A planet orbits a star once every 3.14x10^7 s in a nearly circular orbit of radius 1.51x10^11 m.
(a) With respect to the star, determine the angular speed of the planet
b) With respect to the star, determine the tangential speed of the planet.
c) With respect to the star, determine the magnitude and direction of the planet's centripetal acceleration.
magnitude; direction ---Select--- directed toward the center of the orbit, directed away from the center of the orbit
a) w=2pi/3.14x10^7
=2.0010x10^-7 rad/s
b) v=rw
=(1.51x10^11)(2.0010x10^-7)
=3.0215x10^4 m/s
c) a=v^2/r
=(3.0215x10^4)^2/(1.51x10^11)
=6.0461x10^-3 m/s^2
-directed toward the center of the orbit
To find the answers to these questions, we can use some basic formulas from rotational motion and circular motion.
(a) Angular speed (ω) is given by the equation:
ω = 2π / T
where T is the period of the orbit.
In this case, the period T is equal to 3.14×10^7 s. Hence,
ω = 2π / (3.14×10^7 s)
(b) Tangential speed (v) is given by the equation:
v = r × ω
where r is the radius of the orbit.
In this case, the radius r is equal to 1.51×10^11 m. Hence,
v = (1.51×10^11 m) × ω
(c) Centripetal acceleration (ac) is given by the equation:
ac = v^2 / r
where v is the tangential speed and r is the radius of the orbit.
In this case, the tangential speed v and the radius r are known. Hence,
ac = (v^2) / (1.51×10^11 m)
To determine the direction of the centripetal acceleration, we know that centripetal acceleration is always directed towards the center of the circular motion.
Now, let's calculate the values:
(a) Angular speed:
ω = 2π / (3.14×10^7 s)
(b) Tangential speed:
v = (1.51×10^11 m) × ω
(c) Centripetal acceleration:
ac = (v^2) / (1.51×10^11 m)
The direction of centripetal acceleration is directed towards the center of the orbit.