M phải có giá trị bằng bao nhiêu lần m để lực tổng hợp do m1, m2, m3 tác dụng lên một chất điểm khối

Trả lời

$F_{hd1}=\frac{G{m}_{1}m\text{'}}{r^2};F_{hd2}=\frac{G{m}_{2}m\text{'}}{r^2};F_{hd3}=\frac{G{m}_{3}m\text{'}}{r^2}$

$\sum \overrightarrow{F}={\overrightarrow{F}}_{hd1}+{\overrightarrow{F}}_{hd2}+{\overrightarrow{F}}_{hd3}$

$\sum \overrightarrow{F}=\overrightarrow{0}\Rightarrow {\overrightarrow{F}}_{hd1}+{\overrightarrow{F}}_{hd3}=-{\overrightarrow{F}}_{hd2}$

$\Rightarrow \left\{\begin{array}{l}{\overrightarrow{F}}_{hd13}=-{\overrightarrow{F}}_{hd2}\\ F_{hd13}=F_{hd2}\end{array}\right.$

$\Rightarrow F_{hd13}=F_{hd2}\iff \sqrt{F_{hd1}^2+F_{hd3}^2+2F_{hd1}.F_{hd3}.\cos \left({\overrightarrow{F}}_{hd1};{\overrightarrow{F}}_{hd3}\right)}=F_{hd2}$

$\iff \sqrt{F_{hd1}^2+F_{hd3}^2+2F_{hd1}.F_{hd3}.\cos {120}^0}=F_{hd2}$

$\iff {\left(\frac{G{m}_{1}m\text{'}}{r^2}\right)}^2+{\left(\frac{G{m}_{3}m\text{'}}{r^2}\right)}^2-\left(\frac{G{m}_{1}m\text{'}}{r^2}\right).\left(\frac{G{m}_{3}m\text{'}}{r^2}\right)={\left(\frac{G{m}_{2}m\text{'}}{r^2}\right)}^2$

$\iff m_1^2+m_3^2-m_1{m}_3=m_2^2\iff M^2+m^2-M.m=m^2\iff M=m$