A Proportion Question

The following argumentation has been used in the good old days to teach us about the ideal gas laws of Boyle and Gay Lussac.
  1. Experiments reveal that the pressure $p$ is proportional to the amount of gas $n$
    if temperature $T$ and volume $V$ are held constant. Thus: $\, p \sim n$ .
  2. Experiments reveal that the pressure $p$ is inversely proportional to the volume $V$
    if temperature $T$ and amount of gas $n$ are held constant. Thus: $\, p \sim 1/V$ .
  3. Experiments reveal that the pressure $p$ is proportional to the temperature $T$
    if volume $V$ and amount of gas $n$ are held constant. Thus: $\, p \sim T$ .
  4. Experiments reveal that the volume $V$ is proportional to the temperature $T$
    if pressure $p$ and amount of gas $n$ are held constant. Thus: $\, V \sim T$ .
Then the book (*) tells us that from (2) it follows that $p_1\cdot V_1 = p_2\cdot V_2$ : Boyle's law.
A next step is to introduce absolute temperature $T$ (Kelvin) instead of relative temperature.
Then from (3) we have $p_1/T_1 = p_2/T_2$ : pressure law of Gay Lussac.
The book proceeds with (4) and the volume law of Gay Lussac : $V_1/T_1 = V_2/T_2$.
And then comes a clue. What happens if we change the volume $V$ as well as the temperature $T$ ?
The book says that we should do this in two subsequent steps:
  1. Keep the volume $V$ constant and raise the temperature $T$.
    Then we have according to Gay Lussac: $\;p_1/T_1 = p_2'/T_2 \quad \Longrightarrow \quad p_2' = p_1\cdot T_2/T_1$
  2. Keep the temperature $T$ constant and change the volume $V$.
    Then we have according to Boyle: $\;p_2'\cdot V_1 = p_2\cdot V_2 \quad \Longrightarrow \quad p_1\cdot T_2/T_1 \cdot V_1 = p_2\cdot V_2$
It follows that: $$ \frac{p_1 \cdot V_1}{T_1} = \frac{p_2 \cdot V_2}{T_2} \quad \Longrightarrow \quad p\cdot V = C\cdot T $$ Where $C$ is a constant that depends only on the amount of gas $n$.

It is supposed that the same procedure may be applied for Newton's law of gravitation:

  1. Experiments reveal that the force $F$ is proportional to one of the two masses $m$
    if the other mass $M$ as well as the distance $r$ between the masses is held constant: $\, F \sim m$ .
  2. The same argument hold for the other mass if mass $m$ and distance $r$ are held constant.
    Thus: $\, F \sim M$ .
  3. Experiments reveal that the force $F$ is inversely proportional to the distance $r$ squared
    if the mass $M$ as well as the mass $m$ are held constant. Thus: $\, F \sim 1/r^2$ .
Now we are going to derive Newton's law in three steps. (Don't get confused by the naming!)
  1. Keep the distance constant and change only the mass $m$ : $ F_1/m_1 = F_2/m_4 \quad \Longrightarrow \quad F_2 = F_1 m_4/m_1 $
  2. Keep the distance constant and change only the mass $M$ : $ F_2/M_1 = F_3/M_4 \quad \Longrightarrow \quad F_3 = F_1 m_4 M_4/(m_1 M_1) $
  3. Keep the masses constant and change only the distance $r$ : $ F_3 r_1^2 = F_4 r_4^2 \quad \Longrightarrow \quad F_4 = F_1 m_4 M_4 r_1^2/(m_1 M_1 r_4^2) $
It follows that: $$ \frac{F_4 r_4^2}{m_4 M_4} = \frac{F_1 r_1^2}{m_1 M_1} \quad \Longrightarrow \quad F = \mbox{constant} \times \frac{m M}{r^2} $$ (*) Reference:
DR. SCHWEERS EN DRS. P. VAN VIANEN
NATUURKUNDE
op corpusculaire grondslag
DEEL I
voor de onderbouw van het v.h.m.o.
ZEVENDE DRUK (1960)
L.C.G. MALMBERG 's-HERTOGENBOSCH