yahya515 Posted March 18, 2015 Share Posted March 18, 2015 every system like the planets around the sun will result in centripetal force towards the centre, due to the combination of both the rotation around the centre and the rotation around the axis , in the planets we have extra centripetal force in addition to the gravitational force between the planet and the sun, the force is inversely proportional to the radius of the rotation ®, let say the mass is (m) the linear velocity around the sun is (V) the velocity around the axis is (v), then the force F will be:F= ( m*v*V)/rfor the earth :F= (5.9e24*30000*465.1)/149e9 = 5.5e20 newtonsthe gravitational force is 3.51 e22 newtons , the force due to motion is :F=mv^2/r = 3.56 e22adding my force to the gravitional force will be:5.5e20 + 3.51e22 = 3.56 e22 newtons !!so the force coming from my effect + the gravitional force = the force due to the motion (mv^2/r)I have tried an experiment with a ball hang by string the other side of the string is fixed to a rotating motor, when the motor moves the string moves around to move the ball in a rotaional motion , when moving the ball a round at a particular radius the system will have a centripetal force!! i.e the ball will rotate around its axis and around the radius continuously.the force is coming from the combination of the two spinning motions the one around the axis and the one around the circle, as a rsult that there is not work done for this force , i.e the radius is always constant , the force acting towards the centre without making any work , as a result of that the system creates the centripetal force without an external work done that why I call it self-consistent centripetal force. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2015 Share Posted March 18, 2015 Does this work with planetary orbits and gravity? Link to comment Share on other sites More sharing options...
Robittybob1 Posted March 18, 2015 Share Posted March 18, 2015 (edited) every system like the planets around the sun will result in centripetal force towards the centre, due to the combination of both the rotation around the centre and the rotation around the axis , in the planets we have extra centripetal force in addition to the gravitational force between the planet and the sun, the force is inversely proportional to the radius of the rotation ®, let say the mass is (m) the linear velocity around the sun is (V) the velocity around the axis is (v), then the force F will be: F= ( m*v*V)/r for the earth : F= (5.9e24*30000*465.1)/149e9 = 5.5e20 newtons the gravitational force is 3.51 e22 newtons , the force due to motion is : F=mv^2/r = 3.56 e22 adding my force to the gravitional force will be: 5.5e20 + 3.51e22 = 3.56 e22 newtons !! so the force coming from my effect + the gravitional force = the force due to the motion (mv^2/r) I have tried an experiment with a ball hang by string the other side of the string is fixed to a rotating motor, when the motor moves the string moves around to move the ball in a rotaional motion , when moving the ball a round at a particular radius the system will have a centripetal force!! i.e the ball will rotate around its axis and around the radius continuously. the force is coming from the combination of the two spinning motions the one around the axis and the one around the circle, as a rsult that there is not work done for this force , i.e the radius is always constant , the force acting towards the centre without making any work , as a result of that the system creates the centripetal force without an external work done that why I call it self-consistent centripetal force. I'm sure the Gravitational force should be equal to the Centripetal force so if you can add them together and get another force I'd say you've done something wrong. Could you show all the values you used please? Then I'll recheck your calculations. This would be true in the case of the circularized orbit of the central body. Edited March 18, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
studiot Posted March 18, 2015 Share Posted March 18, 2015 I think you misunderstand the nature of centripetal force. Centripetal force is a real force that has to be applied to a body travelling in a curve, by an external agent, if we can consider that body to be a point particle. Gravity is one such external agent and the gravitational force is centripetal in nature. A point particle means that all the mass of the body can be considered to act as though it was concentrated at one point. Centripetal forces act within a body turning on an axis, but not on the body as a whole. These forces act between the individual particle or parts of the body and are usually manifest as a variation in the cohesive forces holding the body together. There is a further complication in that the centripetal force at any point in the trajectory (which may not be a circle), is directed towards the centre of a circle, which has a common tangent to the trajectory curve at the point of interest. In general, when the trajectory is not a circle, this centre point will vary in position and the distance between the common tangent point and the centre (known as the radius of curvature) will vary. In order to account for this it is usual to work in terms of acceleration, not force, and call this centre the 'instantaneous centre of acceleration'. Working in this way you can add the accelerations (vectorially) on any part of an orbiting planet to generate the true path of that particle on the planet. The accelerations are, as you say, separately due to the centripetal effect of gravity acting on the planet as a whole combined with the internal accelerations due to spin, wobble and other movements, including variation of linear speed. Very tricky indeed. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 Does this work with planetary orbits and gravity? the force increases at small radius of the ellipse and decreases at large radius of the ellipse , however we should take in consideration the spinning speed, the spinning speed may increase to a value that we will not need gravity, the path will take all its centripetal force from spinning, if the speed of spinning equals the speed of circular motion we will not need gravity . if it is larger the object will move towards the centre to reduce its radius, it is a matter of experiment I did not invent that , as it is in my OP experiment , which is an evidence. I did the numbers for the earth in my OP that show how accurate it is !! Link to comment Share on other sites More sharing options...
xyzt Posted March 18, 2015 Share Posted March 18, 2015 in the planets we have extra centripetal force in addition to the gravitational force between the planet and the sun, the force is inversely proportional to the radius of the rotation ®, let say the mass is (m) the linear velocity around the sun is (V) the velocity around the axis is (v), then the force F will be: F= ( m*v*V)/r basic mechanics says that the above is wrong. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 (edited) basic mechanics says that the above is wrong. how ? explain. Edited March 18, 2015 by yahya515 Link to comment Share on other sites More sharing options...
Robittybob1 Posted March 18, 2015 Share Posted March 18, 2015 how ? explain. Could you make up a video of what you are doing and post it on YouTube? Link to comment Share on other sites More sharing options...
studiot Posted March 18, 2015 Share Posted March 18, 2015 (edited) yahya, Did you miss post#4 or are only certain members allowed to comment? Edited March 18, 2015 by studiot Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 Could you make up a video of what you are doing and post it on YouTube? good idea , I forgot to do that , but it may need a week or two , also anyone can do the experiment it is very simple. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2015 Share Posted March 18, 2015 the force increases at small radius of the ellipse and decreases at large radius of the ellipse , however we should take in consideration the spinning speed, the spinning speed may increase to a value that we will not need gravity, the path will take all its centripetal force from spinning, if the speed of spinning equals the speed of circular motion we will not need gravity . if it is larger the object will move towards the centre to reduce its radius, it is a matter of experiment I did not invent that , as it is in my OP experiment , which is an evidence. I did the numbers for the earth in my OP that show how accurate it is !! I don't believe your experiment. There would be no consistency in the laws of physics between the orbits of the moon and Venus, which have very little rotation, and planets like the earth and Mars, which have larger values. You can also trivially falsify this because a fast spinning object will not measurably change its weight reading on a scale. In your numbers for the earth you never compared the numbers for the gravitational force with mv2/r (spoiler: they're equal) Further, if you were right the force would vary according to latitude, because the linear speed varies. This has implications for local gravity and for tides, none of which you've addressed, and should also trivially falsify this claim. how ? explain. The number of ways it fails is too large to easily enumerate. Two equal-mass objects orbiting each other but spinning at different rates would not feel equal and opposite forces. Immediate fail. Link to comment Share on other sites More sharing options...
Robittybob1 Posted March 18, 2015 Share Posted March 18, 2015 (edited) good idea , I forgot to do that , but it may need a week or two , also anyone can do the experiment it is very simple. So are you starting with a ball hanging from a string, say from a drill chuck, and then you make the ball spin by turning on the motor? Edited March 18, 2015 by Robittybob1 Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 In your numbers for the earth you never compared the numbers for the gravitational force with mv2/r (spoiler: they're equal) Further, if you were right the force would vary according to latitude, because the linear speed varies. This has implications for local gravity and for tides, none of which you've addressed, and should also trivially falsify this claim. they are not equal , mv^2/r= 5.9e24*30000*30000/149e9 = 3.563x 10^22 newtons , the gravitational force is F = G M m / r^2 F = (6.674 x 10^-11 )(1.9891 x 10^30 )(5.9736 x 10^24 ) / (1.503 x 10^11 m)^2 F = 3.51 x 10^22 N the weight will not be affected, moving from equator to poles, the object which we measure its weight is not spinning, there will not be any centripetal force it is the same as rotating a ring having a mass at some point , the object should be symmetric. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 So are you starting with a ball hanging from a string, say from a drill chuck, and then you make the ball spin by turning on the motor? a ball hanging with string in order to be free so that you move it around as it is spinning, it will take the mass a while until it rotate , because the motion will be transferred to the ball after a while , and then magic happens!! Link to comment Share on other sites More sharing options...
studiot Posted March 18, 2015 Share Posted March 18, 2015 and then magic happens!! Ah magic, I should have known it was not real science. Sorry I am in the wrong forum. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 I urge everyone in this forum to do the experiment in order to show the truth for the whole world, the time for new mechanics laws has come. Ah magic, I should have known it was not real science. Sorry I am in the wrong forum. if it is real science Newton would know about it , so prepare yourself for the magical physics era. Link to comment Share on other sites More sharing options...
Sensei Posted March 18, 2015 Share Posted March 18, 2015 F= (5.9e24*30000*465.1)/149e9 = 5.5e20 newtons I am confused. Why 465.1? This velocity depends on latitude, longitude, and altitude. For r=6371000m (average Earth radius) d=2*pi*6371000 v=d/t= 463.23947 m/s For r=6378137 m (equatorial Earth radius) d=2*pi*6378137 v=d/t= 463.8312 m/s etc. On Pole it'll be 0. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 I am confused. Why 465.1? This velocity depends on latitude, longitude, and altitude. For r=6371000m (average Earth radius) d=2*pi*6371000 v=d/t= 463.23947 m/s For r=6378137 m (equatorial Earth radius) d=2*pi*6378137 v=d/t= 463.8312 m/s etc. On Pole it'll be 0. more mass is in the equator or near. Link to comment Share on other sites More sharing options...
studiot Posted March 18, 2015 Share Posted March 18, 2015 yahya 5.5e20 + 3.51e22 = 3.56 e22 newtons !! At least show that you can correctly add two forces. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2015 Share Posted March 18, 2015 they are not equal , mv^2/r= 5.9e24*30000*30000/149e9 = 3.563x 10^22 newtons , the gravitational force is F = G M m / r^2 F = (6.674 x 10^-11 )(1.9891 x 10^30 )(5.9736 x 10^24 ) / (1.503 x 10^11 m)^2 F = 3.51 x 10^22 N the weight will not be affected, moving from equator to poles, the object which we measure its weight is not spinning, there will not be any centripetal force it is the same as rotating a ring having a mass at some point , the object should be symmetric. You have used different values for the mass of the earth and for the distance in your two calculations (emphasized in the quote). Considering that you are asserting there is a ~1.4% difference in the result, this is inexcusably sloppy. Just making the distances both be 149e6 km the gravitational answer changes to 3.557e22. You're arguing something based on poor uses of approximations and rounding. There's no physics here. Link to comment Share on other sites More sharing options...
yahya515 Posted March 18, 2015 Author Share Posted March 18, 2015 (edited) You have used different values for the mass of the earth and for the distance in your two calculations (emphasized in the quote). Considering that you are asserting there is a ~1.4% difference in the result, this is inexcusably sloppy. Just making the distances both be 149e6 km the gravitational answer changes to 3.557e22. You're arguing something based on poor uses of approximations and rounding. There's no physics here. you are mistaken .by changing just the quantity 149e24 in the gravitation equation the force will increase to just 3.52e22 leaving the mass 5.9e24 because it appears in both equations Edited March 18, 2015 by yahya515 Link to comment Share on other sites More sharing options...
xyzt Posted March 18, 2015 Share Posted March 18, 2015 how ? explain. i cannot teach you basic mechanics. based on your posts, no one can. Link to comment Share on other sites More sharing options...
Robittybob1 Posted March 18, 2015 Share Posted March 18, 2015 i cannot teach you basic mechanics. based on your posts, no one can. Yahya - try doing what Swansont says, use the figures with no rounding and see if there is any difference first. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2015 Share Posted March 18, 2015 you are mistaken .by changing just the quantity 149e24 in the gravitation equation the force will increase to just 3.52e22 leaving the mass 5.9e24 because it appears in both equations No. I didn't change the mass in the calculation, only r. That you used two different values for mass is a separate problem. It's the wrong value for r (average distance is 149.6 million km), so that may also explain why there is a slight discrepancy in the calculations. (Using the wrong value for r affects v) And we're ignoring perturbations. That gets the discrepancy down to .025, or less than 1%. That's roundoff error. (also less than half of your predicted effect) Regardless, your model fails at a very fundamental level. If it's right then all of Newtonian physics is wrong. And Newtonian physics is not wrong in this regime of calculation. Link to comment Share on other sites More sharing options...
CasualKilla Posted March 19, 2015 Share Posted March 19, 2015 (edited) I urge everyone in this forum to do the experiment in order to show the truth for the whole world, the time for new mechanics laws has come. if it is real science Newton would know about it , so prepare yourself for the magical physics era. Newton was a avid believer in Alchemy so i think he was open to 'magic' Edited March 19, 2015 by CasualKilla Link to comment Share on other sites More sharing options...
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