মহাকর্ষীয় প্রাবল্য ও বিভব

Three particles each of mass m'm' are at the vertices of an equilateral triangle of side length l'l'. Then the work done is moving one particle to infinity is:

হানি নাটস

Solution.

Initial gravitation potentialenergy =Ui=Gmm2Gmm2Gmm2=3Gmm2 =U_{i}=\frac{-G_{m m}}{2}-\frac{G_{m m}}{2}-\frac{G_{m m}}{2}=\frac{-3 G_{m m}}{2} .

 Workdore = of Vi=3Gmm2L(3GmmL)==3Gmm2L+6Gmm2L=3a2L If 2L= Infitity =l. \begin{array}{l} \text { Workdore }= \text { of }-V_{i}=\frac{-3 G m m}{2 L}-\left(-\frac{3 G_{m m}}{L}\right)= \\ =\frac{3 G m m}{2 L}+\frac{6 G_{m m}}{2 L} \\ =\frac{3 a-}{2 L} \\ \text { If } 2 L=\text { Infitity }=l . \end{array}

If 2L= 2 L= Infitity =l =l

Then workdone =3 Gm2l =\frac{3 \mathrm{~Gm}^{2}}{l}

\therefore option A A is correct

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