# Would a falling object have jerk?

1. Oct 4, 2015

### Isaac0427

Hi guys! So I have been learning about jerk/jolt as the third time derivative of distance, and thinking about newtons law of universal gravitation, and I have come to the conclusion that a falling object must have jerk, and even jounce. Here's how I came to that conclusion:
The acceleration of a falling object depends on the mass of the larger body (in this case earth) and the distance from the center of gravity. If an object is accelerating towards the earth, the distance from the center of gravity is changing, therefore changing the acceleration (probably a very minimal amount), which means there is jerk. Then, as the acceleration increases, the faster the object's distance from the center of gravity decreases, therefore changing the rate of change in acceleration, and giving the object jounce. It seems as if it could go on infinitely, like how the infinite derivative of an exponential function is not constant (that may be a really bad way of saying that but you know what I mean). Is this correct or am I missing something?
Thanks!

2. Oct 4, 2015

### Isaac0427

Oh, and also, would that mean the constant 9.8 m/s^2 they teach us is not actually constant?

3. Oct 4, 2015

### RaulTheUCSCSlug

It is important to remember that the "constant" of 9.8m/s^2 only applies when you are near the surface of the earth, and that the farther away you are from earth, the weaker the gravity's pull as you can see from the equation:

F=G(m_1m_2)/r^2

Where G is the actual universal gravity constant and r is the radius between mass one and two. So as an object approaches the center of the earth, there is less force on that object, so you are correct in that regard. Also realize that just because you can continuously take the derivative of an object, does not necessarily give you more information.

4. Oct 4, 2015

### Chandra Prayaga

Yes. The acceleration does change as an object falls toward the earth. So there is a jerk or whatever, the derivative of acceleration with respect to time, though that is not a particularly useful concept. The acceleration due to gravity when an object is close to the surface of the earth, given the symbol g, is indeed only approximately 9.8 m/s^2. It is not "really" a constant. For most problems, the difference from the usually written value is small enough not to matter. There also other reasons for the changes in g, reasons other than just the distance from the center of the earth. I refer you to any introductory textbook of physics for a discussion.

5. Oct 5, 2015

### A.T.

In terms of coordinate acceleration in certain frames of reference, yes. But analyzing those higher derivatives is mainly relevant for proper acceleration, which is constant at zero in free fall.

6. Oct 5, 2015

Thanks guys!

7. Oct 5, 2015

### Nidum

OP - You might like to ponder what happens when a sky diver opens his parachute .

8. Oct 5, 2015