Question Number 212593 by mr W last updated on 18/Oct/24
Commented by mr W last updated on 18/Oct/24
$${a}\:{more}\:{general}\:{case}\:{from}\:{Q}\mathrm{212514} \\ $$
Answered by mr W last updated on 19/Oct/24
Commented by mr W last updated on 20/Oct/24
![F=((kq_1 q_2 )/c^2 ) with k=Coulomb constant c^2 =a^2 +b^2 −2ab cos θ c_1 =((m_2 c)/(m_1 +m_2 )), c_2 =((m_1 c)/(m_1 +m_2 )) c_1 b^2 +c_2 a^2 =c(h^2 +c_1 c_2 ) h^2 =((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )) (F/(m_1 g))=(c_1 /h), or using (F/(m_2 g))=(c_2 /h) ((kq_1 q_2 )/(m_1 gc^2 ))=((m_2 c)/((m_1 +m_2 )(√(((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))))) ((kq_1 q_2 (m_1 +m_2 ))/(m_1 m_2 g))=μ=(c^3 /( (√(((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))))) μ^2 =(c^6 /( ((m_1 a^2 +m_2 b^2 )/(m_1 +m_2 ))−((m_1 m_2 c^2 )/((m_1 +m_2 )^2 )))) c^6 +((μ^2 m_1 m_2 )/((m_1 +m_2 )^2 ))c^2 −((μ^2 (m_1 a^2 +m_2 b^2 ))/(m_1 +m_2 ))=0 ⇒c^2 =(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) ⇒a^2 +b^2 −2ab cos θ=(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) ⇒θ=cos^(−1) {((a^2 +b^2 −(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))+m_1 a^2 +m_2 b^2 ]))^(1/3) +(((μ^2 /(2(m_1 +m_2 )))[(√((m_1 a^2 +m_2 b^2 )^2 +((4μ^2 m_1 ^3 m_2 ^3 )/(27(m_1 +m_2 )^4 ))))−m_1 a^2 −m_2 b^2 ]))^(1/3) )/(2ab))}](https://www.tinkutara.com/question/Q212608.png)
$${F}=\frac{{kq}_{\mathrm{1}} {q}_{\mathrm{2}} }{{c}^{\mathrm{2}} }\:{with}\:{k}={Coulomb}\:{constant} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta \\ $$$${c}_{\mathrm{1}} =\frac{{m}_{\mathrm{2}} {c}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} },\:{c}_{\mathrm{2}} =\frac{{m}_{\mathrm{1}} {c}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} } \\ $$$${c}_{\mathrm{1}} {b}^{\mathrm{2}} +{c}_{\mathrm{2}} {a}^{\mathrm{2}} ={c}\left({h}^{\mathrm{2}} +{c}_{\mathrm{1}} {c}_{\mathrm{2}} \right) \\ $$$${h}^{\mathrm{2}} =\frac{{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }−\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} {c}^{\mathrm{2}} }{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{{F}}{{m}_{\mathrm{1}} {g}}=\frac{{c}_{\mathrm{1}} }{{h}},\:{or}\:{using}\:\frac{{F}}{{m}_{\mathrm{2}} {g}}=\frac{{c}_{\mathrm{2}} }{{h}} \\ $$$$\frac{{kq}_{\mathrm{1}} {q}_{\mathrm{2}} }{{m}_{\mathrm{1}} {gc}^{\mathrm{2}} }=\frac{{m}_{\mathrm{2}} {c}}{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)\sqrt{\frac{{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }−\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} {c}^{\mathrm{2}} }{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} }}} \\ $$$$\frac{{kq}_{\mathrm{1}} {q}_{\mathrm{2}} \left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}{{m}_{\mathrm{1}} {m}_{\mathrm{2}} {g}}=\mu=\frac{{c}^{\mathrm{3}} }{\:\sqrt{\frac{{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }−\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} {c}^{\mathrm{2}} }{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} }}} \\ $$$$\mu^{\mathrm{2}} =\frac{{c}^{\mathrm{6}} }{\:\frac{{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} }{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }−\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} {c}^{\mathrm{2}} }{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$${c}^{\mathrm{6}} +\frac{\mu^{\mathrm{2}} {m}_{\mathrm{1}} {m}_{\mathrm{2}} }{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} }{c}^{\mathrm{2}} −\frac{\mu^{\mathrm{2}} \left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}+{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}−{m}_{\mathrm{1}} {a}^{\mathrm{2}} −{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta=\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}+{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}−{m}_{\mathrm{1}} {a}^{\mathrm{2}} −{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}+{m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]}+\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{2}} }{\mathrm{2}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}\left[\sqrt{\left({m}_{\mathrm{1}} {a}^{\mathrm{2}} +{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} +\frac{\mathrm{4}\mu^{\mathrm{2}} {m}_{\mathrm{1}} ^{\mathrm{3}} {m}_{\mathrm{2}} ^{\mathrm{3}} }{\mathrm{27}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{4}} }}−{m}_{\mathrm{1}} {a}^{\mathrm{2}} −{m}_{\mathrm{2}} {b}^{\mathrm{2}} \right]}}{\mathrm{2}{ab}}\right\} \\ $$
Commented by ajfour last updated on 20/Oct/24
$${too}\:{meticulously}\:{done}! \\ $$