Exact Integration of Newton's Gravitational Law?

[bob012345]

TL;DR Summary

How to do the integral for an object falling under Newton's Gravitational Law without approximating a constant acceleration of gravity $g$.

I realized I never actually derived the kinematic equations of motion for the exact Newtonian gravitational force. For an object falling near the surface of the earth, how do we handle integrating the equation of motion to derive the kinematics equations without using the approximation of constant g?

Starting with $\large \frac{d^2y}{dt^2} = -\frac{GM}{y^2}$, how do we untangle this? Then simplify the solution with the assumption the acceleration is constant to get the usual equation.

 

[Dale]

I wonder if the overall almost-spherical shape is a bigger deviation from the uniform assumption or if local features like mountains are bigger.

 

[Dr.D]

For the eq of motion given, there should be an energy first integral available. Why not give that a shot to see where it takes you?

 

[bob012345]

For the eq of motion given, there should be an energy first integral available. Why not give that a shot to see where it takes you?

Not sure what you mean. Do you mean starting with $\large \frac{dv}{dt}=−\frac{GM}{y^2}$?

 

[kuruman]

No. Start with $$-\frac{GMm}{r_0}=-\frac{GMm}{r}+\frac{1}{2}m\left(\frac{dr}{dt}\right)^2$$ and work on the differential equation $\dfrac{dr}{dt}=\dots$

 

[Delta2]

It is well known fact that this differential equation has no closed form solution for $y(t)$.
Best you can do is solve for the velocity as function of distance y, $v(y)$. Towards this end write $$\frac{d^2y}{dt^2}=\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}=v\frac{dv}{dy}$$, so the differential equation becomes $$v\frac{dv}{dy}=-\frac{GM}{y^2}$$ which you can solve for $v(y)$ by the method of separation of variables.

 

[bob012345]

It looks like I am being offered two paths. Thanks. I'll try both. I actually tried the first path (post #5) but abandoned it when I got confused.

 

[bob012345]

It is well known fact that this differential equation has no closed form solution for $y(t)$.
Best you can do is solve for the velocity as function of distance y, $v(y)$. Towards this end write $$\frac{d^2y}{dt^2}=\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}=v\frac{dv}{dy}$$, so the differential equation becomes $$v\frac{dv}{dy}=-\frac{GM}{y^2}$$ which you can solve for $v(y)$ by the method of separation of variables.

$$v\frac{dv}{dy}=-\frac{GM}{y^2}$$ gives solutions $$v(y) = ±\sqrt{C + \frac{2GM}{y}}$$ Then I can use $$\frac{dy}{dt} = ±\sqrt{C + \frac{2GM}{y}}$$

 

[Delta2]

Yes that's correct.

 

[bob012345]

No. Start with $$-\frac{GMm}{r_0}=-\frac{GMm}{r}+\frac{1}{2}m\left(\frac{dr}{dt}\right)^2$$ and work on the differential equation $\dfrac{dr}{dt}=\dots$

Meanwhile, this is of the form $$ E =\frac{mv^2}{2} + V(y)$$ so going with and changing $r →y$ $$-\frac{GMm}{y_0}=-\frac{GMm}{y}+\frac{1}{2}m\left(\frac{dy}{dt}\right)^2$$ gives us

$$ \frac{dy}{dt} = \sqrt{\frac{2}{m}(-\frac{GMm}{y_0} +\frac{GMm}{y}}$$ or$$ \frac{dy}{dt} = ±\sqrt{-\frac{2GM}{y_0} +\frac{2GM}{y}}$$ which is identical to the first path if $C$ is identified as $-\frac{2GM}{y_0}$

 

[Delta2]

Yes but the point is to go one step further and solve for $y(t)$ if you can, but you can't (at least imo), because after separation of variables and doing the integration you ll get something like $$f(y)=t$$ but$f$ is going to be so complex that you won't be able to find its inverse $f^{-1}$, so that you can write $$y=f^{-1}(t)$$.

 

[bob012345]

Yes but the point is to go one step further and solve for $y(t)$ if you can, but you can't (at least imo), because after separation of variables and doing the integration you ll get something like $$f(y)=t$$ but f is going to be so complex that you won't be able to find its inverse $f^{-1}$, so that you can write $$y=f^{-1}(t)$$.

Of course, I wasn't done. I'm going to try to solve it. Or at least find a published solution path. After all, now that I recognize it, this is just a one-dimensional version of the famous two-body central force problem which has been solved ~400 years ago by Newton himself.

But even if I can't I should be able to see how it reduces to the standard kinematic equations in regions where g is essentially constant over the problem.

 

[bob012345]

Starting with $$ \frac{dr}{dt} = ±\sqrt{\frac{2GM}{r} -\frac{2GM}{r_0}}$$ and rearranging to get

$$±{ \int_{r_0}^{r} }\frac{dr}{ \sqrt{ 2GM( \Large {\frac{1}{r} -\frac{1}{r_0}} )} } = \int_{0}^{t}dt$$

this gives for the definite integral (note $r ≤ r_0)$;

$$±-\dfrac{\sqrt{r_0}\left(\sqrt{r_0-r}\sqrt{r}+r_0\arctan\large\left(\frac{\sqrt{r_0-r}}{\sqrt{r}}\right)\right)}{\sqrt{2}\sqrt{GM}} = t$$

 

[Dr.D]

In #1, the OP made reference to the "kinematic equations of motion" but I suggest to you that is a misnomer. By definition, kinematics is the study of motion without regard to forces. But notice that the starting equation for most of this thread was an application of Newton's 2nd law, a force relation.

I'm well aware that many call the so-called SUVAT equations "kinematic equations" but in reality, they are derived from Newton's 2nd law, a force relation. Calling them kinematics does not really make them so.

 

[bob012345]

Picking up again we have;
$$±{\sqrt{r_0}\left(\sqrt{r_0-r}\sqrt{r}+r_0\arctan\left(\frac{\sqrt{r_0-r}}{\sqrt{r}}\right)\right)}= {(\sqrt{2GM}}) \large t$$

Given some values for $G, M$ and $r_0$ we can compute $t$ for a given $r$ and compare it to the constant $g$ case using that time. It seems the assumption of constant $g$ leads to slightly slower falling if I understand this correctly. (Update: It's backwards. The constant $g$ number in this table are wrong due to using a different value of $M_e$. They should be a bit larger. It is corrected in the table in a later post.)

Height m	Time s	Height @Constant g
1	0.45031	0.99665
2	0.63684	1.99333
3	0.77997	2.99002
4	0.90063	3.98670
5	1.00693	4.98329
10	1.42402	9.96667
20	2.01387	19.9334
50	3.18420	49.8332
100	4.50313	99.666
1000	14.2398	996.61

$G = 6.67x10^{-11}$, M = $5.98x10^{24} kg$, Radius of Earth = $6.37x10^6 m$, Constant $g = 9.8299 \frac{m}{s^2}$

 

[bob012345]

Picking up again I want to reduce this form to the constant g case. ;
$$±{\sqrt{r_0}\left(\sqrt{r_0-r}\sqrt{r}+r_0\arctan\left(\frac{\sqrt{r_0-r}}{\sqrt{r}}\right)\right)}= {(\sqrt{2GM}}) \large t$$

We can use the small angle approximation for the $\arctan$ term since it is a very small term ~$10^{-4}$;
$$±{\sqrt{r_0}\left(\sqrt{r_0-r}\sqrt{r}+r_0\left(\frac{\sqrt{r_0-r}}{\sqrt{r}}\right)\right)}= {(\sqrt{2GM}}) \large t$$

Both $r$ and $r_0$ are very large but only the difference matters. We can approximate $\sqrt{r}$ with $\sqrt{r_0}$ in the range near the surface of the earth;

$$±{\sqrt{r_0}\left(\sqrt{r_0-r}\sqrt{r_0}+r_0\left(\frac{\sqrt{r_0-r}}{\sqrt{r_0}}\right)\right)}= {(\sqrt{2GM}}) \large t$$

This gives;
$$±{{r_0}\left(\sqrt{r_0-r}+\left({\sqrt{r_0-r}}{}\right)\right)}= {(\sqrt{2GM}}) \large t$$
or;
$$±{{r_0}\left(\sqrt{r_0-r}+{\sqrt{r_0-r}}\right)}= {(\sqrt{2GM}}) \large t$$
then;
$$±{{r_0}\left(2{\sqrt{r_0-r}}\right)}= {(\sqrt{2GM}}) \large t$$
Finally we can square both sides;
$$4{r_0}^2 (r_0-r)= {2GM} \large t^2$$

$$ r_0-r = \frac {GM}{2{r_0}^2} \large t^2$$
but since $r_0$ is the radius of the Earth plus a small delta ${r_0}^2 = (R_e + h)^2$ or ${R_e}^2(1+\frac{h}{R_e})^2$≈${R_e}^2$
we can write
$$ r_0-r = \frac {g}{2} \large t^2$$

or $$ r = r_0 - \frac {g}{2} \large t^2$$

since $\large\frac{GM}{{R_e}^2} = g$ and $v_0$ was assumed zero.

 

[kuruman]

What's with the $\pm$ in your equation $$\frac{dr}{dt} = ±\sqrt{\frac{2GM}{r} -\frac{2GM}{r_0}}~?$$When you release a mass from rest in a gravitational field, it moves towards the force center, i.e. $\dfrac{dr}{dt}<0$. Think like a physicist and not like a mathematician and choose the bottom sign in this and all the other equations that follow. This will also fix the entries in your post #15 table which is pure nonsense as it stands. It shows that the height of the dropped mass increases as time increases. Long before Galileo's tower of Pisa experiment at the end of the 16th century people knew that this isn't the case.

 

[bob012345]

What's with the $\pm$ in your equation $$\frac{dr}{dt} = ±\sqrt{\frac{2GM}{r} -\frac{2GM}{r_0}}~?$$When you release a mass from rest in a gravitational field, it moves towards the force center, i.e. $\dfrac{dr}{dt}<0$. Think like a physicist and not like a mathematician and choose the bottom sign in this and all the other equations that follow. This will also fix the entries in your post #15 table which is pure nonsense as it stands. It shows that the height of the dropped mass increases as time increases. Long before Galileo's tower of Pisa experiment at the end of the 16th century people knew that this isn't the case.

Height is the independent variable. Height in the table is not a function of time. I thought the obvious meaning was the height the object was dropped from. It's just a table of different heights.

As for the ±, yes, I was being mathematically precise precisely because there is sometimes a tendency to be criticized which I was trying to avoid.

 

[kuruman]

Height is the independent variable. Height in the table is not a function of time. I thought the obvious meaning was the height the object was dropped from. It's not pure nonsense.

My mistake. I misunderstood what the table displayed. Thank you for setting me straight.

 

[bob012345]

My mistake. I misunderstood what the table displayed. Thank you for setting me straight.

It's ok. I also could have been clearer.

 

[bob012345]

What's with the $\pm$ in your equation $$\frac{dr}{dt} = ±\sqrt{\frac{2GM}{r} -\frac{2GM}{r_0}}~?$$When you release a mass from rest in a gravitational field, it moves towards the force center, i.e. $\dfrac{dr}{dt}<0$.

After further thought, you bring up an interesting point, the boundary between the math and the physics when setting up a problem. In my case, if I had chosen a negative sign it would have been wiped out when I squared both sides. The signs seemed implicitly chosen by having both $r$ and $r_0$ originate from the center of the Earth and the signs of the energies in the original equation. For a constant $g$ case, I could choose the positive direction up or down but ultimately I would have to put the acceleration sign in by choice to agree with that choice.

 

[Delta2]

It seems the assumption of constant g leads to slightly slower falling if I understand this correctly.

I would expect the opposite, because we take g to be the acceleration at the surface of the earth. At an altitude of h from the surface of the Earth this acceleration becomes smaller $G\frac{M}{(R_e+h)^2}$. Smaller acceleration would have to mean slower falling than the case of constant g.
Which brings up the point how exactly did you calculate the table at post #15? Did you put $r_0=R_e$ where $R_e$ the radius of earth, and then put r=1,2,etc and found the corresponding values of t? If yes then r in this case is not the height above the surface of the earth, it is rather the distance the body would go below the surface of the Earth if we assumed all of Earth's mass was concentrated at its center.

 

[bob012345]

I would expect the opposite, because we take g to be the acceleration at the surface of the earth. At an altitude of h from the surface of the Earth this acceleration becomes smaller $G\frac{M}{(R_e+h)^2}$. Smaller acceleration would have to mean slower falling than the case of constant g.
Which brings up the point how exactly did you calculate the table at post #15? Did you put $r_0=R_e$ where $R_e$ the radius of earth, and then put r=1,2,etc and found the corresponding values of t? If yes then r in this case is not the height above the surface of the earth, it is rather the distance the body would go below the surface of the Earth if we assumed all of Earth's mass was concentrated at its center.

I left $r = R_e$ and let $r_0 = R_e + h$ for each height. I think that is correct.

But you are correct. The problem was much simpler. I think I used a very slightly different number for $M_e$ in the constant $g$ part which reduced their numbers. The number I used for this table and in the table above for the non-linear formula is $M_e = 6x10^{24} kg$. The hand calculation for the constant $g$ in the first table I must have used $M_e = 5.98x10^{24} kg$. I apologize.

Height Dropped m	Time of fall (s)	Const. g Height Delta m	% Height Difference
1	0.450315	+2.4202E-06	2.42E-04
2	0.636841	+1.1903E-06	5.95E-05
3	0.779968	+3.7266E-06	1.24E-04
4	0.900629	+7.9818E-07	1.99E-05
5	1.006934	+2.6536E-06	5.31E-05
10	1.424021	+2.3296E-05	2.33E-04
20	2.013873	+1.09926E-04	5.50E-04
50	3.184225	+6.5798E-04	1.32E-03
100	4.503203	+2.5959E-03	2.60E-03
1000	14.24206	+2.62175E-01	2.62E-02

 

[bob012345]

I wonder if the overall almost-spherical shape is a bigger deviation from the uniform assumption or if local features like mountains are bigger.

Probably mountains because the rate of change of an almost spherical shape would likely be smaller than the local change due to mountains because the bulge would be evenly distributed whereas a mountain on an otherwise spherical Earth is abrupt if we consider $\frac{dg}{d\theta}$.

 

[Dale]

Probably mountains because the rate of change of an almost spherical shape would likely be smaller than the local change due to mountains because the bulge would be evenly distributed whereas a mountain on an otherwise spherical Earth is abrupt if we consider ##\frac{dg}{d\theta}.

I haven't actually done the math, but that would be my guess too.

 

[bob012345]

I haven't actually done the math, but that would be my guess too.

Back-of-the envelope for a cone shaped mountain about 2km wide and 1km high with an average density of the crust, I get a disturbance in gravity of about 20ppm $g$ (laterally) measured at the base which goes to nothing a few km away. The flattening of the poles due to rotation gives roughly 1ppm/mile from the equator to the pole due to the shape change of about 21km and about the same from the actual rotation not regarding direction of the disturbance.