The residents of a small planet have bored a hole straight through its center as
ID: 1792800 • Letter: T
Question
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 4450kg/m3, and it has a radius of R=5.73×106m. What is the speed of the message packet as it passes a point a distance of 0.380R 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. b) How long does it take for a message to pass from one side of the planet to the other?
Explanation / Answer
mass M = V = (4/3)r³
M = (4/3) * 4450kg/m³ * (5.73e6m)³ = 3.506e24 kg
a. The GPE at the surface is GPE = -G*m*M / r
and at the specified point is GPE' = -G*m*M / 0.380r
and so the difference is GPE = 0.373*G*m*M/r
where m is the mass of the packet.
That's become KE:
½mv² = 0.373*G*m*M/r mass m cancels, and so
v² = 2 * 0.373 * 6.674e11N·m²/kg² * 3.506e24kg / 5.73e6m = 3.0463e7 m²/s²
v = 5519.32 m/s
b. For "orbit", centripetal acceleration = gravitational acceleration, or
²r = (2/T)²r = 4²r / T² = v²/r = GM/r²
where the first four terms are all expressions for centripetal acceleration
and G = Newton's gravitational constant = 6.674e11 N·m²/kg²
and M = mass of central body
and r = orbit radius
and T = period
and v = orbit velocity
4²r / T² = GM/r² rearranges to
T² = 4²r³ / GM and when we substitute for M we get
T² = (4²r³/G)(3 / 4r³) = 3 / G
T² = 3 / 4450kg/m³*6.674e11N·m²/kg² = 3.173e7 s²
T = 5633 s time for a round trip
so the time to pass from one side to the other is T/2 = 2817 s
or around 47 minutes.