# Centripetal Force and Constant speed

1. Dec 31, 2014

### Scheuerf

I'm confused about how it is possible for speed to be constant when a centripetal force is being applied to an object. If something is orbiting a planet, when it is at what one person considers the top of the planet, all of the objects velocity is in the horizontal direction, and all of the acceleration is in the vertical direction. I would think that this would result in the object gaining speed. Even as it starts to move and gain an acceleration in the direction opposite its motion I would think that the vertical acceleration is increasing the speed faster than the horizontal is decreasing the speed.

2. Dec 31, 2014

### Staff: Mentor

Note that there is no component of acceleration in the direction of the velocity--those two vectors are perpendicular. That remains true at all points in the path. Thus the speed never changes. (But the velocity sure does.)

3. Dec 31, 2014

### PWiz

In centripetal motion, a constant force acting towards a point makes another object circle around it with a constant acceleration acting towards the center. Even though the object is moving along the circumference of a supposed circle, there is a constant force that ALWAYS acts towards the center of the circular orbit. By the $F=ma$ interpretation of Newton's second law of motion, the acceleration is also ALWAYS in the same direction as the force. If an object experiences constant uniform acceleration, then it's velocity should continuously change, and it does - it's always at right angles to the acceleration. It changes instantaneously in direction by always being a tangent to that point of the circle, and that is the reason why the object moves in a circular fashion. If you recall the from the definition of acceleration, either a change in direction, or a change in magnitude of the velocity (or both) is caused by acceleration. Therefore, by changing directions, the velocity continuously changes, but its magnitude (speed) stays constant. The speed only changes when the centripetal force changes(changing acceleration), in which case we categorize it as non-uniform centripetal motion.

4. Jan 1, 2015

### Staff: Mentor

To elaborate on what PWiz said, velocity is a vector quantity, and, as such, it has both magnitude and direction. So the velocity of a particle changes if its magnitude changes and/or if its direction changes. In pure rectilinear (straight line) motion, the direction of the velocity vector is constant, but its magnitude is changing. In pure circular motion, the magnitude of the velocity vector is constant, but its direction is changing. I'm sure you have done projectile problems where the horizontal component of velocity is constant, but the vertical component has a gravitational acceleration downward. So, if the projectile was moving horizontally initially, the acceleration was downward, and thus perpendicular to the velocity vector.

Chet

5. Jan 1, 2015

### HallsofIvy

In fact, for any force acting on a moving object, the component of the force in the same instantaneous direction as the velocity vector changes the speed, the component perpendicular to the velocity vector changes the direction.

6. Jan 1, 2015

### FactChecker

Not true. For any arbitrarily small Δtime, the velocity in that direction changes. The force is instantaneously 0 in that direction but is nonzero at any Δtime before or after.

7. Jan 1, 2015

### rcgldr

As a simple example of this, imaging a car traveling at constant speed, on a winding road, the car is constantly experiencing centripetal forces of various magnitudes, and the speed continues to remain constant.

8. Jan 1, 2015

### Staff: Mentor

Realize that the OP was talking about one instant (the at the "top" of the orbit), not in general.

9. Jan 1, 2015

### FactChecker

The vertical acceleration is increasing the vertical speed (which is ~ 0.0) faster than the horizontal acceleration is decreasing the horizontal speed ( which is ~ total speed)
Yes. Sorry. I guess the real question in the OP is about the rate of change of the horizontal and vertical speeds. The answer to that is that the trade off is exactly so that the total speed remains constant.
Speedtotal = √[Speed2vertical + Speed2horizontal]
Because the component speeds are squared, the vertical speed, which is nearly zero, must increase faster than the horizontal speed, which is nearly the total speed, decreases.
Because the component forces vary proportional to sine and cosine, which trade off exactly right, the total speed remains constant.

Last edited: Jan 1, 2015