太阳的质量Ms没有机会进入太阳系的力学方程
Free 王
2022 8 26
首先，援引一段教材裏有关重力（引力）场的论述。
A gravitational field surrounds all material objects.The magnitude of the force produced upon a test object placed in the fielddefines the gravitational mass of the test object (being proportional to thismass). The term gravitational field is reserved for the force per unit testmass; however，gravitational force is the actualforce produced by the field on the test object, whether of unit mass or not. Themagnitude of the force depends also on the reciprocal of the square of distancefrom the initial mass which gives rise to the field (called the “source”) tothe test mass, and the direction of the force is toward the source. Although itis true that the force is a mutual interaction, in use we generally avoid thepresumption that the Earth be pulled toward a stone by the stone’s field, butrather the converse, or toward the barycenter. On the other hand, if the stonewere sufficiently large, it could distort the Earth’s field,
The study ofparticle motion in a central field readily lends itself to plane polar coordinates,because of the symmetry of the field. to locate the position of the test mass
in space. wedefine a vector, the position vector,that originates at the origin of coordinates the Sunand terminates at the test mass a planet:
r = rrₑ
r= position vector
r =length of r
rₑ= unitpolar vector
ᵠₑ=unit vector of angular motion
Velocityand acceleration in plane polar coordinates
The expressionfor velocity
V=d(rrₑ) /d t=rₑ dr/dt + r drₑ/dt
The expression foracceleration
a = dv/dt =d[ rₑdr/dt + r drₑ/dt]/ dt
= [rₑd²r/ dt² + dr/dtdrₑ/dt]+
[d r/dt drₑ/dt+ r d(drₑ/dt)/ dt]
The velocity vector in plane polarcoordinates is composed of two components: a ra-
dial component, rₑdr/dt,which denotes the planet’s changingdistance from the Sun
as it moves intime; and an angular component, ᵠₑr ᵠ/, which denotes the angu-
lar motion ofthe planet about the Sun, so ᵠₑ rdᵠ/dt =rdrₑ/dt
So that v= rₑ dr/dt+ ᵠₑ rdᵠ/dt (3)
The expression for acceleration
,
a= dv/ dt = [rₑd²r/ dt²+ dr/dt drₑ/dt]+[ᵠₑd(rdᵠ/ dt)/ dt +( rdᵠ/dt ) dᵠₑ/ dt]
Combining terms, we obtain
a = [d²r/ dt² -- r(dᵠ/dt)² ] rₑ +[rd²ᵠ/dt²+2 dr/dt dᵠ/dt] ᵠₑ （4）
The both sides ofEq.1 multiplied respectively by two, and differentiate it with time,
r² d²ᵠ/dt² + 2r (dr/ dt) dᵠ/dt=r( rd²ᵠ/dt²+2 dr/dtdᵠ/dt）=0 ( 5)
Inserting Eq.5 into the second bracket ofEq.4, we have
a = [d²r/ dt² -- r(dᵠ/dt)² ] rₑ （这是行星指向太阳的径向加速度）
就到此吧，再往下有图，网页贴不上。我写了这么多，就是为了说明太阳的质量Ms没有机会进入方程。牛顿的万有引力定律方程F= G MsMp /r²中的Ms是通过一句话的掩护偷换概念混进去的。就是在伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²的後边，通过一句“按牛顿第三定律，既然这个力与行星的质量成正比，那这个力也一定与太阳的质量成正比。”随着这句话，把太阳质量Ms混进伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²裏。他的这句“按牛顿第三定律，既然这个力与行星的质量成正比，那这个力也一定与太阳的质量成正比。”说的没有错。因为：按牛顿第三定律，作用力与反作用力大小相等，所以，这个力无论与等号左边的质量还是右边的质量都成正比，但条件是，这个质量必须跟自己的加速度贴到等号的同一侧，成正比才成立，否则没有任何意义。事实上，伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²裏，根本就没有太阳Ms的加速度，牛顿只是以误导作掩护。牛顿两度偷换概念：一个是通过质点语义上的双关，把被看作是质点的天体切换成了真正的质点小不点的单质分子，所以，才有了“所有的物质总是有着同一个固定的引力常数G”；一个是概念上模棱两可的误导，实施以假乱真。目的只是为了万有的引力常数。
Free 王
2022 8 26
首先，援引一段教材裏有关重力（引力）场的论述。
A gravitational field surrounds all material objects.The magnitude of the force produced upon a test object placed in the fielddefines the gravitational mass of the test object (being proportional to thismass). The term gravitational field is reserved for the force per unit testmass; however，gravitational force is the actualforce produced by the field on the test object, whether of unit mass or not. Themagnitude of the force depends also on the reciprocal of the square of distancefrom the initial mass which gives rise to the field (called the “source”) tothe test mass, and the direction of the force is toward the source. Although itis true that the force is a mutual interaction, in use we generally avoid thepresumption that the Earth be pulled toward a stone by the stone’s field, butrather the converse, or toward the barycenter. On the other hand, if the stonewere sufficiently large, it could distort the Earth’s field,
The study ofparticle motion in a central field readily lends itself to plane polar coordinates,because of the symmetry of the field. to locate the position of the test mass
in space. wedefine a vector, the position vector,that originates at the origin of coordinates the Sunand terminates at the test mass a planet:
r = rrₑ
r= position vector
r =length of r
rₑ= unitpolar vector
ᵠₑ=unit vector of angular motion
Velocityand acceleration in plane polar coordinates
The expressionfor velocity
V=d(rrₑ) /d t=rₑ dr/dt + r drₑ/dt
The expression foracceleration
a = dv/dt =d[ rₑdr/dt + r drₑ/dt]/ dt
= [rₑd²r/ dt² + dr/dtdrₑ/dt]+
[d r/dt drₑ/dt+ r d(drₑ/dt)/ dt]
The velocity vector in plane polarcoordinates is composed of two components: a ra-
dial component, rₑdr/dt,which denotes the planet’s changingdistance from the Sun
as it moves intime; and an angular component, ᵠₑr ᵠ/, which denotes the angu-
lar motion ofthe planet about the Sun, so ᵠₑ rdᵠ/dt =rdrₑ/dt
So that v= rₑ dr/dt+ ᵠₑ rdᵠ/dt (3)
The expression for acceleration
,
a= dv/ dt = [rₑd²r/ dt²+ dr/dt drₑ/dt]+[ᵠₑd(rdᵠ/ dt)/ dt +( rdᵠ/dt ) dᵠₑ/ dt]
Combining terms, we obtain
a = [d²r/ dt² -- r(dᵠ/dt)² ] rₑ +[rd²ᵠ/dt²+2 dr/dt dᵠ/dt] ᵠₑ （4）
The both sides ofEq.1 multiplied respectively by two, and differentiate it with time,
r² d²ᵠ/dt² + 2r (dr/ dt) dᵠ/dt=r( rd²ᵠ/dt²+2 dr/dtdᵠ/dt）=0 ( 5)
Inserting Eq.5 into the second bracket ofEq.4, we have
a = [d²r/ dt² -- r(dᵠ/dt)² ] rₑ （这是行星指向太阳的径向加速度）
就到此吧，再往下有图，网页贴不上。我写了这么多，就是为了说明太阳的质量Ms没有机会进入方程。牛顿的万有引力定律方程F= G MsMp /r²中的Ms是通过一句话的掩护偷换概念混进去的。就是在伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²的後边，通过一句“按牛顿第三定律，既然这个力与行星的质量成正比，那这个力也一定与太阳的质量成正比。”随着这句话，把太阳质量Ms混进伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²裏。他的这句“按牛顿第三定律，既然这个力与行星的质量成正比，那这个力也一定与太阳的质量成正比。”说的没有错。因为：按牛顿第三定律，作用力与反作用力大小相等，所以，这个力无论与等号左边的质量还是右边的质量都成正比，但条件是，这个质量必须跟自己的加速度贴到等号的同一侧，成正比才成立，否则没有任何意义。事实上，伯尔曼方程F=4Mp α²/ a(1- ϵ²) r²裏，根本就没有太阳Ms的加速度，牛顿只是以误导作掩护。牛顿两度偷换概念：一个是通过质点语义上的双关，把被看作是质点的天体切换成了真正的质点小不点的单质分子，所以，才有了“所有的物质总是有着同一个固定的引力常数G”；一个是概念上模棱两可的误导，实施以假乱真。目的只是为了万有的引力常数。
