asked 97.4k views
5 votes
A planet orbits a star, in a year of length 4.34 x 10^7 s, in a nearly circular orbit of radius 1.42 x 10^11 m. With respect to the star, determine

(a) the angular speed of the planet,
(b) the tangential speed of the planet, and
(c) the magnitude of the planet's centripetal acceleration.

asked
User Bautista
by
7.6k points

1 Answer

4 votes

Step-by-step explanation:

It is given that,

Time taken by a planet to orbit a star,
T=4.34* 10^7\ s

Radius of circular orbit,
r=1.42* 10^(11)\ m

(a) Angular speed,
\omega=(2\pi)/(T)


\omega=(2\pi)/(4.34* 10^7)


\omega=1.44* 10^(-7)\ rad/s

(b) Tangential speed of the planet,
v=r* \omega


v=1.42* 10^(11)\ m* 1.44* 10^(-7)\ rad/s

v = 20448 m/s

(c) Centripetal acceleration of the planet,
a=(v^2)/(r\\)


a=((20448)^2)/(1.42* 10^(11))


a=0.0029\ m/s^2

Hence, this is the required solution.

answered
User Moritz Both
by
7.8k points
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