A satellite of mass 1000mathrmkg$1000\mathrm{kg}$ is launched to revolve around the earth in an orbit at a height of 270mathrmkm$270\mathrm{km}$ from the earth's surface. Kinetic energy of the satellite in this orbit is \_ \times 10^{10}\mathrm{J} . (Mass of earth $ . (Mass of earth $= 6\times 10^{24}\mathrm{kg}, Radius of earth $ Radius of earth $= 6.4 \times 10^{6} \mathrm{~m} , Gravitational constant $ , Gravitational constant $= 6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{kg}^{-2}$)

### 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