Problem 3 & The residents of a small planet have bored a hole straight through i

Question

Problem 3 & The residents of a small planet have bored a hole straight through its center as part of a communications system. The hole has been filled with a tube and the air has been pumped out of the tube to virtually eliminate friction. Messages are passed back and forth by dropping packets through the tube. The planet has a density of 4180kg/m3, and it has a radius of R-5.69x10°m. What is the speed of the message packet as it passes a point a distance of 0.500R from the center of the planet? Remember, as we saw in class, this 'oscillator will have a period equal to the period of a satellite in orbit at the surface of the planet 5.325x103 m/s You are correct. Computer's answer now shown above Your receipt is 146-6265 Previous Trics b) How long does it take for a message to pass from one side of the planet to the other? Can you write F-ma? In which case you know the period, Submit Answer Incorrect. Tries 3/6 Previous Tries Due Friday November 24 11:59 am (EST)

Explanation / Answer

mass of the planet M = D*V

D = density

V = volume = (4/3)*Pi*R^3

M = 4180*(4/3)*pi*(5.69*10^6)^3

M = 3.225*10^24 kg

time period of satellite at the surface of planet T = 2*pi*R^(3/2)/sqrt(GM)

T = 2*pi*(5.69*10^6)^(3/2)/sqrt(6.67*10^-11*3.225*10^24)

T = 5814.6 s

angular frequency w(omega) = 2*pi/T

part(a)

speed oa particle at point in oscillatory motion v = w*sqrt(A^2 - x^2)

v = (2pi/5814.6)*sqrt((5.69*10^6)^2-(0.5*5.69*10^6)^2)

T = (2*pi*(5.69*10^6)^(3/2))/(sqrt(6.67*10^-11*4180*(4/3)*pi*(5.69*10^6)^3) = 5814.12 s

v = (2*pi/5814.6)*sqrt((5.69*10^6)^2-(0.50*5.69*10^6)^2))

speed v = 5325 m/s = 5.325*10^3 m/s

====================

part(b)

time taken t = T/2 = 2907.3 s <<<----------ANSWER

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