B Mathematics involving a gravitational force

[rb75]

TL;DR Summary

Calculation of the non-uniform displacement, velocity, acceleration and jerk produced by a gravitational force at any instantaneous point in time.

It's been more than 60 years since I attended high school, and I am trying to learn and understand concepts in Newton's physics that were not taught and were not part of the school curriculum during the 1950's.

It is my understanding that the current mathematics taught and used in our universities and schools to calculate the quantity of the non-uniform displacement, velocity, acceleration and jerk that is produced by an object under the influence of a gravitational force, involves the use of differential calculus and partial differential equations derived from the functions of the expressions in the polynomial equation $d = v_it + 1/2at^2 +1/6jt^3$.

If my understanding is correct, what are the partial differential equations that are taught to students enrolled in our universities, that correctly and accurately calculates the quantity of the displacement, velocity, acceleration, and jerk that is produced by an object under the influence of a gravitational force at any instantaneous point in time?

 

[Ibix]

Depends. Newtonian gravity is fairly straightforward if you know what a vector is. You'll probably need a computer for anything more than a two-body problem.

If you need speeds near $c$ or need to get close to black holes you need general relativity. Unless it's a really simple scenario, you need to be comfortable with differential geometry and have enough programming skill to use fairly sophisticated numerical methods to solve them (and possibly pay for time on a powerful cluster, depending how complex your scenario is and much precision you need).

There are some half-way houses like the Parameterised Post-Newtonian model, about which I know little more than the name. What problem are you trying to solve?

 

[rb75]

Depends. Newtonian gravity is fairly straightforward if you know what a vector is. You'll probably need a computer for anything more than a two-body problem.

If you need speeds near $c$ or need to get close to black holes you need general relativity. Unless it's a really simple scenario, you need to be comfortable with differential geometry and have enough programming skill to use fairly sophisticated numerical methods to solve them (and possibly pay for time on a powerful cluster, depending how complex your scenario is and much precision you need).

There are some half-way houses like the Parameterised Post-Newtonian model, about which I know little more than the name. What problem are you trying to solve?

 

[Ibix]

Was there supposed to be a question there, @rb75, or was that an accidental post?

 

[rb75]

Hello Ibex, thanks for taking the time to reply to my question concerning the mathematics used to calculate the continually changing non-uniform displacement, velocity, acceleration and jerk that is produced by a gravitational attractive force at any instantaneous point in time.

I'm glad you stated that Newtonian gravity is fairly straight forward, and with your input I hope I might be able to learn something. I do have a computer, and by trial and error I have managed to learn some of the fundamentals in AutoLisp programing.

My current knowledge of vectors is limited. I am familiar with geometric vectors as applied to Euclidean geometry, and problems in mechanics, where quantities such as displacement, velocity, acceleration and jerk have both a magnitude and a direction.

In answer to your question of what problem I'm trying to solve?, the problem is as stated below, and is what I assume to be a simple two-body problem involving motion with speeds that are attained by bodies in our solar system.

Problem:
Newton's physics states, an object in freefall due to the influence of a gravitational attractive force, produces a non-uniform rate of change in the quantity of the objects non-uniform displacement, velocity, acceleration and jerk, with each and every change in the radial distance between the centers of the two gravitational attractive bodies, and with each and every change in the interval of time.

Given the following information, It would be appreciated if you would demonstrate the mathematics that correctly and accurately calculates the quantity of the continually changing non-uniform displacement, velocity, acceleration and jerk that is produced by an object in freefall due to the influence of the Earth's gravitational attractive force at the end of the 1st, 2nd, and 3rd, one second intervals in time.

Initial conditions:
1. The motion of the object occurs in the vacuum of outer space and no other forces are involved.
2. The Earth is considered to be a stationary object.
3. The values given below are accurate to 5 decimal places.

Given:
Polynomial equation total gravitational displacement $d_{i}=v_{i}t+1/2g_{i}t^2+1/6j_{I}t^3$
Newton's 2nd Law of motion $F=ma$ and/or $p=mvt$
Newton's equation for gravitational force $F_{g}=$ GM$_{1}m_{2}/d^2$
Newton's equation for gravitational acceleration $g=$ GM$_{1}/d^2$
Gravitational constant G =$6.67408$ x $10^{-11}N.m^2/kg^2$
Objects mass = $1kg$
Earth mass = $5.97420$ x $10^{24}kg$
Earth radius = $6.37800$ x $10^{6}meters$
Objects initial height above the Earth's surface = $1000000meters$
Objects initial radial distance from the Earth's center = $7378000meters$
Objects initial velocity = $0.0m/s$
Objects initial acceleration due to gravity = $7.32476m/s^{2}$
Objects initial jerk due to gravity = $0.0m/s^{3}$
Objects total time in freefall to the Earth = $3sec.$

Thanks for the time and interest, it is greatly appreciated

rb75

 

[Ibix]

You've already written down the acceleration as a function of the radial distance from the center of the earth, so you have $\ddot{r}=GM/r^2$. That's the only equation you need if you are restricting yourself to radial motion and neglecting the motion of the Earth.

The actual solution is a bit messy, but has been discussed here previously.

You don't need to include higher order time derivatives like jerk in the model. Once you've solved for the velocity as a function of time you can differentiate it as many times as you want to get those if you want, but you don't really need them for anything. There may be infinitely many of them, by the way, not just velocity, acceleration, and jerk.

 

[Ibix]

By the way, once you've done the full calculation you might like to compare it to assuming a constant acceleration of $g=GM/r^2$, which gives you a distance traveled in time $t$ of $\frac 12gt^2$. Given the initial conditions you specified it's barely going to fall 30m in 3s, which isn't going to make a huge difference to $g$. Is the extra precision you gain really worth the effort in this case?

 

[bob012345]

Given:

Objects initial height above the Earth's surface = $1000000meters$
Objects initial velocity = $0.0m/s$
Objects initial acceleration due to gravity = $7.32476m/s^{2}$
Objects total time in freefall to the Earth = $3sec.$

Those numbers don't add up. If you drop an object 1000km high it is not going to reach the ground in 3 seconds!

 

[Ibix]

Those number don't add up. If you drop an object 1000km high it is not going to reach the ground in 3 seconds!

I assume OP means "towards the Earth" rather than literally "to" the Earth's surface. But I agree with your point - from rest at 1000km up, the fall time to the surface is somewhere between 450s and 525s.

 

[rb75]

I assume OP means "towards the Earth" rather than literally "to" the Earth's surface. But I agree with your point - from rest at 1000km up, the fall time to the surface is somewhere between 450s and 525s.

You've already written down the acceleration as a function of the radial distance from the center of the earth, so you have $\ddot{r}=GM/r^2$. That's the only equation you need if you are restricting yourself to radial motion and neglecting the motion of the Earth.

The actual solution is a bit messy, but has been discussed here previously.

You don't need to include higher order time derivatives like jerk in the model. Once you've solved for the velocity as a function of time you can differentiate it as many times as you want to get those if you want, but you don't really need them for anything. There may be infinitely many of them, by the way, not just velocity, acceleration, and jerk.

 

[rb75]

I had originally thought that Newton’s equation for gravitational acceleration was the only equation that was needed for this problem. The only reason I included Newton’s equation for gravitational force and the 2nd Law of motion and momentum, was because I had received a reply regarding a solution to my original question from Frabjous, a Gold Member, that involved Newton’s 2nd Law of Motion and momentum to calculate the value for the jerk created by a gravitational force.
I included these equations in my problem in case they might have some relevance to your solution to this problem.

As you suggested, I reviewed the previously discussed information of this problem, and am still not sure of the mathematics that is used to determine the value for velocity as a function of time. As you stated, knowing this value is the only requirement that is needed for the solution to this problem. Could you please demonstrate and explain the mathematics used to determine the value of the velocity as a function of time for this problem in the simplest terms possible.
Thanks rb75

 

[rb75]

You've already written down the acceleration as a function of the radial distance from the center of the earth, so you have $\ddot{r}=GM/r^2$. That's the only equation you need if you are restricting yourself to radial motion and neglecting the motion of the Earth.

The actual solution is a bit messy, but has been discussed here previously.

You don't need to include higher order time derivatives like jerk in the model. Once you've solved for the velocity as a function of time you can differentiate it as many times as you want to get those if you want, but you don't really need them for anything. There may be infinitely many of them, by the way, not just velocity, acceleration, and jerk.

I assume OP means "towards the Earth" rather than literally "to" the Earth's surface. But I agree with your point - from rest at 1000km up, the fall time to the surface is somewhere between 450s and 525s.

 

[bob012345]

I suggest you first learn how to apply Newton's laws to the simpler case of constant acceleration. Also, if you are still mainly interested in the first 3 seconds as in the post 5, you could assume the acceleration due to gravity is basically constant over the short distance the object falls in the first 3 seconds.

 

[erobz]

As was posted by @Ibix if you are interested in the mathematics, to find the velocity we need to first find $r(t)$. And as @bob012345 points out gravity isn't going to change that much over that short of a fall? But anyhow...

I could try to work out the math with you.

Starting with Conservation of Energy:

$$ -\frac{GMm}{r_o} = -\frac{GMm}{r} + \frac{1}{2} m \left( \frac{dr}{dt}\right)^2 $$

We have to solve this ODE by separation of variables. Start by isolating $\frac{dr}{dt}$:

$$ \frac{dr}{dt} = \sqrt{2GM} \sqrt{ \frac{1}{r} - \frac{1}{r_o} } $$

Now we use the method of Separation of Variables:

$$ \int \frac{dr}{ \sqrt{ \frac{1}{r} - \frac{1}{r_o} } } = \sqrt{2GM} \int dt $$

Next, we are going to use something called trigonometric substitution. The objective is to try to transform the integral on the left using a variable substitution (in this case using trigonometry)

Define a right triangle with:

hypotenuse $\sqrt{\frac{1}{r}}$, side length opposite the angle $\theta$ is $\sqrt{\frac{1}{r_o}}$, and adjacent side $\sqrt{ \frac{1}{r} - \frac{1}{r_o} }$

We note that the integrand in question can now be transformed into $\tan \theta$ by the multiplication of a constant $\sqrt{\frac{1}{r_o}}$

$$ \int \frac{dr}{ \sqrt{ \frac{1}{r} - \frac{1}{r_o} } } = \sqrt{r_o} \int \tan \theta \, dr $$

Using the method above (the triangle developed) would you like to transform the remaining $dr \to f( \theta ) d\theta$, before we continue?

 

[rb75]

You've already written down the acceleration as a function of the radial distance from the center of the earth, so you have $\ddot{r}=GM/r^2$. That's the only equation you need if you are restricting yourself to radial motion and neglecting the motion of the Earth.

The actual solution is a bit messy, but has been discussed here previously.

You don't need to include higher order time derivatives like jerk in the model. Once you've solved for the velocity as a function of time you can differentiate it as many times as you want to get those if you want, but you don't really need them for anything. There may be infinitely many of them, by the way, not just velocity, acceleration, and jerk.

I assume OP means "towards the Earth" rather than literally "to" the Earth's surface. But I agree with your point - from rest at 1000km up, the fall time to the surface is somewhere between 450s and 525s.

 

[rb75]

@erobz

Thanks for the interest.

I am just in the act posting a question.

can I get back to you about working out the math later on.

Thanks

rb75

 

[rb75]

As was posted by @Ibix if you are interested in the mathematics, to find the velocity we need to first find $r(t)$. And as @bob012345 points out gravity isn't going to change that much over that short of a fall? But anyhow...

I could try to work out the math with you.

Starting with Conservation of Energy:

$$ -\frac{GMm}{r_o} = -\frac{GMm}{r} + \frac{1}{2} m \left( \frac{dr}{dt}\right)^2 $$

We have to solve this ODE by separation of variables. Start by isolating $\frac{dr}{dt}$:

$$ \frac{dr}{dt} = \sqrt{2GM} \sqrt{ \frac{1}{r} - \frac{1}{r_o} } $$

Now we use the method of Separation of Variables:

$$ \int \frac{dr}{ \sqrt{ \frac{1}{r} - \frac{1}{r_o} } } = \sqrt{2GM} \int dt $$

Next, we are going to use something called trigonometric substitution. The objective is to try to transform the integral on the left using a variable substitution (in this case using trigonometry)

Define a right triangle with:

hypotenuse $\sqrt{\frac{1}{r}}$, side length opposite the angle $\theta$ is $\sqrt{\frac{1}{r_o}}$, and adjacent side $\sqrt{ \frac{1}{r} - \frac{1}{r_o} }$

We note that the integrand in question can now be transformed into $\tan \theta$ by the multiplication of a constant $\sqrt{\frac{1}{r_o}}$

$$ \int \frac{dr}{ \sqrt{ \frac{1}{r} - \frac{1}{r_o} } } = \sqrt{r_o} \int \tan \theta \, dr $$

Using the method above (the triangle developed) would you like to transform the remaining $dr \to f( \theta ) d\theta$, before we continue?

According to the thread posted by Ibex the only mathematics need or required for the solution to this problem is Newton's equation for gravitational acceleration, and the calculus that will solve for velocity as a function dependent on time. Ibex states after you solve for velocity as a function of time, you can then use the 2nd derivative of velocity to solve for jerk.

The calculus and the solution to solve velocity as a function of time was provided in a previous post by Perok, and involved the problem of a star falling with varying acceleration. Perok's calculus seems pretty straight forward, but appears to have some limitations, in that it requires and depends on knowing the values for two predetermined quanties,

1. The initial predetermined radial distance between the centers of two gravitational attractive bodies $(r_0)$ and,
2. The final predetermined radial distance between the centers of two gravitational attractive bodies $(r_1)$ .

To use Perok's calculus, you must know the values for these two predetermined quantities, $(r_0)$ , and $(r_1)$ , in order to calculate the displacement $(r)$, which then permits solving for velocity as a function of time.

Would it be correct to say, that Perok's calculus, even though it does calculate velocity as a function of time, the calculus is still basically a calculus that based and dependent on a function of distance?

After examination, I don't think Perok's calculus would be the appropriate calculus to use for calculating the quantity of the non-uniform displacement, velocity, acceleration and jerk that is produced by a gravitational force at any instantaneous point in time.

Also wouldn't the accuracy of the calculated values for velocity vary in accuracy with different distances, and depend on the values that are given for the initial and final radial distances?

Using the equation $v = at$, only one predetermined value is required to calculate the velocity at any instantaneous point in time, "acceleration".

Is there a method taught in calculus, that will correctly and accurately calculate the quantity of the non-uniform displacement, velocity, acceleration and jerk, that is produced by an object under the influence of a gravitational attractive force at any instantaneous point in time?
2.

 

[bob012345]

According to the thread posted by Ibex the only mathematics need or required for the solution to this problem is Newton's equation for gravitational acceleration, and the calculus that will solve for velocity as a function dependent on time. Ibex states after you solve for velocity as a function of time, you can then use the 2nd derivative of velocity to solve for jerk.

The calculus and the solution to solve velocity as a function of time was provided in a previous post by Perok, and involved the problem of a star falling with varying acceleration. Perok's calculus seems pretty straight forward, but appears to have some limitations, in that it requires and depends on knowing the values for two predetermined quanties,

1. The initial predetermined radial distance between the centers of two gravitational attractive bodies $(r_0)$ and,
2. The final predetermined radial distance between the centers of two gravitational attractive bodies $(r_1)$ .

To use Perok's calculus, you must know the values for these two predetermined quantities, $(r_0)$ , and $(r_1)$ , in order to calculate the displacement $(r)$, which then permits solving for velocity as a function of time.

Would it be correct to say, that Perok's calculus, even though it does calculate velocity as a function of time, the calculus is still basically a calculus that based and dependent on a function of distance?

No, @PeroK 's solution is perfectly sufficient and valid. I think you do not quite understand what it represents. Those initial and final values are arbitrary.

After examination, I don't think Perok's calculus would be the appropriate calculus to use for calculating the quantity of the non-uniform displacement, velocity, acceleration and jerk that is produced by a gravitational force at any instantaneous point in time.

No, it is completely appropriate.

Also wouldn't the accuracy of the calculated values for velocity vary in accuracy with different distances, and depend on the values that are given for the initial and final radial distances?

No, why would you think that?

Using the equation $v = at$, only one predetermined value is required to calculate the velocity at any instantaneous point in time, "acceleration".

That only works for a constant acceleration. I think you should try to learn how to use Newton's equations assuming constant acceleration first.

Is there a method taught in calculus, that will correctly and accurately calculate the quantity of the non-uniform displacement, velocity, acceleration and jerk, that is produced by an object under the influence of a gravitational attractive force at any instantaneous point in time?

Yes, the method you were already shown. The general case of two bodies attracted by gravity leads to orbital mechanics but this freefall problem is a special case.

 

[erobz]

I don't know where you are getting $r_1$? $r_o$ is the information you need to begin ( along with the mass of the large attracting body ).

I was being hasty in thinking I could get $v(t)$ explicitly...it's not going to play out like I had hoped.

 

[bob012345]

I don't know where you are getting $r_1$? $r_o$ is the information you need to begin ( along with the mass of the large attracting body ).

I was being hasty in thinking I could get $v(t)$ explicitly...it's not going to play out like I had hoped.

I think $r_1$ is just the surface of the earth.

 

[erobz]

You plug in a time $t$, and you must solve for the radius $r$ in PeroK's derivation.

The velocity at that time $t$ will follow from what I started with ( with what @PeroK started with in his second approach). You will plug in that radius and solve for the velocity ( that velocity is the velocity at time $t$ )

The acceleration same thing. Plug in that radius to what Ibix had provided. That will be the acceleration at time $t$.

If you want the "jerk" we take the derivative of the acceleration with respect to $r$ , plug in the radius at time $t$ and multiply it by the velocity at time $t$.

If you want the "jounce" take the derivative of the "jerk", plugging in the necessary quantities

etc...

Oh, and if you are really only interested in the fist 3 seconds of the fall you should do what @bob012345 says. Calculate $g$ in the vicinity of the initial position, and assume it’s constant. You won’t have any “jerk” or higher derivatives with that approach.

 

[malawi_glenn]

You could also just do a numerical calculation.

 

[Ibix]

I was being hasty in thinking I could get $v(t)$ explicitly...it's not going to play out like I had hoped.

Yeah. You can solve for $t(r)$ but looks like you can't get $r(t)$, which is annoying. However, if I'm not mistaken you can get $dt/dr$ from it, which is $1/v(r)$. You've already got $a(r)$. I think that gives you enough to get all the higher derivatives, dependant only on a numerical solver to get $r$ from $t(r)$.

 

[rb75]

I still fail to see the logic behind the differential and integral calculus that was demonstrated on PF using Isaac Newton’s non-uniform gravitational acceleration function $f \left (r \right) = \frac {2GM} {r^2}$, and its derived non-uniform velocity function $f^{'} \left (r \right) = \frac {2GM} {r}$, as the solution to the physics problem, Falling to a Star with Varying Acceleration, and to the problem that I had posted concerning the calculus that would correctly calculate the quantity of the non-uniform displacement and the non-uniform velocity that is produced by the earth’s gravitational force at the end of any specific interval in time.

The following is what I hope, my correct understanding of what is required, and when it is appropriate to use Isaac Newton’s differential and integral calculus for solving problems in physics and mathematics. Comments regarding the correctness or the incorrectness of any of the following would be appreciated, and would help in my attempt to understand concepts in Isaac Newton’s physics and differential and integral calculus.

The Concept of Continuity:
In mathematics continuity means the fact of not stopping or not changing. Continuity refers to something that is occurring on an ongoing uninterrupted state or something that is occurring on an ongoing regular basis.

Definition of Differentiability:
$F\left(x \right)$ is said to be differentiable at the point $x = a$ if the derivative $f^{'} \left (a \right)$ exists at every point in its domain. For a function to be differentiable at any point $x = a$ in its domain, it must be continuous at that particular point.
If a function is continuous at a particular point $x = a$ in its domain, a right and a left hand limit exists at that particular point in its domain. If the right and left hand limits are equal at a particular point $x = a$ in the domain for that function, this means the function is differentiable at that particular point, and Isaac Newton’s methods in differential and integral calculus are an appropriate and correct mathematics to use when solving problems in physics and mathematics involving that function.

When acceleration due to a gravitational force is falsely assumed to have a constant unchanging value, the displacement time graph produced by the uniform velocity function $f^{'} \left ( t \right) = vt$, results in a smooth straight line having a constant slope. The slope of a straight line drawn to each and every adjacent point on the graph of this uniform velocity function has the same constant ongoing uninterrupted state in the slope of the straight line drawn to each and every adjacent point over the entire domain for this function. The ongoing uninterrupted state in the slope of a straight line drawn to each and every adjacent point, demonstrates this uniform velocity function is continuous at each and every point over its entire domain, and that equal right and left hand limits exist at each and every point $x = a$ over the entire domain for this velocity function.

The displacement time graph produced by the uniform acceleration function $f^" \left (t \right) = \frac {1} {2} at^2$ results in a smooth curved line. A straight line drawn to each and every adjacent point on the graph of this uniform acceleration function has the same ongoing regular rate of increase in its slope at each and every point over its entire domain. This ongoing regular rate of increase in the slope of a line drawn to each and every adjacent point, demonstrates the uniform acceleration function is continuous at each and every point over its entire domain and that equal right and left hand limits exist at each and every point $x = a$ over the entire domain for this acceleration function.

The displacement time functions, $f^{’} \left (t \right) = vt$, and $f^" \left (t \right) = \frac {1} {2} at^2$ represent a false concept in physics. These two functions do not represent Isaac Newton physics concerning the quantity of the non-uniform displacement and the non-uniform accelerated motion that is produced by a gravitational attractive force.

Newton’s physics states; the magnitude of the acceleration due to a gravitational attractive force changes in value with each and every change in the radial distance between the centers of the two gravitational attractive bodies and with each and every change in the interval of time. A gravitational force produces a different quantity of non-uniform displacement, and a different quantity of non-uniform accelerated motion, with each and every change in the radial distance between the centers of the two gravitational attractive bodies, and with each and every change in the interval of time.

The graph of the non-uniform velocity function $f^{’} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, both result in a plot line that is not, smooth, curved, or straight. The slope of a straight line drawn to each and every adjacent point on the graphs for these two functions has a different slope and different rate of increase in their slope at each and every point over the entire domain for these two functions. This erratic difference in the slope, and the change in the slope of a line at each and every point, demonstrates that these two functions are not continuous, and that equal right and left hand limits do not exist at any point over the entire domains for these two functions. This means the non-uniform velocity function $f^{’} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, are not differentiable at any point over their entire domains, and that a discontinuity exists at each and every point over the entire range and domain for these two functions.

If my above understanding of a derivative, continuity, equal right and left hand limits, differentiability, and discontinuity is correct, then the non-uniform velocity function $f^{'} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, are not differentiable at any point over their entire domains. This would mean the differential and integral calculus that was demonstrated on PF as a solution to the physics problem, Falling to a Star with Varying Acceleration is inappropriate and incorrect mathematics.

Comments regarding the correctness or the incorrectness of any of the above would be greatly appreciated. Thanks, rb75.

 

[bob012345]

The displacement time functions, $f^{’} \left (t \right) = vt$, and $f^" \left (t \right) = \frac {1} {2} at^2$ represent a false concept in physics. These two functions do not represent Isaac Newton physics concerning the quantity of the non-uniform displacement and the non-uniform accelerated motion that is produced by a gravitational attractive force.

These are approximations used when the gravitational force is almost constant such as near the surface of the earth. It's not constant but changes with height over human scale distances so slowly we can treat it as constant and that is how those simplified functions come about. They are approximations.

 

[erobz]

I still fail to see the logic behind the differential and integral calculus that was demonstrated on PF using Isaac Newton’s non-uniform gravitational acceleration function $f \left (r \right) = \frac {2GM} {r^2}$, and its derived non-uniform velocity function $f^{'} \left (r \right) = \frac {2GM} {r}$, as the solution to the physics problem, Falling to a Star with Varying Acceleration, and to the problem that I had posted concerning the calculus that would correctly calculate the quantity of the non-uniform displacement and the non-uniform velocity that is produced by the earth’s gravitational force at the end of any specific interval in time.

The following is what I hope, my correct understanding of what is required, and when it is appropriate to use Isaac Newton’s differential and integral calculus for solving problems in physics and mathematics. Comments regarding the correctness or the incorrectness of any of the following would be appreciated, and would help in my attempt to understand concepts in Isaac Newton’s physics and differential and integral calculus.

The Concept of Continuity:
In mathematics continuity means the fact of not stopping or not changing. Continuity refers to something that is occurring on an ongoing uninterrupted state or something that is occurring on an ongoing regular basis.

Definition of Differentiability:
$F\left(x \right)$ is said to be differentiable at the point $x = a$ if the derivative $f^{'} \left (a \right)$ exists at every point in its domain. For a function to be differentiable at any point $x = a$ in its domain, it must be continuous at that particular point.
If a function is continuous at a particular point $x = a$ in its domain, a right and a left hand limit exists at that particular point in its domain. If the right and left hand limits are equal at a particular point $x = a$ in the domain for that function, this means the function is differentiable at that particular point, and Isaac Newton’s methods in differential and integral calculus are an appropriate and correct mathematics to use when solving problems in physics and mathematics involving that function.

When acceleration due to a gravitational force is falsely assumed to have a constant unchanging value, the displacement time graph produced by the uniform velocity function $f^{'} \left ( t \right) = vt$, results in a smooth straight line having a constant slope. The slope of a straight line drawn to each and every adjacent point on the graph of this uniform velocity function has the same constant ongoing uninterrupted state in the slope of the straight line drawn to each and every adjacent point over the entire domain for this function. The ongoing uninterrupted state in the slope of a straight line drawn to each and every adjacent point, demonstrates this uniform velocity function is continuous at each and every point over its entire domain, and that equal right and left hand limits exist at each and every point $x = a$ over the entire domain for this velocity function.

The displacement time graph produced by the uniform acceleration function $f^" \left (t \right) = \frac {1} {2} at^2$ results in a smooth curved line. A straight line drawn to each and every adjacent point on the graph of this uniform acceleration function has the same ongoing regular rate of increase in its slope at each and every point over its entire domain. This ongoing regular rate of increase in the slope of a line drawn to each and every adjacent point, demonstrates the uniform acceleration function is continuous at each and every point over its entire domain and that equal right and left hand limits exist at each and every point $x = a$ over the entire domain for this acceleration function.

The displacement time functions, $f^{’} \left (t \right) = vt$, and $f^" \left (t \right) = \frac {1} {2} at^2$ represent a false concept in physics. These two functions do not represent Isaac Newton physics concerning the quantity of the non-uniform displacement and the non-uniform accelerated motion that is produced by a gravitational attractive force.

Newton’s physics states; the magnitude of the acceleration due to a gravitational attractive force changes in value with each and every change in the radial distance between the centers of the two gravitational attractive bodies and with each and every change in the interval of time. A gravitational force produces a different quantity of non-uniform displacement, and a different quantity of non-uniform accelerated motion, with each and every change in the radial distance between the centers of the two gravitational attractive bodies, and with each and every change in the interval of time.

The graph of the non-uniform velocity function $f^{’} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, both result in a plot line that is not, smooth, curved, or straight. The slope of a straight line drawn to each and every adjacent point on the graphs for these two functions has a different slope and different rate of increase in their slope at each and every point over the entire domain for these two functions. This erratic difference in the slope, and the change in the slope of a line at each and every point, demonstrates that these two functions are not continuous, and that equal right and left hand limits do not exist at any point over the entire domains for these two functions. This means the non-uniform velocity function $f^{’} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, are not differentiable at any point over their entire domains, and that a discontinuity exists at each and every point over the entire range and domain for these two functions.

If my above understanding of a derivative, continuity, equal right and left hand limits, differentiability, and discontinuity is correct, then the non-uniform velocity function $f^{'} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, are not differentiable at any point over their entire domains. This would mean the differential and integral calculus that was demonstrated on PF as a solution to the physics problem, Falling to a Star with Varying Acceleration is inappropriate and incorrect mathematics.

Comments regarding the correctness or the incorrectness of any of the above would be greatly appreciated. Thanks, rb75.

And how exactly does $f'(t) = vt$, while $f''(t) = \frac{1}{2}at^2$?

 

[PeroK]

If my above understanding of a derivative, continuity, equal right and left hand limits, differentiability, and discontinuity is correct, then the non-uniform velocity function $f^{'} \left (r \right) = \frac {2GM} {r}$, and the non-uniform gravitational acceleration function $f \left(r\right) = \frac {2GM} {r^2}$, are not differentiable at any point over their entire domains. This would mean the differential and integral calculus that was demonstrated on PF as a solution to the physics problem, Falling to a Star with Varying Acceleration is inappropriate and incorrect mathematics.

Comments regarding the correctness or the incorrectness of any of the above would be greatly appreciated. Thanks, rb75.

It's not correct at all. The gravitational force function is differentiable for all $r \ne 0$.

 

[rb75]

Hello Perok

I thought I was beginning to understand some basic concepts in Newton’s physics and differential and integral calculus i.e. the concept of continuity and how it affects differentiability, but apparently my understanding is totally wrong. If you have the time, could you clarify and explain what is wrong in my understanding regarding each of the following?

1. In mathematics continuity means the fact of not stopping or not changing. Continuity refers to something that is occurring on an ongoing uninterrupted state or something that is occurring on an ongoing regular basis.
2. F(x) is said to be differentiable at the point x = a, if the derivative $f^{'} \left (a \right)$ exists at every point in its domain.
3. If the derivative of a function is continuous at a particular point x =a in its domain, equal right and a left hand limits exist at that particular point, and the function is differentiable at that particular point x = a.
4. The slope of a tangent line at any point on the graphs of the non-uniform acceleration function f(r) = GM/r2, $f \left (r \right) = \frac {2GM} {r^2}$, and its derivative non-uniform velocity function f’(r) = GM/r, function $f^’\left (r \right) = \frac {2GM} {r}$, does not have an ongoing uninterrupted state in the value of its slope, nor does it have an ongoing regular rate of change in the value of its slope. The slope of the tangent line at each point on the graphs for these two functions has a different slope and a different rate of change in its slope over the entire range and `domain for these two functions. The irregular value and the irregular rate of change in the value of the slope of the tangent line at every point on the graphs for these two functions means, these two functions are not continuous at any point x =a in their domain, therefore they are not differentiable at any point x = a in their domain.

Thanks. rb75

 

[PeroK]

Hello Perok

I thought I was beginning to understand some basic concepts in Newton’s physics and differential and integral calculus i.e. the concept of continuity and how it affects differentiability, but apparently my understanding is totally wrong. If you have the time, could you clarify and explain what is wrong in my understanding regarding each of the following?

1. In mathematics continuity means the fact of not stopping or not changing. Continuity refers to something that is occurring on an ongoing uninterrupted state or something that is occurring on an ongoing regular basis.

Those are just words that could mean anything or nothing.

2. F(x) is said to be differentiable at the point x = a, if the derivative $f^{'} \left (a \right)$ exists at every point in its domain.

That's a confusion of ideas and notation. You can check online for the definitionof derivative.

3. If the derivative of a function is continuous at a particular point x =a in its domain, equal right and a left hand limits exist at that particular point, and the function is differentiable at that particular point x = a.

that makes no sense either.

4. The slope of a tangent line at any point on the graphs of the non-uniform acceleration function f(r) = GM/r2, $f \left (r \right) = \frac {2GM} {r^2}$, and its derivative non-uniform velocity function f’(r) = GM/r, function $f^’\left (r \right) = \frac {2GM} {r}$, does not have an ongoing uninterrupted state in the value of its slope, nor does it have an ongoing regular rate of change in the value of its slope. The slope of the tangent line at each point on the graphs for these two functions has a different slope and a different rate of change in its slope over the entire range and `domain for these two functions. The irregular value and the irregular rate of change in the value of the slope of the tangent line at every point on the graphs for these two functions means, these two functions are not continuous at any point x =a in their domain, therefore they are not differentiable at any point x = a in their domain.

That's unintelligible.

Thanks. rb75

I'm not sure where you're learning calculus, but generally you have a confusion of ideas.

 

[jbriggs444]

I am not @PeroK, but can answer for him.

1. In mathematics continuity means the fact of not stopping or not changing. Continuity refers to something that is occurring on an ongoing uninterrupted state or something that is occurring on an ongoing regular basis.

This sounds fairly accurate. A function is "continuous" if its graph has no sudden jumps. The usual way of introducing the idea is that if you can draw the graph without lifting your pencil from the paper then the function is continuous.

The formal definition for continuity at a point is that the function must be defined at that point. Further, the function values on either side must approach the value of the function at that point.

https://en.wikipedia.org/wiki/Continuous_function

Edit: However, it seems that you take the notion of "continuing or uninterrupted regular basis" differently than I'd expected. By that you mean "is constant or is linear". But that is absolutely not what continuity is about.

2. F(x) is said to be differentiable at the point x = a, if the derivative $f^{'} \left (a \right)$ exists at every point in its domain.

For $F(x)$ to be differentiable at a point only requires that the derivative be defined at that point. Not everywhere.

3. If the derivative of a function is continuous at a particular point x =a in its domain, equal right and a left hand limits exist at that particular point, and the function is differentiable at that particular point x = a.

This gets into finicky details of definition. If the derivative is undefined throughout a neighborhood of point $x = a$ except that is defined at $x=a$ exactly then strictly speaking, the derivative is automatically continuous at that point.

All functions are "continuous" at isolated points in their domain.

However, if we demand that the derivative be defined everywhere in the domain of $f$ and if $x=a$ is not an isolated point in the domain of $f$ on either the right or left hand sides then yes, continuity for $f'(x)$ at $x=a$ demands that the right and left hand limits for $f'$ at $x=a$ must match $f'(a)$.

4. The slope of a tangent line at any point on the graphs of the non-uniform acceleration function $f \left (r \right) = \frac {2GM} {r^2}$, and its derivative non-uniform velocity function $f’\left (r \right) = \frac {GM} {r}$, does not have an ongoing uninterrupted state in the value of its slope, nor does it have an ongoing regular rate of change in the value of its slope.

[$\LaTeX$ repaired, and hopefully corrected]

I am having trouble understanding what you are saying here.

I think that you mean that the acceleration function $f(r) = \frac{2GM}{r^2}$ does not have a constant derivative (with respect to $r$). Further, it does not have a derivative with respect to $r$ which is a linear function of $r$.

Indeed, that much is true. The first derivative of $\frac{2GM}{r^2}$ would be $-\frac{4GM}{r^3}$ which is neither constant nor constantly changing.

You also mean that the "velocity function" $f'(r) = \frac{GM}{r}$ [which is actually the negative of the integral, not the derivative] also is neither constant nor is it a linear function of $r$. Nor do you get velocity when you integrate acceleration over distance. Instead, the path integral of acceleration over distance is something very much like "work" and delivers something akin to energy, but without the factor of $m$.

In any case, the first derivative of $-\frac{GM}{r}$ with respect to $r$ is $\frac{2GM}{r}$ which is, as you have correctly pointed out neither constant nor linear in $r$.

You seem to be well off the rails now, but let us continue.

The slope of the tangent line at each point on the graphs for these two functions has a different slope and a different rate of change in its slope over the entire range and `domain for these two functions. The irregular value and the irregular rate of change in the value of the slope of the tangent line at every point on the graphs for these two functions means, these two functions are not continuous at any point x =a in their domain, therefore they are not differentiable at any point x = a in their domain.

This is completely wrong. Non-constant functions with non-constant and non-linear derivatives can be continuous and continuously differentiable.

For example, $f(x) = \sin x$ is continuous and continuously differentiable to all orders. But none of those infinitely many n'th level derivatives (or integrals for that matter) are ever constant or linear.