Calculating the angular acceleration.

Calculating radial acceleration?

• An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 24.0 days on average to complete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 2.060 × 10^5 km and the planet has a radius of 3775 km, calculate the moon\'s radial acceleration. Hints: First, find the total distance from the center of the moon to the center of the planet. Then calculate the radial acceleration after converting the given quantities to standard SI units. Start this problem by finding the total radius of rotation (this is the distance from the CENTER of the moon to the CENTER of the planet). Next, convert this distance into meters and the period into seconds. Then substitute the converted quantities into the equation for the radial acceleration of an object in uniform circular motion. From what I read it seems like you have to add 2.060 × 10^5 km +3775 km and convert to meters by multiplying by a thousand. Then you must convert the period 24 days into seconds, and I do. However, when i substitutes these answers into the equation like it says I get it wrong. Can anyone help? The equation needed is v^2/r right?

• Answer:

  Distance between centres of mass, R = 2.06^8m + 3.775^6m .. R = 2.098^8 m Periodic time, T = 24d x 24h x 3600s .. T = 2.07^6 s Angular velocity, ω = 2π / T rad/s .. .. 2π/2.07^6s .. .. ω = 3.03^-6 rad/s Radial acceleration, a = Rω² .. .. (2.098^8m) x (3.03^-6rad/s)² .. .. ►a = 1.90^-3 m/s² If you prefer .. a = v²/R .. a = (d/T)²/R a = (2πR/T)² / R = 4π².R / T².. .. 4π² x 2.098^8m / (2.07^6s)² .. .. a = 1.90^-3 m/s²