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 Variation of g$g$ with height: g' = gleft(fracRR+hright)^2$$g' = g\left(\frac{R}{R+h}\right)^2$$ Time period of a simple pendulum: T = 2pisqrtfraclg'$$T = 2\pi\sqrt{\frac{l}{g'}}$$ ### Core Logic Given height h = 2R$h = 2R$, the effective acceleration due to gravity becomes: g' = gleft(fracRR+2Rright)^2 = fracg9$$g' = g\left(\frac{R}{R+2R}\right)^2 = \frac{g}{9}$$ For a second's pendulum, the time period is defined exactly as T = 2mathrm~s$T = 2\mathrm{~s}$. ### Step 1: Calculate Length Substitute T = 2mathrm~s$T = 2\mathrm{~s}$ and g' = fracg9$g' = \frac{g}{9}$ into the time period formula: 2 = 2pisqrtfraclg/9$$2 = 2\pi\sqrt{\frac{l}{g/9}}$$ 1 = pisqrtfrac9lg$$1 = \pi\sqrt{\frac{9l}{g}}$$ Squaring both sides: 1 = pi^2 cdot frac9lg$$1 = \pi^2 \cdot \frac{9l}{g}$$ Since g = pi^2 mathrm~m/s^2$g = \pi^2 \mathrm{~m/s^2}$: 1 = 9l implies l = frac19mathrm~m$$1 = 9l \implies l = \frac{1}{9}\mathrm{~m}$$ ### Pattern Recognition Second's pendulum always has T = 2mathrm~s$T = 2\mathrm{~s}$. Height 2R$2R$ from the surface means a total distance of 3R$3R$ from the center, which yields a frac19$\frac{1}{9}$ drop in gravity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation Class 11 Physics: Oscillations

### Related Formula According to Kepler's Third Law (Law of Periods): T^2 propto r^3$T^2 \propto r^3$ where: * T$T$ is the time period of revolution. * r$r$ is the orbital radius of the planet. ### Core Logic Using the proportionality relationship for two states: fracT_2^2T_1^2 = left( fracr_2r_1 right)^3$$\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} \right)^3$$ Given: * T_1 = 200text days$T_1 = 200\text{ days}$ * r_2 = fracr_14$r_2 = \frac{r_1}{4}$ ### Step 1: Calculate the New Time Period Substitute the values into the proportionality relation: fracT_2^2(200)^2 = left( fracr_1 / 4r_1 right)^3 = left( frac14 right)^3 = frac164$$\frac{T_2^2}{(200)^2} = \left( \frac{r_1 / 4}{r_1} \right)^3 = \left( \frac{1}{4} \right)^3 = \frac{1}{64}$$ Taking the square root on both sides: fracT_2200 = sqrtfrac164 = frac18$$\frac{T_2}{200} = \sqrt{\frac{1}{64}} = \frac{1}{8}$$ T_2 = frac2008 = 25text days$$T_2 = \frac{200}{8} = 25\text{ days}$$ ### Pattern Recognition If orbital distance scales by x$x$, the period scales by x^3/2$x^{3/2}$. Here, distance scales by frac14$\frac{1}{4}$, so the period scales by left(frac14right)^3/2 = frac18$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$. Thus, 200 times frac18 = 25text days$200 \times \frac{1}{8} = 25\text{ days}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula g = fracGMR^2 implies g propto frac1R^2$$g = \frac{GM}{R^2} \implies g \propto \frac{1}{R^2}$$ ### Core Logic Since diameter reduces to half, the radius R_2$R_{2}$ also reduces to half of its initial value R_1$R_{1}$: R_2 = fracR_12$$R_{2} = \frac{R_{1}}{2}$$ Setting up the ratio: fracg_2g_1 = left(fracR_1R_2right)^2 = left(fracR_1R_1/2right)^2 = 4$$\frac{g_{2}}{g_{1}} = \left(\frac{R_{1}}{R_{2}}\right)^{2} = \left(\frac{R_{1}}{R_{1}/2}\right)^{2} = 4$$ ### Step 1: Final Calculation g_2 = 4g_1 = 4g$$g_{2} = 4g_{1} = 4g$$ ### Pattern Recognition Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of 2^2 = 4$2^2 = 4$ matching constant mass bounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula Acceleration due to gravity at a height h$h$ above the Earth's surface: g_p = fracg R^2(R + h)^2$$g_p = \frac{g R^2}{(R + h)^2}$$ Acceleration due to gravity at a depth h$h$ below the Earth's surface: g_q = g left(1 - frachRright)$$g_q = g \left(1 - \frac{h}{R}\right)$$

### Core Logic We need the weights to be identical, meaning g_p = g_q$g_p = g_q$ at the exact same distance value h$h$: fracg R^2(R + h)^2 = g left(1 - frachRright)$$\frac{g R^2}{(R + h)^2} = g \left(1 - \frac{h}{R}\right)$$ Dividing by g$g$ and simplifying the left side denominator fraction: frac1left(1 + frachRright)^2 = 1 - frachR$$\frac{1}{\left(1 + \frac{h}{R}\right)^2} = 1 - \frac{h}{R}$$ left(1 - frachRright)left(1 + frachRright)^2 = 1$$\left(1 - \frac{h}{R}\right)\left(1 + \frac{h}{R}\right)^2 = 1$$ ### Step 1: Set Up Algebraic Equation Let frachR = x$\frac{h}{R} = x$. Then: (1 - x)(1 + x)^2 = 1$$(1 - x)(1 + x)^2 = 1$$ (1 - x)(1 + 2x + x^2) = 1$$(1 - x)(1 + 2x + x^2) = 1$$ 1 + 2x + x^2 - x - 2x^2 - x^3 = 1$$1 + 2x + x^2 - x - 2x^2 - x^3 = 1$$ x - x^2 - x^3 = 0$$x - x^2 - x^3 = 0$$ ### Step 2: Solve for x Since x neq 0$x \neq 0$ (distance cannot be zero), divide by x$x$: 1 - x - x^2 = 0 implies x^2 + x - 1 = 0$$1 - x - x^2 = 0 \implies x^2 + x - 1 = 0$$ Solving via quadratic formula: x = frac-1 pm sqrt1^2 - 4(1)(-1)2 = frac-1 pm sqrt52$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$ Since distance parameter x gt 0$x \gt 0$, we take the positive root: x = fracsqrt5 - 12$$x = \frac{\sqrt{5} - 1}{2}$$ ### Step 3: Find Height h Substitute back x = frachR$x = \frac{h}{R}$: frachR = fracsqrt5 - 12 implies h = fracR2 (sqrt5 - 1) = fracsqrt5R - R2$$\frac{h}{R} = \frac{\sqrt{5} - 1}{2} \implies h = \frac{R}{2} (\sqrt{5} - 1) = \frac{\sqrt{5}R - R}{2}$$ Therefore, the required distance is fracsqrt5R - R2$\frac{\sqrt{5}R - R}{2}$. ### Pattern Recognition Do not use the linear approximation formula g_h approx g(1 - frac2hR)$g_h \approx g(1 - \frac{2h}{R})$ unless the problem explicitly states h ll R$h \ll R$. Equating the approximated form to depth gives h_textheight = frac12 h_textdepth$h_{\text{height}} = \frac{1}{2} h_{\text{depth}}$, which fails when looking for a single unified distance value h$h$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula V_e = sqrtfrac2GMR$$V_e = \sqrt{\frac{2GM}{R}}$$ ### Core Logic For Earth: V_e = sqrtfrac2GM_ER_E = 11.2 mathrm~km/s$V_e = \sqrt{\frac{2GM_E}{R_E}} = 11.2 \mathrm{~km/s}$ For the planet: mathrmR_mathrmP = fracmathrmR_mathrmE3$\mathrm{R}_{\mathrm{P}} = \frac{\mathrm{R}_{\mathrm{E}}}{3}$ and mathrmM_mathrmP = fracmathrmM_mathrmE6$\mathrm{M}_{\mathrm{P}} = \frac{\mathrm{M}_{\mathrm{E}}}{6}$ We can express the escape velocity of the planet V_p$V_p$ as a ratio of the Earth's escape velocity. ### Step 1: Ratio of Velocities fracV_pV_e = sqrtfracM_pM_E times fracR_ER_p$$\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_E} \times \frac{R_E}{R_p}}$$ fracV_pV_e = sqrtleft(frac16right) times left(frac31right) = sqrtfrac12$$\frac{V_p}{V_e} = \sqrt{\left(\frac{1}{6}\right) \times \left(\frac{3}{1}\right)} = \sqrt{\frac{1}{2}}$$ ### Step 2: Calculate Escape Velocity V_p = fracV_esqrt2$$V_p = \frac{V_e}{\sqrt{2}}$$ V_p = frac11.21.414 approx 7.92 mathrm~km/s$$V_p = \frac{11.2}{1.414} \approx 7.92 \mathrm{~km/s}$$ ### Pattern Recognition Any scaling of a planet's mass by factor alpha$\alpha$ and radius by factor beta$\beta$ scales the escape velocity by a factor of sqrtalpha / beta$\sqrt{\alpha / \beta}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation