### 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

### 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