• Will Flannery
In summary, The freefall wiki entry wiki Freefall has an analytic solution for freefall distance in a gravitational field, but ... it doesn't seem to work ... at least i can't get it to work ... here is my MATLAB program to test it ...clear
Will Flannery
The freefall wiki entry wiki Freefall has an analytic solution for freefall distance in a gravitational field, but ... it doesn't seem to work ... at least i can't get it to work ... here is my MATLAB program to test it ...
clear
G=6.7e-11; % gravitational constant m^3/(kg*s^2)
mEarth = 5.9742e24; % mass of Earth in kg
rEarth = 6.378e6; % radius of Earth in m

% drop a 1kg ball from 100m above the surface of the earth
mu = G*(mEarth + 1);
y0 = rEarth + 100;

t = 1;
x = (3/2*(pi/2-t*sqrt(2*mu/y0^3)))*2/3;
y1 = y0*(x - x^2/5 - 3*x^3/175 - 23*x^4/7875 - 1894*x^5/3931875 - 3293*x^6/21896875 - 2418092*x^7/62077640625);

y0-y1

ans =

9.5639e+04

So, the apple fell 95000 m in the 1st second ... (or I made a mistake)
I tried to check the references, the first has a lot of formulas but not the one above, and the second is behind paywall (The Physics Teacher) and the abstract does not look promising.

Is there an analytic formula anywhere ?

Last edited:
Your definition for x has a typo in the exponent at the end. As an aside, I recall a lot of these series having terrible convergence properties. You might try doing the calculation truncating the inverted series at different powers to see if the answer varies wildly.

Thanks, corrected ...
x = (3/2*(pi/2-t*sqrt(2*mu/y0^3)))^2/3;

However, now the apple is falling upwards ...

y0-y1 = ans =

-2.2703e+04

The last term in the series ... 2418092*x^7/62077640625
ans =

0.0028

but ... *y0 = 1.8177e+04

So, it's a long way from converging ... However, as t increases, x decreases and convergence improves, so when t = 500 the last term (including multiplication by y0) is 0.1935 ! So, is it converging to the true value ?

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Try ^(2/3). a^2/3 is interpreted as ##a^2/3##, but you want ##a^{2/3}##.

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davenn
there are a couple of old threads here where an analytical solution for position vs time is derived. It is messier than one might imagine.

Ibix said:
Try ^(2/3). a^2/3 is interpreted as ##a^2/3##, but you want ##a^{2/3}##.
By Jove ! ... and I've only been Matlabbing for 100 years !

G=6.7e-11; % gravitational constant m^3/(kg*s^2)
mEarth = 5.9742e24; % mass of Earth in kg
rEarth = 6.378e6; % radius of Earth in m

mu = G*(mEarth + 1);
y0 = rEarth;

t = 1;
x = (3/2*(pi/2-t*sqrt(2*mu/y0^3)))^(2/3);
y1 = y0*(x - x^2/5 - 3*x^3/175 - 23*x^4/7875 - 1894*x^5/3931875 ...
- 3293*x^6/21896875 - 2418092*x^7/62077640625);

y0-y1 = -2.9134e+04
y0*2418092*x^7/62077640625 = 1.3488e+04

so it's still not converging on the high end ...
for t = 500
y0 -y1 = 1.3217e+06
y0*2418092*x^7/62077640625 = 296.6925

So, it's better but still not so good.
Conclusion - there are not enough terms in this thing to tell if it works or not (or there is yet another bug ... )

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vanhees71 said:
Just use LaTeX, and one can read you math!

https://www.physicsforums.com/help/latexhelp/
That's not math, that's MATLAB.

In any case, now the question is, why does every book ignore the problem of an analytical solution for the 1-D gravity problem (as noted in the ref. in the wiki article, the guy looked at 100 physics books) and what is the solution?

From the reference in the wiki article, the approach to a solution is thru Kepler's eq. Note that when eccentricity e approaches 1 in Kepler's eq, the orbit does not approach a parabola, Kepler's original eq. only works for ellipses, and as e -> 1 the orbit gets flatter and flatter and when e = 1 you get a straight line !, I just plotted one, So, apparently we can use either a numeric or analytic solution to Kepler's eq, with e = 1, to plot a 1-d trajectory. And so the questions for the day is ... how?

So, if we drop a ball from a height of 100m, a = 100+rEarth and we can plot the trajectory ...
(1.5 hours later) ... that seems to work ...
clear
G=6.7e-11; % gravitational constant m^3/(kg*s^2)
mEarth = 5.9742e24; % mass of Earth in kg
rEarth = 6.378e6; % radius of Earth in m

a = 100+rEarth;
T = sqrt(a^3 /(G*mEarth/(4*pi^2))); % from Kepler's 3rd Law
e = 1;

N=100;
for i = 1:4
E = 2*pi*(i-1)/N + pi; % Eccentric anomaly, start at apogee
M = E - e*sin(E); % Mean anomaly
t(i) = M*T/(2*pi)-T/2; % M(t) = 2*pi*t/T
x(i) = a*cos(E);
end

plot(t,-x-rEarth) % invert graph

weirdoguy
Why are you writing when you don't want to be read? SCNR.

vanhees71 said:
Why are you writing when you don't want to be read? SCNR.
? That doesn't make sense, I want it to be read... even debugged if necessary ...

Generally, things become much clearer when you actually have to program them and check if they work ... For example, it took me over an hour, and many corrections, to get that program to work.

If it's code, put it in code tags.
Matlab:
X=1 % It's even got a syntax highlighter

Nugatory and vanhees71
If it's code, put it in code tags.
Code:
SCNR

EDIT: after thinking for a few more minutes, I think my concern is bogus. I think my post isn't worth reading, but don't like to delete posts since it feels dishonest...

I know this is a very old thread, but I just stumbled across it. If you still care about this it seems to me that you have a precision problem. Matlab uses double precision by default, so mEarth+1=mEarth. The smallest mass you could use and not get this problem will be the order of 1e9 kg. Now the value of ##\mu## doesn't change the result, but you may have similar problems with other portions of this calculation as well. Have you looked at the orders of magnitude of each of the terms in your sum?

I have certainly run into cases like this in my work on numerous occasions, where a pretty analytical formula is completely useless for practical calculations. Sometimes using appropriate numerical techniques is much more practical. jason

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Delta2

An analytical solution for freefall is a mathematical equation that describes the motion of an object under the influence of gravity. It takes into account the initial position, velocity, and acceleration of the object to determine its position at any given time during the fall.

## 2. How is an analytical solution for freefall different from a numerical solution?

An analytical solution for freefall provides a precise equation that can be used to calculate the position of an object at any time during the fall. A numerical solution, on the other hand, uses a computer program to approximate the position of the object at specific time intervals.

An analytical solution for freefall assumes that the object is falling in a vacuum, there is no air resistance, and the acceleration due to gravity is constant. It also assumes that the object is a point mass with no dimensions.

## 4. How can an analytical solution for freefall be applied in real-world situations?

An analytical solution for freefall can be used to calculate the trajectory of objects in freefall, such as a skydiver or a falling object. It can also be used to determine the time of impact and the velocity at impact, which can be important in engineering and safety applications.

## 5. Are there any limitations to using an analytical solution for freefall?

Yes, there are limitations to using an analytical solution for freefall. It assumes ideal conditions and does not take into account factors such as air resistance, wind, or the shape and size of the falling object. In real-world situations, these factors can affect the accuracy of the solution.

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