### Related Formula v_textescape = sqrtfrac2GMR$v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$ ### Core Logic For the planet: v = sqrtfrac2GM_pR_p$v = \sqrt{\frac{2GM_p}{R_p}}$ For the moon: M_m = fracM_p144$M_m = \frac{M_p}{144}$ and R_m = fracR_p16$R_m = \frac{R_p}{16}$. ### Step 1: Setup the Ratio v_m = sqrtfrac2G M_mR_m$v_m = \sqrt{\frac{2G M_m}{R_m}}$ v_m = sqrtfrac2G left(fracM_p144right)left(fracR_p16right)$v_m = \sqrt{\frac{2G \left(\frac{M_p}{144}\right)}{\left(\frac{R_p}{16}\right)}}$ v_m = sqrtfrac2G M_pR_p times frac16144$v_m = \sqrt{\frac{2G M_p}{R_p} \times \frac{16}{144}}$ ### Step 2: Simplification v_m = sqrtfrac2G M_pR_p times sqrtfrac19$v_m = \sqrt{\frac{2G M_p}{R_p}} \times \sqrt{\frac{1}{9}}$ v_m = v times frac13 = fracv3$v_m = v \times \frac{1}{3} = \frac{v}{3}$ ### Pattern Recognition Escape velocity scales as sqrtM/R$\sqrt{M/R}$. If M$M$ scales by x$x$ and R$R$ scales by y$y$, velocity scales by sqrtx/y$\sqrt{x/y}$. Here, sqrt(1/144)/(1/16) = sqrt16/144 = sqrt1/9 = 1/3$\sqrt{(1/144)/(1/16)} = \sqrt{16/144} = \sqrt{1/9} = 1/3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula F = fracG m_1 m_2r^2$F = \frac{G m_1 m_2}{r^2}$ ### Core Logic

Let the side length of the square be a$a$. Considering one corner mass, it experiences forces from the adjacent two masses (distance a$a$) and the diagonally opposite mass (distance sqrt2a$\sqrt{2}a$). The forces from the two adjacent masses are at 90^circ$90^\circ$ to each other: F = fracGm^2a^2$F = \frac{Gm^2}{a^2}$ The resultant of these two is sqrt2F = sqrt2 fracGm^2a^2$\sqrt{2}F = \sqrt{2} \frac{Gm^2}{a^2}$, directed along the diagonal. ### Step 2: Total Force Equation The force from the diagonal mass is: F' = fracGm^2(sqrt2a)^2 = fracGm^22a^2$F' = \frac{Gm^2}{(\sqrt{2}a)^2} = \frac{Gm^2}{2a^2}$ Total resultant force F_textnet = sqrt2F + F'$F_{\text{net}} = \sqrt{2}F + F'$: F_textnet = sqrt2 fracGm^2a^2 + fracGm^22a^2 = fracGm^2a^2 left( sqrt2 + frac12 right)$F_{\text{net}} = \sqrt{2} \frac{Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right)$ F_textnet = fracGm^2a^2 left( frac2sqrt2 + 12 right)$F_{\text{net}} = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$ Equating this to the given force value: left(frac2sqrt2 + 132right)fracGm^2L^2 = fracGm^2a^2 left( frac2sqrt2 + 12 right)$\left(\frac{2\sqrt{2} + 1}{32}\right)\frac{Gm^2}{L^2} = \frac{Gm^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$ frac132 L^2 = frac12 a^2$\frac{1}{32 L^2} = \frac{1}{2 a^2}$ a^2 = 16 L^2$a^2 = 16 L^2$ a = 4L$a = 4L$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation