A newly discovered planet has a mean radius of 7380 km. A vehicle on the planet/'s surface is moving in the same direction as the planet/'s rotation, and its speedometer reads 121 km/h. If the angular velocity of the vehicle about the planet/'s center is 9.78 times as large as the angular velocity of the planet, what is the period of the planet/'s rotation?

292796435 seconds ≈ 300 million seconds

Explanation:

First of all, the speed of the car is 121km/h = 33.6111 m/s

The radius of the planet is given to be 7380 km = 7380000 m

From the relationship between linear velocity and angular velocity i. e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec

If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have

w (vehicle) = 9.78 x w (planet)

w (planet) = w (vehicle) / 9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec

To find the period of the planet's rotation; we use the equation

w (planet) = 2π:T

Where w (planet) is the angular velocity of the planet and T is the period

From the equation T = 2π:w = 2 * (22/7) : 4.66 x 10⁻⁷ = 292796435 seconds

Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds