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

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 3910kg/m3, and it has a radius of R-5.27x10 m. What is the speed of the message packet as it passes a point a distance of 0.460R 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 planct SHM it turns out Y ou can write the e uation of motion Newton#2 equation. and find the time at which the acket s a the indicated position e speed can be obtained fromheder at 20 he osition Submit Answer Incorrect. Tries 1/6 Previous Tries 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 1/6 Previous Tries

Explanation / Answer

a ] Increase in KE = decrease in PE

0.5 mv^2 = - GMm/R - -GMm/2R *(3-0.460^2)

v^2 = 2.7884GM/R - 2GM/R = 0.7884GM/R

v = sqrt(0.7884GM/R) = sqrt(0.7884*6.67e-11*4/3*pi*5.27e6^2*3910)

= 4891 m/s answer

b] F = ma

-GMmr/R^3 = ma

a = [-GM/R^3]r = [-G*4/3*pi*rho] r

it is similar to SHM equation, a = -w^2 x

where w = sqrt(G*4/3*pi*rho)

time required = T/2 = 0.5*2pi / sqrt(G*4/3*pi*rho) = pi / sqrt(6.67e-11*4/3*pi*3910)

= 3005.76 s answer

  

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