Two planets, A and B are orbiting a common star in circular orbits of radii R_mathrmA$R_{\mathrm{A}}$ and R_mathrmB$R_{\mathrm{B}}$ , respectively, with R_mathrmB = 2R_mathrmA$R_{\mathrm{B}} = 2R_{\mathrm{A}}$ . The planet B is 4sqrt2$4\sqrt{2}$ times more massive than planet A. The ratio left(fracL_mathrmBL_mathrmAmathrmright)$\left(\frac{L_{\mathrm{B}}}{L_{\mathrm{A}}\mathrm{}}\right)$ of angular momentum (L_mathrmB)$(L_{\mathrm{B}})$ of planet B to that of planet mathrmA(L_mathrmA)$\mathrm{A}(L_{\mathrm{A}})$ is closest to integer ______.

### Related Formula V = -fracGM_ER_E + h$$V = -\frac{GM_E}{R_E + h}$$ g' = fracGM_E(R_E + h)^2$$g' = \frac{GM_E}{(R_E + h)^2}$$ ### Core Logic The gravitational potential (V$V$) and acceleration due to gravity (g'$g'$) at a distance r = R_E + h$r = R_E + h$ from the center of the earth can be related by dividing their magnitudes: |V| / g' = r$|V| / g' = r$. ### Step 1: Set Up Equations From the given data: -fracGM_ER_E + h = -5.12 times 10^7 quad dots (i)$$-\frac{GM_E}{R_E + h} = -5.12 \times 10^7 \quad \dots (i)$$ fracGM_E(R_E + h)^2 = 6.4 quad dots (ii)$$\frac{GM_E}{(R_E + h)^2} = 6.4 \quad \dots (ii)$$ ### Step 2: Isolate Variable Divide equation (i) by (ii) (taking magnitudes): R_E + h = frac5.12 times 10^76.4$$R_E + h = \frac{5.12 \times 10^7}{6.4}$$ R_E + h = 0.8 times 10^7 mathrm~m = 8000 mathrm~km$$R_E + h = 0.8 \times 10^7 \mathrm{~m} = 8000 \mathrm{~km}$$ ### Step 3: Solve for h Given mean radius of the earth R_E = 6400 mathrm~km$R_E = 6400 \mathrm{~km}$: 6400 + h = 8000$6400 + h = 8000$ h = 1600 mathrm~km$$h = 1600 \mathrm{~km}$$ ### Pattern Recognition Always exploit the V/g = r$V/g = r$ relationship to extract distances cleanly without having to substitute large values for G$G$ or M_E$M_E$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### 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