the answer needs to be in hrs.
the solution i tried was.
G (Me)/ r^3 = w^2
then f = w/2(pi)
T= 1/f
this is how the instructor showed us and it isnt working.
2006-10-01 11:47:30 · 3 answers · asked by robtrink 1 in Science & Mathematics ➔ Physics
GxMe = (Earth's standard gravitational parameter) = 398,600 km³/s²
r = (Earth radius)+ 615 km = 6,378.135+ 615 = 6,993.135 km
omega² = GxMe/r³ = 398,600/(6,993.135)³ = 1.1655 x10^-6 1/s²
(2xpi/T)² = 1.1655x10^-6 1/s²
2xpi/T = 1.0800x10^-3 1/s
T = 2xpi/(1.0800x10^-3) = 5,820.0 sec = 1,6167 h
Answer: 1h 37min
2006-10-01 12:01:43 · answer #1 · answered by Illusional Self 6 · 0⤊ 0⤋
at first, the area Shuttles have all been retired, so they're not component of the tale from now on. The Hubble area Telescope is in a stable orbit, so is basically in loose fall, with the forces of gravity and inertia balancing one yet another out.
2016-10-18 07:56:35 · answer #2 · answered by ? 4 · 0⤊ 0⤋
5million times the time it Takes for you to eat an extra hot soup in a hurry.
2006-10-01 11:54:14 · answer #3 · answered by solange' 1 · 0⤊ 1⤋