possible Problem with the Equations of Classical Physics Relating to Work and Energy.
Author: Greg Alexander
16th June 2003.
Consider a rocket travelling in the frictionless environment of space. According to the equation for kinetic energy, K.E. = ½ mv², when the rocket doubles its velocity (by increasing v to 2v for example), four times as much kinetic energy would be obtained. However considering that the law of conservation of energy must always apply would it really take four times the amount of fuel to increase the velocity from v to 2v as it took to increase the velocity from 0 to v? Surely the energy input supplied by the burning fuel would at all times equal the kinetic energy obtained by the space rocket? Such a situation would surely suggest that the formula would actually be K.E. = mv, where the relation between kinetic energy and velocity is a linear one.
Such an equation simply states that to increase your velocity by a set amount in the frictionless environment of space the same proportion of energy would be required each time no matter what speed you were travelling at to begin with. The law of conservation would clearly dictate that the energy supplied to the craft by its power source would at all times equal the resulting kinetic energy, taking into account the various efficiency ratios involved of course. If the kinetic energy did equal ½ mv² then this would result in increasingly larger amounts of fuel being needed to accelerate the craft which in turn would produce a limiting velocity requiring its own separate equation.
Of course we have all been taught since day one that it is actually momentum which equals mv and not kinetic energy. But is it really that straight forward? For example is momentum a force or is it actually a form of energy? Indeed the units it is measured in are rather peculiar, for example kilograms per metre per second. But which do these denote, force or energy?
Having discovered that there is a possible problem with the current equation for kinetic energy and that the exact nature of the physical property of momentum is open to question, it is now necessary to enquire where the underlying fault would appear to lie. In my opinion the best place to start is with the definition of the joule itself which we are told equals a force of one Newton applied through a distance of 1 metre. An objection which can immediately be raised to this apparently straight forward definition is that the work done by one Newton through one metre is surely dependant upon the mass it is acting upon? For example would it not take more work for one Newton to move a 2 kg mass through one metre than it would the same force to move 1 kg through the same distance? The unit mass is entirely absent from the definition.
To further illustrate my point, using F = ma and s = ½ at², 1 Newton acting on a 1 kg mass would take 1.41 seconds to reach one metre (i), whereas 1 Newton acting on a 2 kg mass would take 2 seconds to reach one metre (ii). Surely it would be the case that if one Newton were applied for a longer duration of time as with case (ii) more work would have been done as a consequence? However if the equations v = at and K.E. = ½ mv² are applied to the accelerations and velocities obtained in (i) and (ii), we do in fact end up with 1 Joule of energy in each case as is predicted. In fact this is the very basis of the definition.
But surely it has to be the case that 1 Newton applied for 2 seconds in case (ii) would have carried out an extra 40% more work than in case (i) where 1 Newton was only applied for 1.41 seconds? This point is surely indisputable? If it is then a large part of classical physics has to be in error and possibly as a consequence many other areas of the same science would also be undermined?
The above demonstration would clearly suggest therefore, that the formula Work Done = Force x Distance, would appear to be not at all valid. Indeed it is this equation that the definition of the Joule is based upon. Instead the correct equation would perhaps be Work Done = Force x Time. Since the formula for kinetic energy is also derived from the former apparently incorrect equation, it too must be unsound. Indeed it is this formula which is actually responsible for calculating the erroneous 1 Joule of work done for Cases (i) and (ii). Interestingly if it is instead assumed that K.E. is proportional to mv, then we obtain the energies of 1.41 Joules and 2 Joules respectively for cases (i) and (ii). This is entirely consistent with a 40% difference in work done between the two cases.
Indeed it would even appear possible to prove by integration that kinetic energy equals mv if it is assumed that Work Done = Force x Time, as follows: K.E. = ∫ F.dt = ∫ m dv/dt . dt = ∫ m dv = mv or alternatively : K.E. = Work Done = Force x Time = F x t = m a x t (and because v = at)= m v
Further evidence that would suggest that kinetic energy were equal to mv is the observation that when two objects of known masses collide, it is the conservation of momentum mv which would apply and not the conservation of kinetic energy according to the currently accepted equation which covers it, ½mv². Indeed the two cases produce quite different results : Consider two bowls each with a mass of 1.5 kg. The first bowl has a velocity of 2 m/sec and impacts at a slight angle with the second which is stationary. If the velocity of the first bowl after impact is 0.5 m/sec, the velocity of the second bowl based upon the conservation of momentum would be 1.5 m/sec (i.e. 2 – 0.5 = 1.5 m/sec). But if it were assumed that the conservation of kinetic energy applied instead, the velocity of the second bowl would be as follows : v = √(2² - 0.5²) = 1.94 m/sec. A different value altogether.
Since conservation would only occur experimentally where mv is used, perhaps it can be assumed therefore that the kinetic energy also equals mv?
There would also appear to be corresponding problems with the formulae covering potential energy. At short distances of elevation the equation V = mgr would apply but in the case of larger distances of elevation that of gravitational potential, V = -GM/r, would apply. The reason that is stated as to why two separate equations are needed is because V = mgr only works over short heights as a result of the fact that the value for g varies with the height r. However on a basic level the two equations are exactly the same anyway. If you take V = mgr and substitute into in g = GM/r², you end up with V = GM/r, which is the exact same equation apart from a missing negative sign. But again g is fixed at r and does not vary as r decreases leading one to suspect that it also would give erroneous answers over large distances just as would V = mgr.
However returning to V = mgr it would appear that in addition to this the equation, like that for kinetic energy, is also flawed for a quite separate reason. It has already been demonstrated that the equation Work Done = Force x Distance would not apply in the horizontal as a result of the relation between force and distance being non-linear with respect to an accelerating body. However it is also possible to demonstrate that much the same applies in the vertical. The primary cause of this is the relation s = ½ at² which in the case of the vertical would of course be s = ½ gt². The relation between force and distance in the vertical, gravitational potential, as has been stated previously, is said to be mgr. However at a height of 10 metres it would take an object, according to s = ½ gt², about 1.43 seconds to hit the ground, whereas at 20 metres it would take the same object around 2.02 seconds. The ratio between the two times is 1 : 1.41 or 1 : √2. It is apparent therefore that the ratio rather than being 1 : 2 with respect to time as you might have expected when the height is doubled, it is actually a square root relation with respect to time. This is quite evidently the result of the equation s = ½ gt².
But wouldn’t it be correct to assume that since a 1 kg mass, say, produces a force of about 10 Newtons, which is applied for 40% longer at 20 metres in height, actually does 40% more work as a consequence? If such is a logical deduction, and it is difficult to dispute this, then the equation V = mgr would not appear to apply as it would have predicted 100% more work done. Again the reason why the latter is not the case is because the relation between force and distance is non-linear as a result of the acceleration that the former produces.
If we introduce this relation into a new equation for potential energy, we obtain the following : Potential Energy = Force x Time = mg x √2r/g (from r = ½ gt²). And therefore P² = m²g² x 2r/g = m²g x 2r. Therefore P = m √2gr
This equation basically tells us that the potential energy stored in a height of r is far less than one might have expected. This is a result of the fact that the object moves with a velocity of v as a result of work already done which would cause it to hit the ground anyway even if it were possible to remove the force accelerating it.
It would therefore appear that in combination the equations for kinetic energy and potential energy only appear to work as a result of the fact that the v² relation in the formula for kinetic energy compensates for the lack of a square root relation with respect to height in the formula for potential energy.
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