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A satellite of mass m is fired vertically upward from the Earth's surface with initial speed u. On reaching a height equal to R (R = Earth's radius), it ejects a rocket of mass m/10 so that the remaining satellite then moves in a circular orbit at that height. Find the kinetic energy of the ejected rocket. (G = gravitational constant, M = mass of Earth.)

1. (m/20)*(u - sqrt(2*G*M/(3*R)))²
2. 5m*(u² - 119*G*M/(200*R))
3. (3m/8)*(u + sqrt(5*G*M/(6*R)))²
4. (m/20)*(u² + 113*G*M/(200*R))

Correct answer: 5m*(u² - 119*G*M/(200*R))

Solution

Use energy conservation from surface to height R to find the satellite's speed v there (radially upward). For the 9m/10 fragment to orbit circularly at radius 2R, it needs tangential speed v_c = sqrt(GM/(2R)) and zero radial speed. Momentum conservation (total momentum was radial, m*v) splits between the orbiting fragment (tangential) and the rocket; computing the rocket's KE yields 5m*(u² - 119*G*M/(200*R)).

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