# Jerk due to Gravity

1. Dec 21, 2009

### danielatha4

I've been doing A LOT of brainstorming, algebra, and applying physics to a simple question that I simply can't find the answer to.

If an object is in free-fall towards a fixed point of mass due to gravity then its acceleration at a given point of distance is GM/d^2 but this for some reason doesn't cut it for me (probably the OCD). As the object falls or "approaches" then the distance decreases thus increasing the acceleration at the given point.

I've been constantly trying to find the change of acceleration due to gravity with respect to time (Jerk due to Gravity), and my only assumption is that I have to find the changing distance as a function of time. All my efforts have failed as I seem to hit a looping paradox with the changing distance being a function of both time and time derivatives of position, such as acceleration making it hard to find any equation easily explaining (or explaining anything at all) acceleration due to gravity as a function of time.

I won't even begin to tell all the different ways I've approached this problem. I'll wait and see what response's I get.

Thank you

2. Dec 21, 2009

### Nabeshin

Ah yes. I seem to recall myself thinking exactly the same thing you are right now a few years ago. The solution is the branch of mathematics called differential equations which deals with these types of problems. The original thread I posted is found here: https://www.physicsforums.com/showthread.php?t=265195

Anyways, perhaps you've heard of differential equations and can figure out how they apply to this situation. I'm not going to give a whole longwinded explanation -- I'm sure some exist on these forums and several members would be happy to provide one, so that's all my input for now.

3. Dec 21, 2009

### danielatha4

Unfortunately I have no education on differential equations.

The first step I took was to assume the acceleration due to gravity at a given point is given by a = Gm / (Xo - ΔX)^2 where ΔX = Vot + 1/2Aot^2 + 1/6Jot^3... (I don't even know how many derivatives to use to define the change in distance). I don't even know where to begin with applying differential equations.

4. Dec 21, 2009

### Nabeshin

Well, the change in distance term is just an expansion. So it actually has infinitely many terms in this case.
Consider newton's second law:
$$F=m\frac{d^2x}{dt^2}$$
And with the force given by newton's universal law we have
$$-G\frac{Mm}{x^2}=m\frac{d^2x}{dt^2}$$
Or equivalently,
$$\ddot{x}+\frac{GM}{x^2}=0$$
Where the dot denotes a time derivative. This is the differential equation you want to solve (namely, you want to solve for the position as a function of time. Once you have this, getting the velocity, accel, jerk, etc., terms is just a simple matter of differentiation), and any textbook on differential equations would give you the techniques necessary to solve it. However, one does need to specify initial conditions, specifically the position and velocity at a given time (t=0 is as good as any), so let's say the particle is initially a distance D away from the large mass M and moving with a velocity v0.

I'm a bit lazy to solve this analytically, but the solution should be an conic section which is simply a line that passes straight through the sun and turns around at the same distance on the other side. Someone else might want to demonstrate how to do that but I just did it numerically and found the function x(t) and then plotted it and its derivatives. If you can find a numerical DE solver somewhere you could do the same, maybe someone else will post an analytical solution.

5. Dec 22, 2009

### danielatha4

I hate to simply ask how... but I have no idea how to solve the differential equation. How would you get any sort of time variable into the equation?

6. Dec 22, 2009

### Monocles

That's not something that can be explained in a single post. You'd need to take a course on differential equations. The time variable doesn't appear explicitly in that equation, but it will appear in the solution.

7. Dec 30, 2009

### tiny-tim

Hi danielatha4!

(try using the X2 tag just above the Reply box )

If you know the Chain Rule of calculus, it's easy …

the Chain Rule says that "fractions" work: da/dt = da/dx dx/dt …

in other words, the rate of change of a wrt t is the rate of change of a wrt x times the rate of change of x wrt t (or the "time change" of a is the "space change" of a, times the speed).

dx/dt is the speed, v, so da/dt = v da/dx, and a = Gm/x2 so da/dx is easy, and you can get v from conservation of energy, with PE = -GmM/x.

8. Dec 31, 2009

### danielatha4

Thanks tiny-tim. I actually thought of it that way, but had a hard time finding dx/dt as I was using the displacement formula that was nothing but an expansion Xf=Xo+vot+1/2aot2+...

da/dx is easy. da/dx = -2GM/X3 or should GM/X2 be negative before finding the derivative? I get confused with the proper signs.

Do I set potential energy PE=-GMm/x equal to kinetic energy KE=1/2mv2 to find velocity?

I do understand the concept of the chain rule though da/dx by dx/dt (velocity) equals da/dt

9. Dec 31, 2009

### hover

Perhaps this thread would interest you.

This is a thread I created to find out how to represent a jerk in an equation form. I show how I derived the equation of a jerk and so forth. Basically it comes down to being able to take an indefinite integral. If you know how to do that then it isn't difficult to come up with some equation to represent a jerk. The equation I recommend you use is

$$j=\frac{a_f^2-a_i^2}{2(v_f-v_i)}$$

simply because it doesn't deal with a time variable which is more difficult to calculate. So figuring out the initial and final acceleration of an object would be no more difficult than using Newton's law of gravitation. Which would be

$$a=\frac{GM}{r^2}$$

The only other tid-bit of info you need is what the final velocity of the object. That too can be found with integration. By the way, when you put all the math into that equation you will see that the average jerk will also change similar to what happened to the acceleration. The jerk isn't constant at all and so are the other derivatives of position.

$$x, \frac{dx}{dt} , \frac{d^2x}{dt^2}, \frac{d^3x}{dt^3}...$$

I hope I helped

Last edited by a moderator: Apr 24, 2017
10. Dec 31, 2009

### tiny-tim

Yup!

∆PE = - ∆KE, ie -GMm(1/x - 1/x0) = 1/2 m(v02 - v2)

11. Jan 1, 2010

### danielatha4

I'm assuming I should find the final velocity from the change in energy?

If so

Vf= sqrt[Vo2 + 2GM/Xf - 2GM/Xo

Since this equation represents dx/dt I should be able to multiply it by da/dt, which is -2GM/X3

Another thing to think about, does anyone know which derivative of position is a constant when free falling?

12. Jan 2, 2010

### ideasrule

Well, da/dt=dx/dt * da/dx, so you have to multiply speed (dx/dt) by -2GM/x^3.

13. Jan 3, 2010

### danielatha4

Alright I'm getting close here. Thank you everyone who has helped.

First off, was I suppose to find the FINAL velocity from the conservation of energy? and secondly should the X variable in the da/dx be the final position or initial position? I'm thinking it should be final... but I'm not positive.

Everything else makes sense:
da/dt = dx/dt(velocity) * da/dx = sqrt[Vo2 + 2GM/Xf - 2GM/Xo] * -2GM/Xf^3

14. Jan 3, 2010

### ideasrule

Whenever you see a variable in a derivative (e.g. x, v, a, or t in dx/dt, dv/dt, da/dt, da/dx, etc), that variable always represents "value at a certain moment". dx/dt, for example, is the infinitesimal change in position at a certain moment over the infinitesimal change in time at a certain moment. So yes, the "x" in da/dx and the dx/dt in da/dt=dx/dt*da/dx are both "final".

15. Jan 6, 2010

### danielatha4

thanks ideasrule. That's what I thought. I just felt the need to double check myself.

From the way I understand it, the jerk due to gravity at a given instant of time can be calculated by multiplying the instantaneous velocity by -2GM/X^3. However, the only sensible way to calculate velocity would be using the conservation of energy which is dependent on the initial position and velocity.

How would one find the fourth derivative of position with respect to time due to gravity? and so on? Is there, in theory, a derivative of position with respect to time that becomes an absolute constant? OR so infinitesimally small that the prior derivative can be considered a constant even when free falling from great distances?

Or can I go ahead and put this question to rest by assuming that an object starting at rest and gaining any position at all is a sign that all derivatives of position with respect to time take on a value that is not 0?