Initially a satellite of 100 kg is in a circular orbit of radius 1.5R_E$1.5R_{E}$. This satellite can be moved to a circular orbit of radius 3R_E$3R_{E}$ by supplying alpha times 10^6J$\alpha \times 10^{6}J$ of energy. The value of alpha$\alpha$ is ____. (Take Radius of Earth R_E = 6 times 10^6text m$R_{E} = 6 \times 10^{6}\text{ m}$ and g = 10text m/s^2$g = 10\text{ m/s}^{2}$)

### 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