the angular speed of the earth?

Question

The earth orbits the sun once a year (3.16 107 s) in a nearly circular orbit radius 1.50 1011 m. With respect to the sun, determine the following.
(a) the angular speed of the earth
______rad/s

(b) the tangential speed of the earth
______m/s

(c) the magnitude and direction of the earth's centripetal acceleration
magnitude _______m/s2
and the direction???

Answer 1

attractive force of gravity provides exactly for required centripetal force needed to be in orbit of (R)

G Ms * Me/R^2 = Me v^2/R = Me R w^2
where w = v/R= ang speed = 2pi/T (T =period)

w = 2 pi / 3.16*10^7 = 2*3.14 / 3.16*10^7
angular speed w = 1.987 *10^-7 rad/s
--------------------------------------------------
tangential speed
v = R w = 1.5*10^11 * 2*3.14 / 3.16*10^7 m/s
v= 2.98*10^4 m/s
-------------------
centri accel = v^2/R = R w^2
centri accel = 1.5*10^11 [1.987*10^-7]^2
centri accel = 5.922*10^-3 m/s^2

its direction is always pointing from revolving object - earth-towards the controlling mass - sun

Answer 2

(a) Angular speed= form of rotations in preserving with time Angular speed = a million x 2 pi/3.16E+7 =a million.98E-7 rad/sec (b) Tangential speed= 2 pi R/t Tangential speed= 2 pi a million.50 E+11 / 3.16E+7= Tangential speed= 29,800 m/sec (c) Centripetal acceleration = (Tangential speed)^2 / R Centripetal acceleration = (29,800)^2 / a million.50 E+11= 5.ninety 2 E-3 m/s^2

Answer 3

(a) Angular speed= number of rotations per time


Angular speed = 1 x 2 pi/3.16E+7 =1.98E-7 rad/sec

(b) Tangential speed= 2 pi R/t

Tangential speed= 2 pi 1.50 E+11 / 3.16E+7=
Tangential speed= 29,800 m/sec

(c) Centripetal acceleration = (Tangential speed)^2 / R

Centripetal acceleration = (29,800)^2 / 1.50 E+11=
5.92 E-3 m/s^2