Question

A planet orbits a star, in a year of length 2.27 x 107 s, in a nearly circular orbit of radius 3.99 x 1011 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.

Answer 1

Explanation:

It is given that,

Radius of circular orbit, $r=3.99\times 10^{11}\ m$

Time taken, $t=2.27\times 10^{7}\ m$      

(a) Angular speed of the planet is given by :

$\omega=\dfrac{2\pi}{t}$

$\omega=\dfrac{2\pi}{2.27\times 10^{7}\ m}$

$\omega=2.76\times 10^{-7}\ rad/s$

(b) The tangential speed of the planet is given by :

$v=r\times \omega$

$v=3.99\times 10^{11}\times 2.76\times 10^{-7}$

v = 110124 m/s

(c) The centripetal acceleration of the planet,

$a=\dfrac{v^2}{r}$

$a=\dfrac{(110124)^2}{3.99\times 10^{11}}$

$a=0.0303\ m/s^2$

Hence, this is the required solution.