User:Nedim Ardoğa/Free fall

I want to share an interesting question about Celestial Mechanics. I deeply feel gratitute to two Wikipedians who helped me to solve the problem. Long ago I've posted on the talk page of Gravitation accelaration the following question:

''In acceleration problems acceleration is always given to be uniform. However, for a falling particle from a remote point this is not so. Gravitational acceleration ( g ) depends on the distance ( r ), the distance depends on the velocity which in turn depends on acceleration. For example what is the falling time of a small mass in free fall ( Vin = 0 ) from 150 000 000 km (earth orbit radius) to Sun ? ( G • M for Sun is approximatelly 1.327• 1020 in SI units) Certainly, $$\mathbf{r}= g \cdot t^2 /2$$ is not applicable in this case.(non uniform acceleration) This problem can be solved by numerical methods. But I'd like to know if there is an analytic solution for such distance dependent acceleration case (maybe with the help of NL differential equations) ?''

Solution
Two solutions were suggested. To save time and space I'll link to discussions.


 * Norbeck who contributed to Page Free fall proposed to use his equation. Well I did and I obtained a simple answer in his Talk Page.
 * $$\mathbf {t} = \sqrt{\frac{1}{32}} \cdot T$$ where t is the free fall time and T is the orbital period.


 * Quite unexpectedly on the same page, I saw a message from SBharris which relates the orbital period to falling time using only geometry.

I’ll try to explain it:

The orbit of the Earth (like all other heavenly bodies) is not exactly circular. It is elliptical and the Sun is in one of the foci of the ellipse. Kepler’s 3rd law relates the orbital period to semimajor axis of the elipse


 * $$T^2=A^3\cdot\frac{4\cdot\pi^2} {\mu}$$

Where T is the orbital period, A is the semi major axis and µ is the product of gravitation constant and the mass of the Sun.

This relation holds no matter what the eccentricity is. Now suppose that the eccentricty approaches to unity and the orbit is nothing but a straight line between the Earth and the Sun. The relation is still valid. But in this case the semimajor axis is the half of the case where the orbit is circular. Using lowercase letters for e=1 case;


 * $$a=\frac{A}{2}$$

Now the orbital period is


 * $$t^2=a^3\cdot\frac{4\cdot\pi^2} { \mu}= \frac{A^3}{8}\cdot\frac{4\cdot\pi^2} { \mu}$$

Using the above relations ratio of $$t/T$$''' is


 * $$ \frac{t}{T}=\sqrt \frac{1}{8}$$

Since falling is half of a complete period


 * $$ \frac{t}{T}=\frac {1}{2}\sqrt \frac{1}{8}= \sqrt \frac{1}{32}$$

Well, what is the answer ? Since the orbital period of the Earth is about 31 557 000 seconds, the fall time can be found to be
 * 5 578 000 seconds or 64 days 13 hours .

Equivalent acceleration
The above relation for constant acceleration can also be used.
 * $$\mathbf{r}= g \cdot t^2 /2$$

But in this case, instead of the acceleration calculated at the 0 distance (g0) an equivalent acceleration (gav) must be used such that


 * $$ g_{av}=\frac{g_0 \cdot 16}{\pi ^2}$$

Note that this relation is independent of Solar mass and can be used for all stelar systems.