- #1
dedaNoe
- 51
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Newton must respect the law of lever as we all must.
The law of lever is the 1st law ever in physics and every thing
must agree with it.
\documentclass{article}
\begin{document}
\section{Newton must respect the law of lever}
\hspace{1 cm}I know most of you will object me as I'm the first
with such approach. But, I do have a point so please let me make
it. Assume that body A and body B have masses $M_a$ and $M_b$,
such that $M_a>M_b$. They are initially at distance R and without
initial push they start to fall on each other only because their
mutual gravity pull. We view only the part of the interaction from
the release until the collision. The gravity force for each body
according to Newton is given with:
\begin{equation}
F=G\frac{M_a M_b}{R^2}
\end{equation}
Newton's gravity law respects his third law, so:
\begin{equation}\label{Newton3}
F_a=-F_b
\end{equation}
Newton's second law claims that force equals to mass times
acceleration so \ref{Newton3} will turn into:
\begin{equation}
a_a M_a = -a_b M_b
\end{equation}
This means that $a_a < a_b$ because $M_a > M_b$ so within same
time body A will pass smaller distance from body B i.e.
\begin{equation}\label{condit}
\Delta X_a<\Delta X_b
\end{equation}
The work done by some force while making displacement is given
with:
\begin{equation}\label{Work_done}
W=\int_{X1}^{X2} Fdx
\end{equation}
In our system we have only two forces making displacement - only
two works done and they must cancel because of conservation of
energy for such a closed system given with:
\begin{equation}\label{Cosevr_energy}
\sum W_i=0
\end{equation}
From \ref{Cosevr_energy} we have $W_a=-W_b$. Widen up it should
give:
\begin{center}
$\int_{X_{1a}}^{X_{2a}}F_a dx=-\int_{X_{1b}}^{X_{2b}}F_b dx$
$\int_{X_{1a}}^{X_{2a}}Fdx=\int_{X_{1b}}^{X_{2b}}Fdx$
$F(X_{2a}-X_{1a})=F(X_{2b}-X_{1b})$
\end{center}
So it must be:
\begin{equation}\label{must}
F\Delta X_a=F\Delta X_b
\end{equation}
But it is not so because of \ref{condit}. By the way the equation
\ref{must} is very similar to the law of lever only though the
forces must have inverse the ratio of their distances or:
\begin{equation}\label{Lever}
F_a D_a = D_b F_b
\end{equation}
Therefore, the conflict between Newton's gravity and conservation
of energy is due to Newton's disobedience for the law of lever.
\end{document}
tex doesn't work that well. Just copy pase into your TeX editor
The law of lever is the 1st law ever in physics and every thing
must agree with it.
\documentclass{article}
\begin{document}
\section{Newton must respect the law of lever}
\hspace{1 cm}I know most of you will object me as I'm the first
with such approach. But, I do have a point so please let me make
it. Assume that body A and body B have masses $M_a$ and $M_b$,
such that $M_a>M_b$. They are initially at distance R and without
initial push they start to fall on each other only because their
mutual gravity pull. We view only the part of the interaction from
the release until the collision. The gravity force for each body
according to Newton is given with:
\begin{equation}
F=G\frac{M_a M_b}{R^2}
\end{equation}
Newton's gravity law respects his third law, so:
\begin{equation}\label{Newton3}
F_a=-F_b
\end{equation}
Newton's second law claims that force equals to mass times
acceleration so \ref{Newton3} will turn into:
\begin{equation}
a_a M_a = -a_b M_b
\end{equation}
This means that $a_a < a_b$ because $M_a > M_b$ so within same
time body A will pass smaller distance from body B i.e.
\begin{equation}\label{condit}
\Delta X_a<\Delta X_b
\end{equation}
The work done by some force while making displacement is given
with:
\begin{equation}\label{Work_done}
W=\int_{X1}^{X2} Fdx
\end{equation}
In our system we have only two forces making displacement - only
two works done and they must cancel because of conservation of
energy for such a closed system given with:
\begin{equation}\label{Cosevr_energy}
\sum W_i=0
\end{equation}
From \ref{Cosevr_energy} we have $W_a=-W_b$. Widen up it should
give:
\begin{center}
$\int_{X_{1a}}^{X_{2a}}F_a dx=-\int_{X_{1b}}^{X_{2b}}F_b dx$
$\int_{X_{1a}}^{X_{2a}}Fdx=\int_{X_{1b}}^{X_{2b}}Fdx$
$F(X_{2a}-X_{1a})=F(X_{2b}-X_{1b})$
\end{center}
So it must be:
\begin{equation}\label{must}
F\Delta X_a=F\Delta X_b
\end{equation}
But it is not so because of \ref{condit}. By the way the equation
\ref{must} is very similar to the law of lever only though the
forces must have inverse the ratio of their distances or:
\begin{equation}\label{Lever}
F_a D_a = D_b F_b
\end{equation}
Therefore, the conflict between Newton's gravity and conservation
of energy is due to Newton's disobedience for the law of lever.
\end{document}
tex doesn't work that well. Just copy pase into your TeX editor
