B suppose the satellite launches a probe into a

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(a) At what height above Earth’s surface is the satellite orbiting? (b) Suppose the satellite launches a probe into a circular orbit that requires 8 hours for one revolution. With what speed was the probe launched? (c) How much work does the satellite do on the probe from part (b) in order to launch it? (d) Suppose we wish for the satellite to launch a 300-kg probe that escapes into outer space. In order to minimize the required launch speed, the satellite launches the probe forward along the direction of its instantaneous orbital motion. What must be the probe’s minimum speed with respect to the satellite ? (a) The satellite’s orbital radius is related as follows to its period T s : T s = 2 R 3 / 2 s p GM E R s = T s p GM E 2 2 / 3 = 1 . 48 10 7 m The corresponding height above Earth’s surface is h s = R s - R E = 8 . 47 10 6 m = 8468 km (b) The radius of the probe’s orbit is R p , where T p = 2 R 3 / 2 p p GM E R p = T p p GM E 2 2 / 3 = 2 . 03 10 7 m We use energy conservation to find the required launch speed v p : - GM E m p R s + 1 2 m p v 2 p = - GM E m p 2 R p - GM E R s + 1 2 v 2 p = - GM E 2 R p v p = s 2 GM E 1 R s - 1 2 R p = 5840 m / s 14
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(c) The amount of work required to launch the probe is equal to its change in kinetic energy, where its initial speed is the orbital speed of the satellite: W = 1 2 m p ( v 2 p - v 2 s , orb ) = 1 2 m p v 2 p - GM E R s = 1 2 (300) (5840) 2 - (6 . 67 10 - 11 )(5 . 98 10 24 ) 1 . 48 10 7 = 1 . 08 10 9 J (d) In order to give the probe the largest possible launch speed, the satellite should launch the probe in the direction of its orbital motion. The required escape speed for the probe is v esc = r 2 GM E R s = r (2)(6 . 67 10 - 11 )(5 . 98 10 24 ) 1 . 48 10 7 = 7330 m / s The required launch speed with respect to the satellite is v esc - v orb = r 2 GM E R s - r GM E R s = v esc 1 - 1 p 2 = 2147 m / s 15
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