Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T=4$T=4$ seconds. If the sides of the triangle are increased to length 2a and also the masses of the spheres are made 2m, then they will collide after ________ seconds.

### Related Formula E = frac-GM_E m2r$$E = \frac{-GM_E m}{2r}$$ Delta E = E_f - E_i$$\Delta E = E_f - E_i$$ g = fracGM_ER_E^2 implies GM_E = g R_E^2$$g = \frac{GM_E}{R_E^2} \implies GM_E = g R_E^2$$ ### Core Logic Energy of a satellite in a circular orbit is given as E = frac-GM_E m2r$E = \frac{-GM_E m}{2r}$ where r is the radius of the circular orbit. Required energy to be supplied Delta E = E_f - E_i$\Delta E = E_f - E_i$: Delta E = left( frac-GM_E m2(3R_E) right) - left( frac-GM_E m2(1.5R_E) right)$$\Delta E = \left( \frac{-GM_E m}{2(3R_E)} \right) - \left( \frac{-GM_E m}{2(1.5R_E)} \right)$$ Delta E = frac-GM_E m6R_E + fracGM_E m3R_E = fracGM_E m6R_E$$\Delta E = \frac{-GM_E m}{6R_E} + \frac{GM_E m}{3R_E} = \frac{GM_E m}{6R_E}$$ ### Step 1: Evaluate Delta E Substitute GM_E = g R_E^2$GM_E = g R_E^2$ into the expression: Delta E = frac(g R_E^2) m6R_E = frac16 m g R_E$$\Delta E = \frac{(g R_E^2) m}{6R_E} = \frac{1}{6} m g R_E$$ = frac16 times 100 times 10 times (6 times 10^6)$$= \frac{1}{6} \times 100 \times 10 \times (6 \times 10^6)$$ = 1000 times 10^6text J$$= 1000 \times 10^6\text{ J}$$ Comparing with alpha times 10^6text J$\alpha \times 10^6\text{ J}$, we get alpha = 1000$\alpha = 1000$. ### Pattern Recognition Orbital transition energy is always Delta E = fracGMm2 (frac1r_i - frac1r_f)$\Delta E = \frac{GMm}{2} (\frac{1}{r_i} - \frac{1}{r_f})$. Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### 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 Conservation of Total Mechanical Energy: E_i = E_f$E_i = E_f$ U_i + K_i = U_f + K_f$$U_i + K_i = U_f + K_f$$ ### Core Logic The initial distance from the center of the earth is r = R + 3R = 4R$r = R + 3R = 4R$. Initial mechanical energy: E_i = -fracGMm4R + frac12mv^2$$E_i = -\frac{GMm}{4R} + \frac{1}{2}mv^2$$ To just escape to infinity, the final mechanical energy at infinity must be at least zero: E_f = 0$E_f = 0$ ### Step 1: Apply Energy Conservation Setting initial energy equal to zero: -fracGMm4R + frac12mv^2 = 0$$-\frac{GMm}{4R} + \frac{1}{2}mv^2 = 0$$ frac12v^2 = fracGM4R implies v = sqrtfracGM2R$$\frac{1}{2}v^2 = \frac{GM}{4R} \implies v = \sqrt{\frac{GM}{2R}}$$

### Pattern Recognition Be extremely careful with the phrase 'above the earth's surface'. Distance from center r = R + h$r = R + h$. Escape condition always sets net mechanical energy ge 0$\ge 0$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation