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In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 1.18 × 102 minutes = 7.08 × 103s and orbital speed of 3.40 × 103m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106m. 

a. Calculate the radius of the GS orbit. 

b. Calculate the mass of Mars. 

c. Calculate the total mechanical energy of the GS in this orbit. 

d. If the GS was to be placed in a lower circular orbit (closer to the surface of Mars), would the new orbital period of the GS be greater than or less than the given period? 

_________Greater than _________ Less than Justify your answer. 

e. In fact, the orbit the GS entered was slightly elliptical with its closest approach to Mars at 3.71 × 105m above the surface and its furthest distance at 4.36 × 105m above the surface. If the speed of the GS at closest approach is 3.40 × 103m/s, calculate the speed at the furthest point of the orbit.

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d. From Kepler’s third law r3/T2 =  constant so if r decreases, then T must also.

e. Conservation of angular momentum gives mv1r1 = mv2r2 so v2 = r1v1/r2, but the distances above the surface are given so the radius of Mars must be added to the given distances before plugging them in for each r. This gives v2 = 3.34 × 103m/s. 

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