 Thus far, we've derived the acceleration and the velocity of objects undergoing uniform circular motion, but how does this behaviour occur? In this video, we'll investigate the centripetal force that acts on objects undergoing uniform circular motion, and we'll also look at some useful mathematical relationships of uniform circular motion. Thus far, we've found that the velocity of uniform circular motion is equal to 2 pi r, the radius, over big t, the period of the revolution. We've also found the acceleration of uniform circular motion, the centripetal acceleration, is equal to V squared, the velocity squared, on r, the radius. So what is the force acting on the object? Well, we know that force is equal to mass times acceleration, and we know that for an object undergoing uniform circular motion, the acceleration is the centripetal acceleration. So if we substitute in the values we have for the centripetal acceleration, we'll see that the centripetal force is equal to mass times velocity squared divided by the radius mv squared on r. Now we'll look at the equation for velocity, and as we've previously discussed, this is equal to 2 pi r divided by big t, which is the time taken for an object's complete one period of the uniform circular motion. If we substitute in this value of velocity, we get that the centripetal force is equal to 4 pi mr on t squared, where m is the mass, r is the radius, and t is the period. We also know that the centripetal force must be pointing radially inwards, because it is pointing in the same direction as the centripetal acceleration. We also know that this must be perpendicular to the tangential velocity of uniform circular motion. We can now see that an object undergoing uniform circular motion is experiencing a force radially inwards that is equal to mv squared on r, and that it is experiencing a velocity that is perpendicular to this force as it orbits the center of the circle. So let's look at an object undergoing uniform circular motion. Let's look at the Earth orbiting the Sun. While in reality, this is not a perfectly circular orbit, we can approximate its motion as uniform circular motion. Because we know that the Earth is undergoing uniform circular motion, the net force must be inward and exactly the right amount to keep the Earth orbiting, or else it wouldn't be following a circular path at a constant speed. Something must be providing this net force, and the only force acting on the Earth is gravity, so gravity is providing just the right amount of force to keep the Earth undergoing uniform circular motion. In other words, the centripetal force is equal to the gravitational force. We have just derived that the centripetal force is equal to 4π squared mr on t squared. We also know that the gravitational force is equal to big g, big m, small m on r squared, which is the gravitational constant, the mass of the object being orbited, the mass of the object that is in orbit divided by the radius squared. If we set the gravitational force and the centripetal force equal to one another, if we cancel out the mass of the Earth as it appears on both sides of the equation, and we move the r squared over to the other side, we can see that big g, big m, is equal to 4π squared r cubed on big t squared. Rearranging this equation, we can see that big t squared on r cubed is equal to 4π squared on big g, big m. As we haven't actually substituted any values for this particular problem in, we know that the solution is applicable to Earth orbiting the Sun, as well as any satellite orbiting in space. Finally, let's review our equations for uniform circular motion. Firstly, we have the tangential velocity v is equal to 2π r the radius divided by big t the period. The centripetal acceleration is equal to v or the velocity squared divided by r or the radius. The force on an object undergoing uniform circular motion acts radially inwards and is equal to the mass times the velocity squared divided by the radius of the uniform circular motion. We also have the special case for satellites which uses Newton's universal law of gravitation. If we substitute in these values, we find that the period of an orbit squared, or big t squared, divided by the radius of that orbit cubed, or r cubed, is equal to 4π squared divided by big g, big m, which is the universal constant of gravitation multiplied by the mass of the object that is being orbited.