Part F please (answer is not 1684.27m/s) Oops! An unmanned spacecraft s in a cir

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Part F please (answer is not 1684.27m/s)

Oops! An unmanned spacecraft s in a circular orbit around a distant moon, observing the moon's surface from an altitude 0 6 0 cm The mass of this particular moon s·55 10 2 gandits radius s 10 the mass of the spacecraft is 841.0 kg. Agroup of scientists and engineers is controlling the spacecraft from a great distance; to their dismay, a programming error causes an on-board thruster to fire. The misprogrammed thruster fires in a direction opposite to the velocity of the spacecraft; the resulting thrust decreases the speed of the spacecraft by 21.298 m/s. m NOTE: This problem has a tolerance of only 0.5%, instead of the usual tolerance of 2%. Please enter your answers to this question to four significant figures rather than the usual three. Use 6.673x10 11 N m2/kg2 for the universal constant of gravitation in this problem. Part A What was the speed of the spacecraft before the unfortunate programming error? Also, what is the speed of the spacecraft immediately after the unfortunate programming error? Hints speed before, speed just after = Submit 1624, 1603 m/s, m/s My Answers Give Up Correct Whereas the spacecraft had been in a circular orbit before the error occured, the controllers are able to determine that the spacecraft is now in an elliptical orbit. For this orbit, the maximum distance from the center of the moon is the orbital radius at which the programming error occurred, and the distance of closest approach (assuming all the mass of the moon were concentrated at a single point at the center of the moon) is alarmingly only only 1813.00 km, which is less than the radius of the moon! Imagine that all the mass of the moon were concentrated at a single point at the center of the moon. If nothing is done to correct the orbit of the spacecraft, what would be the resulting speed of the spacecraft at its distance of closest approach to the imaginary concentrated mass? Your eventual task is to solve for this speed by two different methods. Part B What would be the kinetic energy of the spacecraft when at its distance of closest approach? Also, what would be the magnitude of the angular momentum of the spacecraft when at its distance of closest Give your answers as an ordered pair, with the kinetic energy first, followed by a comma, followed by the angular momentum. Hints kinetic energy, angular momentum 1.199x109 2.575x1012 J, kg-m2 is Submit approach My Answers Give Up

Explanation / Answer

F] By energy conservation,

0.5mu^2 - GMm/r = 0.5mv^2 -GMm/R

0.5*1603^2 - 6.673e-11*7.55e22/[1.85e6+60e3] = 0.5v^2 -  6.673e-11*7.55e22/[1.85e6]

v = 1655.5 m/s answer

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