A newly discovered planet is in a circular orbit around a distant star with an o

Question

A newly discovered planet is in a circular orbit around a distant star with an orbital period of 300 Earth days. The planet also rotates on its axis, making one full rotation every 3.00 Earth days. The radius of the planet is rp = 2.50 106 m and the radius of the planet's orbit about the star is rs = 1.50 1011 m.

Determine the ratio of the radial acceleration, due to the rotation of the planet, of an object at the equator of the planet (acp) to the radial acceleration of the planet in its orbit about the star (acs).

=

Explanation / Answer

a) let angular velocity of the planet about the star be w1.

time taken 300*24*60*60 = 2.592*107 s, angle rotated 2pi

Hence w1 = 2*3.14/ 2.592*107 = 2.42*10-7 rad/s.

radial accleration acs = (angular velocity)2*radius = (2.42*10-7 )2 * 1.50 10^11 = 8.8*10^-3 rad/s2

b) angular velocity of planet about its own axis be w2

Then time taken to rotate about its own axis t = 3*24*3600 = 2.592*105 s

angle rotated 2pi

Hence w1 = 2*3.14/ 2.592*105 = 2.42*10-5 rad/s.

radial accleration acp = (angular velocity)2*radius = (2.42*10-5 )2 * 2.5*106 = 1.46*10-3  rad/s2

acp/acs =1.46*10-3  /8.8*10^-3 = 0.166

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