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 Escape kinetic energy requirement formulation: textKEtextescape = fracG M mR = m g R$$\text{KE}{\text{escape}} = \frac{G M m}{R} = m g R$$ Gravitational potential energy field definition: U = -fracG M mr$$U = -\frac{G M m}{r}$$ At r to infty$r \to \infty$, Utextmax = 0$U{\text{max}} = 0$. ### Core Logic * **Assertion Check:** The work required to move an object from R$R$ to infty$\infty$ is Delta U = 0 - left(-fracGMmR ight) = fracGMmR = mgR$\Delta U = 0 - \left(-\frac{GMm}{R} ight) = \frac{GMm}{R} = mgR$. The Assertion states it is frac12mgR$\frac{1}{2}mgR$, which represents only the orbital velocity metric or binding equivalent. Thus, Assertion A is **false**. * **Reason Check:** Since gravitational energy is attractive and negative, it strictly ascends asymptotically up to a ceiling limit value of zero at infinity. Hence, Reason R is **true**. ### Pattern Recognition Escape energy = mgR$mgR$. Binding energy magnitude = frac12mgR$\frac{1}{2}mgR$ (for circular orbiters). Zero represents the reference point ceiling of a negative bound system state. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula fracdAdt = fracL2m$$\frac{dA}{dt} = \frac{L}{2m}$$ [cite: 684] ### Core Logic Kepler's Second Law state that the areal velocity fracdAdt$\frac{dA}{dt}$ is directly proportional to the angular momentum L$L$ of the planet[cite: 55, 684]. Because the gravitational force between the Sun and the planet acts strictly along the line joining their centers (a central force field), its torque vectau = vecr times vecF = 0$\vec{\tau} = \vec{r} \times \vec{F} = 0$[cite: 56, 686]. Since torque is zero, the angular momentum L$L$ remains completely constant over time[cite: 686]. As a consequence, fracdAdt = textconstant$\frac{dA}{dt} = \text{constant}$[cite: 55, 684]. Both statements are true and (R) is the correct explanation of (A)[cite: 63]. ### Pattern Recognition Areal velocity constancy is a direct geometric manifestation of the conservation of angular momentum under any central force field[cite: 55, 56, 686]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula Surface gravitational acceleration: g = fracGMR^2$$g = \frac{GM}{R^2}$$ ### Core Logic Since diameter drops to 1/3$1/3$, the radius R'$R'$ also scales down to 1/3$1/3$ its original value (R' = R/3$R' = R/3$), while mass M$M$ remains constant. New acceleration value g'$g'$ calculation: g' = fracGM(R/3)^2 = 9 cdot fracGMR^2 = 9g$$g' = \frac{GM}{(R/3)^2} = 9 \cdot \frac{GM}{R^2} = 9g$$ The scaling factor is 9. ### Pattern Recognition Gravity follows an inverse-square law with respect to radius. Shrinking the radius by a factor of n$n$ increases surface gravity by n^2$n^2$ if the mass is unchanged. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula By Kepler's Third Law of Planetary Motion, the square of the orbital period T$T$ is proportional to the cube of the orbital radius R$R$: T^2 = K cdot R^3$$T^{2} = K \cdot R^{3}$$ ### Core Logic Taking logs and differentiating to find fractional errors for small changes: 2 fracDelta TT = 3 fracDelta RR$$2 \frac{\Delta T}{T} = 3 \frac{\Delta R}{R}$$ fracDelta TT = frac32 left(fracDelta RR ight)$$\frac{\Delta T}{T} = \frac{3}{2} \left(\frac{\Delta R}{R} ight)$$ ### Step 1: Computing Percentage Change The change in radius is Delta R = 1.03R - R = 0.03R$\Delta R = 1.03R - R = 0.03R$, which means fracDelta RR = 0.03$\frac{\Delta R}{R} = 0.03$ or 3\%$3\%$ [cite: 106, 708]. Substitute this into our fraction scaling relation: fracDelta TT = frac32 times 0.03 = 0.045 = 4.5\%$$\frac{\Delta T}{T} = \frac{3}{2} \times 0.03 = 0.045 = 4.5\%$$ ### Pattern Recognition Power factors act as direct linear multipliers for small percentage shifts. Here, the scaling factor is simply frac32$\frac{3}{2}$ times the radius change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula The expression for escape velocity from a spherical planetary body is given by: v_textescape = sqrtfrac2GMR$$v_{\text{escape}} = \sqrt{\frac{2GM}{R}}$$ ### Core Logic Let the planet's mass be M_P$M_P$ and its radius be R_P$R_P$. According to the problem statement : * Earth's mass, M_E = 8 M_P implies fracM_PM_E = frac18$M_E = 8 M_P \implies \frac{M_P}{M_E} = \frac{1}{8}$ * Earth's radius, R_E = 2 R_P implies fracR_ER_P = 2$R_E = 2 R_P \implies \frac{R_E}{R_P} = 2$ Taking the ratio of escape velocities : fracv_Pv_E = sqrtleft(fracM_PM_Eright) times left(fracR_ER_Pright)$$\frac{v_{P}}{v_{E}} = \sqrt{\left(\frac{M_P}{M_E}\right) \times \left(\frac{R_E}{R_P}\right)}$$ fracv_Pv_E = sqrtfrac18 times 2 = sqrtfrac14 = frac12$$\frac{v_{P}}{v_{E}} = \sqrt{\frac{1}{8} \times 2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$ Given that v_E = 11.2 text km/s$v_E = 11.2 \text{ km/s}$: v_P = frac12 times 11.2 = 5.6 text km/s$$v_P = \frac{1}{2} \times 11.2 = 5.6 \text{ km/s}$$ ### Pattern Recognition Setting up quick ratios prevents substitution mistakes. For any planetary variant, notice how scaling properties scale inside the root operator directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation