martillo 1,831 Posted August 11, 2019 Report Share Posted August 11, 2019 (edited) This is an extract of "Appendix A" in my manuscript on a new theory which can be accessed through the link in my profile if someone would be interested.The real Equation of Force is F = ma Today's Physics is stating that the Equation of Force is F = dp/dt. We will analyze the equation of motion of rockets to see that the real Equation of Force is: F = ma A rocket has variable mass in its trajectory and it's important to see its motion equation. Let m be its variable mass at any instant in its movement composed by the mass of the rocket plus the mass of its contained fuel. I have made a search in the internet about rocket motion equations and all the sites agree in the equation: F = m(dv/dt) = –ve(dm/dt) where ve is the speed of the fuel expelled relative to the rocket. One web site:http://www.braeunig.us/space/propuls.htm They all agree that the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt) and that the equation of motion is F = m(dv/dt) = ma. I assume the equation have been completely verified experimentally with enough precision from a long time ago. It is evident that it is used the equation: F = ma for the force and not: F = dp/dt ... ... … This indicates that today's Physics is wrong stating the Equation of Force as F = dp/dt. The right equation for force is F = ma even when mass varies. Note that the natural derivation of the famous equation E = mc2 by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt (http://www.emc2-explained.info/Emc2/Deriving.htm#.XVBNlvZFyM8). Relativity Theory becomes a wrong theory since it is based on a wrong law. Edited August 11, 2019 by martillo Quote Link to post Share on other sites

martillo 1,831 Posted August 11, 2019 Author Report Share Posted August 11, 2019 It must be noted that by definition: p = mv Then: dp/dt = mdv/dt + vdm/dt = ma + vdm/dt While F = ma (demonstrated by rockets' dynamics) dp/dt = F + vdm/dt It follows that F = dp/dt only if dm/dt = 0 meaning constant mass only Quote Link to post Share on other sites

Flummoxed 220 Posted August 17, 2019 Report Share Posted August 17, 2019 F = ma, hum so what happens when acceleration changes F/a, does your mass suddenly change or perhaps are you accelerating an inertia. Your hypothesis is wrong, very wrong !!. Quote Link to post Share on other sites

Dubbelosix 139 Posted August 26, 2019 Report Share Posted August 26, 2019 This is an extract of "Appendix A" in my manuscript on a new theory which can be accessed through the link in my profile if someone would be interested.The real Equation of Force is F = ma Today's Physics is stating that the Equation of Force is F = dp/dt. We will analyze the equation of motion of rockets to see that the real Equation of Force is: F = ma A rocket has variable mass in its trajectory and it's important to see its motion equation. Let m be its variable mass at any instant in its movement composed by the mass of the rocket plus the mass of its contained fuel. I have made a search in the internet about rocket motion equations and all the sites agree in the equation: F = m(dv/dt) = –ve(dm/dt) where ve is the speed of the fuel expelled relative to the rocket. One web site:http://www.braeunig.us/space/propuls.htm They all agree that the force acting on the rocket is due to the expelled mass and is F = –ve(dm/dt) and that the equation of motion is F = m(dv/dt) = ma. I assume the equation have been completely verified experimentally with enough precision from a long time ago. It is evident that it is used the equation: F = ma for the force and not: F = dp/dt ... ... … This indicates that today's Physics is wrong stating the Equation of Force as F = dp/dt. The right equation for force is F = ma even when mass varies. Note that the natural derivation of the famous equation E = mc2 by Relativity Theory has no sense since it is based in the wrong relation F = dp/dt (http://www.emc2-explained.info/Emc2/Deriving.htm#.XVBNlvZFyM8). Relativity Theory becomes a wrong theory since it is based on a wrong law. Why not use [math]\frac{dp}{dt}[/math]? I've used the same relativistic description for force a number of times, most widely time I used it was for the new luminosity calculations of moving black holes. Quote Link to post Share on other sites

isaac 1 Posted April 11 Report Share Posted April 11 Dear friends, you must repeat elementary mathematics and physics because: F = m a = m dv/dt = d(mv)/dt = dp/dt, so F = m a, and F = dp/dt are one and the same equation !!! Quote Link to post Share on other sites

exchemist 730 Posted April 11 Report Share Posted April 11 Dear friends, you must repeat elementary mathematics and physics because:F = m a = m dv/dt = d(mv)/dt = dp/dt, soF = m a, and F = dp/dt are one and the same equation !!! Not always. In special relativity, p = γmv, where γ = 1/√(1 - (v²/c²) ). But F = dp/dt is always true, whether in Newtonian or relativistic situations. So, dear friend, it is you that needs to learn a bit more. Quote Link to post Share on other sites

isaac 1 Posted April 12 Report Share Posted April 12 Dear friend, The basic principle of Relativity is that in the Universe, the same laws of physics always apply, no matter what inertial system we observe! Therefore, there are no "relativistic" or "Newtonian" situations in the Universe. The Special theory of relativity is much simpler than you think! If you, for (y), have "relativistically" corrected the moment in the numerator then for the same amount (y) you had to correct the time in the denominator, and again it will be: F = m a = m dv/dt = d(mv)/dt = ydp/ydt = dp/dt So don't waste your precious time on charlatan theories of "science popularizers" and/or fake "physics professors"! Life is short! Be healthy! Quote Link to post Share on other sites

Alon 10 Posted April 14 Report Share Posted April 14 Dear friends, you must repeat elementary mathematics and physics because:F = m a = m dv/dt = d(mv)/dt = dp/dt, soF = m a, and F = dp/dt are one and the same equation !!! No.d(mv)/dt = dm/dt v + mdv/dt if you assume that dm/dt=0 then they are the same, but otherwise they aren't the same. Quote Link to post Share on other sites

isaac 1 Posted April 17 Report Share Posted April 17 Dear Alon, You would certainly be right if this was a post about math and the derivation of the product of two independent functions! But this is a post about Physics! In Physics, mass and velocity are not two independent functions, but the opposite! Mass and velocity are the parameters (variables) of one unique function, which we call Moment! That's why the Force is just a simple derivation of Moment by Time! But suppose your derivation is "physically correct"? So what in Physics would be the first member of your equation and what would you call it? And how much would that member be, in most cases in Physics, which is m = constant? (well, you've already answered that!) But again, if we obey Newton and his "Law of Force" which states that each "force of action" is opposed by a symmetric "reaction force" (F_{a} = F_{r}), then it means that each Force, indeed, has two identical members F = F_{a} + F_{r} , so due to F_{r} = F_{a}, the Force is F = 2F_{a}! Maybe this is this "one and the same formula" that we are discussing here: F = 2 m a = 2 m d^{2}x/dt^{2} = 2 m dv/dt = 2 d(mv)/dt = 2 dp/dt !!! Alon, thank you so much for your interesting remark! Maybe someone learns something from this! Quote Link to post Share on other sites

maheshkhati 5 Posted April 30 Report Share Posted April 30 Not always. In special relativity, p = γmv, where γ = 1/√(1 - (v²/c²) ). But F = dp/dt is always true, whether in Newtonian or relativistic situations. So, dear friend, it is you that needs to learn a bit more. This is true. Quote Link to post Share on other sites

martillo 1,831 Posted June 23 Author Report Share Posted June 23 No.d(mv)/dt = dm/dt v + mdv/dt if you assume that dm/dt=0 then they are the same, but otherwise they aren't the same.Seems you got the point… Quote Link to post Share on other sites

martillo 1,831 Posted June 23 Author Report Share Posted June 23 (edited) I took a look at my profile…I wonder how someone with just 152 posts can get the bad negative -1831 likes!Amazing!!! What else could be expected from the mainstream to someone proposing, not a new theory, but a totally new Physics debunking Relativity, Quantum Physics and even the Standard Model of basic particles at once, isn't it?I understand… :cool:Now I'll wait for the shower of silly comments about this… Edited June 23 by martillo Quote Link to post Share on other sites

hazelm 109 Posted June 23 Report Share Posted June 23 F = ma, hum so what happens when acceleration changes F/a, does your mass suddenly change or perhaps are you accelerating an inertia. Your hypothesis is wrong, very wrong !!. Doesn't it, Flummoxed? Doesn't acceleration make a mass lighter? Or does it only seem to make it lighter? Once it gets going, it goes with less energy behind it. Hmmm??? Quote Link to post Share on other sites

Flummoxed 220 Posted June 24 Report Share Posted June 24 Doesn't it, Flummoxed? Doesn't acceleration make a mass lighter? Or does it only seem to make it lighter? Once it gets going, it goes with less energy behind it. Hmmm??? Er yes Mass is related to acceleration and gravity. However it is the inertia of the object that resists the change in movement. Inertia is proportional to mass when accelerated. In space with no gravity, it is the inertia of the object that requires force to change its movement, not the objects mass. A mass only becomes apparent when an object is accelerated, and as the acceleration increases the apparent mass increases, the inertia remains the same. Quote Link to post Share on other sites

hazelm 109 Posted June 24 Report Share Posted June 24 Er yes Mass is related to acceleration and gravity. However it is the inertia of the object that resists the change in movement. Inertia is proportional to mass when accelerated. In space with no gravity, it is the inertia of the object that requires force to change its movement, not the objects mass. A mass only becomes apparent when an object is accelerated, and as the acceleration increases the apparent mass increases, the inertia remains the same. All right. I should not have jumped in here when I did but that underscored part (my underscoring) is confusing me. I'll back out and think on it. Sorry for the interruption. Carry on. Quote Link to post Share on other sites

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