### Related Formula 1. Orbital speed (v_0$v_0$) of a satellite at distance r$r$ from earth's center: v_0 = sqrtfracG M_er$v_0 = \sqrt{\frac{G M_e}{r}}$ 2. Orbital Radius: r = R_e + h$r = R_e + h$ 3. Kinetic Energy of the orbiting satellite: mathrmKE = frac12 m v_0^2 = fracG M_e m2(R_e + h)$\mathrm{KE} = \frac{1}{2} m v_0^2 = \frac{G M_e m}{2(R_e + h)}$ ### Core Logic Given parameters: - Mass of satellite m = 1000 \ mathrmkg = 10^3 \ mathrmkg$m = 1000 \ \mathrm{kg} = 10^3 \ \mathrm{kg}$ - Orbit altitude h = 270 \ mathrmkm = 0.27 times 10^6 \ mathrmm$h = 270 \ \mathrm{km} = 0.27 \times 10^6 \ \mathrm{m}$ - Earth Radius R_e = 6.4 times 10^6 \ mathrmm$R_e = 6.4 \times 10^6 \ \mathrm{m}$ - Earth Mass M_e = 6 times 10^24 \ mathrmkg$M_e = 6 \times 10^{24} \ \mathrm{kg}$ - Gravitational constant G = 6.67 times 10^-11 \ mathrmN cdot m^2 / kg^2$G = 6.67 \times 10^{-11} \ \mathrm{N \cdot m^2 / kg^2}$ ### Step 1: Calculate kinetic energy First, compute the orbital radius r$r$: r = R_e + h = 6.4 times 10^6 \ mathrmm + 0.27 times 10^6 \ mathrmm = 6.67 times 10^6 \ mathrmm$r = R_e + h = 6.4 \times 10^6 \ \mathrm{m} + 0.27 \times 10^6 \ \mathrm{m} = 6.67 \times 10^6 \ \mathrm{m}$ Substitute r = 6.67 times 10^6 \ mathrmm$r = 6.67 \times 10^6 \ \mathrm{m}$ into the kinetic energy equation: mathrmKE = fracG M_e m2 r$\mathrm{KE} = \frac{G M_e m}{2 r}$ mathrmKE = frac6.67 times 10^-11 times 6 times 10^24 times 10^32 times 6.67 times 10^6$\mathrm{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^3}{2 \times 6.67 \times 10^6}$ Notice that the value 6.67$6.67$ cancels out directly: mathrmKE = frac6 times 10^162 times 10^6 = 3 times 10^10 \ mathrmJ$\mathrm{KE} = \frac{6 \times 10^{16}}{2 \times 10^6} = 3 \times 10^{10} \ \mathrm{J}$ Thus, the kinetic energy coefficient is 3$3$. ### Pattern Recognition Sees: Kinetic energy of a satellite orbiting at an altitude above earth's surface. Trap: Doing long division calculation for 6.67/2$6.67/2$. Check for clean cancellations in formulas first! Shortcut: Notice that R_e + h = 6.4 times 10^6 + 0.27 times 10^6 = 6.67 times 10^6$R_e + h = 6.4 \times 10^6 + 0.27 \times 10^6 = 6.67 \times 10^6$, which matches the value of the Gravitational constant G = 6.67 times 10^-11$G = 6.67 \times 10^{-11}$ perfectly. This clean cancellation leaves behind simple integer math to give 3 times 10^10 mathrm~J$3 \times 10^{10} \mathrm{~J}$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula v_0 = sqrtfracGM_textstarR$v_0 = \sqrt{\frac{GM_{\text{star}}}{R}}$ L = m v_0 R = m sqrtG M_textstar R$L = m v_0 R = m \sqrt{G M_{\text{star}} R}$ ### Core Logic The orbital angular momentum scales as L propto m sqrtR$L \propto m \sqrt{R}$, where m$m$ is the mass of the orbiting planet and R$R$ is its orbital radius. Setting up the ratio for planet B to planet A: fracL_BL_A = left(fracm_Bm_Aright) cdot sqrtfracR_BR_A$\frac{L_B}{L_A} = \left(\frac{m_B}{m_A}\right) \cdot \sqrt{\frac{R_B}{R_A}}$ Substitute the relative constraints provided by the text: - m_B = 4sqrt2 m_A$m_B = 4\sqrt{2} m_A$ - R_B = 2 R_A$R_B = 2 R_A$ fracL_BL_A = (4sqrt2) times sqrt2 = 4 times 2 = 8$\frac{L_B}{L_A} = (4\sqrt{2}) \times \sqrt{2} = 4 \times 2 = 8$ ### Pattern Recognition Orbital velocity goes down as 1/sqrtR$1/\sqrt{R}$, but angular momentum features an explicit distance product multiplier (m v R$m v R$), shifting the baseline radius factor to a clean numerator scaling profile: sqrtR$\sqrt{R}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula By Dimensional Analysis or Scaling of Kepler's Third Law: T propto m^x G^y a^z$T \propto m^x G^y a^z$ ### Core Logic Let's perform a dimensional matching to express the collision time T$T$ in terms of physical scaling variables m$m$, G$G$, and a$a$: [T] = [M]^x [M^-1L^3T^-2]^y [L]^z$[T] = [M]^x [M^{-1}L^3T^{-2}]^y [L]^z$ Equating dimensions on both sides: - Mass (M$M$): x - y = 0 implies x = y$x - y = 0 \implies x = y$ - Length (L$L$): 3y + z = 0 implies z = -3y$3y + z = 0 \implies z = -3y$ - Time (T$T$): -2y = 1 implies y = -1/2$-2y = 1 \implies y = -1/2$ Solving these equations: x = -1/2, quad y = -1/2, quad z = 3/2$x = -1/2, \quad y = -1/2, \quad z = 3/2$ ### Step 1: Scaling Formula of Time Therefore, the scaling relationship for time T$T$ is: T propto m^-1/2 G^-1/2 a^3/2 implies T propto sqrtfraca^3m$T \propto m^{-1/2} G^{-1/2} a^{3/2} \implies T \propto \sqrt{\frac{a^3}{m}}$ Let's write the ratio for two cases: fracT_2T_1 = sqrtleft(fraca_2a_1right)^3 cdot left(fracm_1m_2right)$\frac{T_2}{T_1} = \sqrt{\left(\frac{a_2}{a_1}\right)^3 \cdot \left(\frac{m_1}{m_2}\right)}$ ### Step 2: Calculating Final Time Given values: - a_1 = a$a_1 = a$, a_2 = 2a$a_2 = 2a$ - m_1 = m$m_1 = m$, m_2 = 2m$m_2 = 2m$ - T_1 = 4mathrm~s$T_1 = 4\mathrm{~s}$ fracT_24 = sqrtleft(frac2aaright)^3 cdot left(fracm2mright) = sqrt2^3 cdot frac12 = sqrt4 = 2$\frac{T_2}{4} = \sqrt{\left(\frac{2a}{a}\right)^3 \cdot \left(\frac{m}{2m}\right)} = \sqrt{2^3 \cdot \frac{1}{2}} = \sqrt{4} = 2$ T_2 = 4 times 2 = 8mathrm~seconds$T_2 = 4 \times 2 = 8\mathrm{~seconds}$ ### Pattern Recognition Kepler's Third Law / free-fall collapse scaling: whenever a orbit or a direct gravitational collapse scale is involved, the time scales as T propto sqrtfracR^3GM$T \propto \sqrt{\frac{R^3}{GM}}$. Thus doubling R$R$ multiplies time by sqrt8$\sqrt{8}$ and doubling M$M$ divides time by sqrt2$\sqrt{2}$. The combination results in a clean doubling of time: sqrt8/sqrt2 = 2$\sqrt{8}/\sqrt{2} = 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation Class 11 Physics: Units and Measurements: Dimensional Analysis

### 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