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After launch from Earth orbit, a robot spacecraft of mass 5400 kg is coasting at constant speed halfway through its six- month flight to Mars when a NASA engineer discovers that, instead of heading for a 100-km-high orbit above the Martian surface, it is headed on a collision course directly toward the center of the planet. To correct the course, the engineer orders a short burst from the spacecraft's thrusters transverse to the direction of its motion. The thrust engines provide a constant force of 1200 N. For how long a time must the thrusters fire to achieve the correct course? Take needed data from Appendix C, and assume the distance between Earth and Mars to re- main constant at its smallest possible value. After launch from Earth orbit, a robot spacecraft of mass 5400 kg is coasting at constant speed halfway through its six- month flight to Mars when a NASA engineer discovers that, instead of heading for a 100-km-high orbit above the Martian surface, it is headed on a collision course directly toward the center of the planet. To correct the course, the engineer orders a short burst from the spacecraft's thrusters transverse to the direction of its motion. The thrust engines provide a constant force of 1200 N. For how long a time must the thrusters fire to achieve the correct course? Take needed data from Appendix C, and assume the distance between Earth and Mars to re- main constant at its smallest possible value.
The radius of Mars is 3396 km. Rounding off to two significant figures, the radius of the Mars will be 3400 km.A transverse direction signifies the angle is the right angle. So the distance that the thrusters have imparted a momentum sufficient to direct the space craft to the side of the original path would be,d = 100 km+3400 km = 3500 kmAs the space craft is half way through the six month journey, thus it has 3 months to move the 3500 km to the side.So, time, t = 3 monthsTo obtain the corresponding transverse speed v, substitute 3500 km for d and 3 months for t in the equation v = d/t,v = d/t =(3500 km/3 months) =(3500 km/3 months) (103 m/1 km) (1 month/30 days) (1 day/24 h) (1 h /3600 s) = 0.45 m/s.To find the time Δt for the rocket to fire, substitute 5400 kg for the mass of spacecraft m, 0.45 m/s for speed v and 1200 N for F in the equation Δt = mv/F,Δt = mv/F = (5400 kg) (0.45 m/s)/(1200 N) = (5400 kg) (0.45 m/s)/(1200 N) )(1 kg.m/s2/1 N) = 2.0sFrom the above observation we conclude that, the required time for the rocket to fire would be 2.0s.
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