Centripetal Acceleration
TECTechnics Classroom   TECTechnics Overview

Centripetal Acceleration

(a) Is an object moving in a circle with constant speed accelerating?
(b) What is the acceleration of the moon around earth if the period of the moon's path around earth is
27(1/3) days, and the distance of the moon from the earth is 240,000 miles?

The string:
(a) S7P5A51 (Linear Motion).
The math:
(a) The Pj Problem of Interest (PPI) is of type change. Problems of speed, velocity, acceleration and duration are change problems.
Newton's first law of motion states that an object remains at rest if it is at rest or it moves with constant speed in linear motion if it is in motion, unless a force acts on it to change its state.
Newton's second law of motion indicates that the force that changes the linear motion of an object is equal to the product of its mass and acceleration
So, an object in circular motion does accelerate. This acceleration is called cetripetal acceleration and is given by:
v2/r (where v is the constant speed and r is the radius of the circular path).
(b) Distance of earth to moon 240,000 miles = 240,000 x 5280 = 1.27 x 109 ft.
Circumference of moon's path = 2πr = 2π(240,000 x 5280) = 2π(1.27 x 109) ft
Time it takes moon to complete one revolution around earth
= 27(1/3) days = (27.33 x 24 x 3600) = 2.36 x 106 secs
So, moon's speed around the earth, v = [2π(1.27 x 109)]/(2.36 x 106)
approximately = 3.30 x 103 ft/sec
So, acceleration of moon around earth:
= v2/r
= (10.89 x 106)/(1.27 x 109) = 8.6 x 10-3 = 0.0086 ft/sec2 approximately.

Blessed are they that have not seen, and yet have believed. John 20:29