Calculation of Speed for Message Packet on Small Planet

[dantechiesa]

Homework Statement

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 3790kg/m3, and it has a radius of R=5.25×106m. 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

a) What is the speed of the message packet as it passes a point a distance of 0.380R from the center of the planet?

Homework Equations

x = A cos wt
v = -A sin wt * 2pi/t

The Attempt at a Solution

I have T = 6106 s

Utilizing x = A cos wt to find t at .380R (A = R)

.380R = R Cos(2pi*t/T)
R's cancel

Cos^-1(.380) = 1.181
1.818 * 6106 / 2pi

t = 1147.695 (can someone confirm this, I've done it a few time and get the same answer

Then I plug t in the derived velocity equation

v = -R sin(2pit/T) * 2pi/T

and the answer I get is -1.08x10^5m/s

Any thoughts?, Thanks

 

[RedDelicious]

Redo your calculation of v. I get a different value using your numbers. Everything else looks fine. Also, make sure you give your final answer as the speed.

 

[haruspex]

1.818 * 6106 / 2pi

Typo. It was 1.181.

v = -R sin(2pit/T) * 2pi/T

That looks ok, but I calculate from that something of the order of 1000 to 10000.

 

[dantechiesa]

Typo. It was 1.181.

That looks ok, but I calculate from that something of the order of 1000 to 10000.

Thank you!

 

[dantechiesa]

Redo your calculation of v. I get a different value using your numbers. Everything else looks fine. Also, make sure you give your final answer as the speed.

Thats what it was, Thanks!