Physics help PLEASE !!?
A rock is thrown straight upward with an initial velocity of 9.6 m/s in a location where the acceleration due to gravity has a magnitude of 9.81 m/s2.
What is period of its elliptical orbit?
2007-06-28 08:46:23 · 4 answers · asked by Alexander 6 in Science & Mathematics ➔ Physics
Bekki:
what is r ?
2007-06-28 09:04:41 · update #1
Assuming the earth weren't there...
You would calculate the height, add that to the earth's radius if it were significant, which it isn't.
According to Kepler, the period of that orbit is the same as that of a circular orbit with radius r.
Centrifugal force = gravity
m(omega)^2 r = GmM/r^2
omega = sqrt (GM/r)
T = 2 pi / omega = 2 pi / sqrt (GM/r)
2007-06-28 09:00:33 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If you're near the Earth's surface, the gravity is essentially the same everywhere (and it points down). The rock will go straight up, and come straight back down. It won't go into orbit, and the motion has no period.
2007-06-28 15:52:14 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋
If it is thrown straight up, it has no elliptical orbit.
2007-06-28 15:49:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Download the free physics Vertical motion software from http://www.goldenkstar.com
I think it can solve this problem for you.
2007-07-01 08:48:59 · answer #4 · answered by Anonymous · 0⤊ 0⤋