The earth orbits the sun once a year (3.16e10^7 s) in a nearly circular orbit radius 1.50e10^11 m. With respect to the sun, determine the angular speed of the Earth in rad/s.
OK, so angular speed = change in theta/ change in time.
for theta I got 2(pi)r. Is that not right?
Help, please.
2007-11-23 02:59:00 · 3 answers · asked by frosty 2 in Science & Mathematics ➔ Physics
jg-
Thanks for the clarification! Looks like I got it know! Thanks again, everyone!
2007-11-23 03:16:22 · update #1
OK, I'm not trying to get the centripetal acceleration. I tried it =rw^2 (w= angular acceleration). What am I missing here?
2007-11-23 04:18:18 · update #2
Sorry, mistype! I AM trying to get centripetal acceleration now.......
2007-11-23 04:20:52 · update #3
Mainly, what you're doing wrong is posting physics problems on Black Friday when you should be shopping.
But here I am, answering them...
Angular speed = change in angle / change in time
There are 2 pi radians in a circle and the Earth makes one orbit in a year, so
Angular speed of Earth = 2 pi / 1 year
Convert 1 year to seconds and divide that result into 2 pi to finish the problem.
2007-11-23 03:05:07 · answer #1 · answered by jgoulden 7 · 0⤊ 1⤋
The circumference is 2πr. The angle is just 2π. So the angular velocity is 2π/T. You don't need the radius to calculate the angular velocity. The earth's speed IS 2πr/T.
2007-11-23 03:05:46 · answer #2 · answered by anobium625 6 · 1⤊ 0⤋
Theta is 2pi.
Linear speed is 2(pi)r
So omega (angular speed) is 2pi/3.16E7, in radians/second.
2007-11-23 03:08:54 · answer #3 · answered by ysk 4 · 0⤊ 0⤋