### 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

### 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