# Why should the trick solution to the problem of how long it takes to fall into the sun work?

There is a famous problem which asks:

Suppose an object is held at 1 au from the Sun and released from rest. How long does it take to fall into the Sun (neglecting the size of the sun)?

The trick solution is to imagine the trip into the sun as the limit of an elliptical orbit where the semi-minor axis of the orbit goes to zero. In that limit, the sun is at one focus of the ellipse, but also at the extreme end of the ellipse, and so it's an orbit with semi-major axis of 1/2 au, so the time taken is $2^{-3/2}$ years.

This is correct, but why?

If we drop the object straight at the sun (and imagine that it can pass through the sun), then it should pass through the sun in a straight line, continue out the other side, and eventually wind up 1 au away from the sun on the opposite side.

In the thin elliptical orbit, the behavior is totally different. The object only just barely makes it past the sun at all before returning in the direction it came.

Additionally, the time to return to the starting point is different in the two scenarios. If we drop the object straight towards the sun, then when it gets to the sun it has completed 1/4 a period, whereas for the thin elliptical orbit it has covered half a period.

So without doing a more direct calculation of the time for the object to fall into the sun, why should we believe that the "limit of a thin orbit trick" works?

• At every point on the two trajectories the velocity of the object is the same. This follows from conservation of energy. Hence the time evolution of the two trajectories must be the same. – John Rennie Sep 10 '18 at 19:04
• @JohnRennie My question claimed that one of them takes twice as long as the other to go back to the starting point. Are you denying it in your comment? – Mark Eichenlaub Sep 10 '18 at 20:22
• Oops, yes, sorry I read your question too hastily. The Sun in effect reflects the trajectory, so the ellipse is precisely half the trajectory you'd get if the falling object could pass straight through the Sun. – John Rennie Sep 11 '18 at 4:17

If you let the sun have its actual size and density structure (or its actual size but a uniform density), but imagine that your object can pass through without interacting, then it is true that it would shoot out the other side, but this would not be a Keplerian orbit--the object spends time inside the sun, and thus not in a $\frac{1}{r}$ gravitational potential. So you can't use Kepler's laws to calculate the period of the oscillation.

However, the entire part of the journey you care about takes place outside the sun, where the "orbit" is Keplerian. Until it reaches the surface of the sun, the motion of the object is the same as it would have been if all the sun's mass were concentrated at one point, in which case (ignoring GR) you'd get the "thin orbit" you describe. Neglecting the tiny fraction of the total time of the hypothetical thin orbit around the point sun which would occur inside the boundaries of the actual sun, you can approximate the time from release to reaching the sun's surface as half a period of this thin orbit.

Also: the period of the thin orbit is, as you say, $2^{-3/2}$ years, making the time to fall into the sun half that, or $2^{-5/2}$ years.

If we drop the object straight at the sun (and imagine that it can pass through the sun), then it should pass through the sun in a straight line, continue out the other side, and eventually wind up 1 au away from the sun on the opposite side.

You need to be more precise about how you "imagine that it can pass through the Sun". If we are compressing the Sun's mass down to an ideal point, then the object's velocity and kinetic energy diverge as it hits the Sun, so its trajectory ceases to be well-defined. So it's not actually the case that it passes through, slows down symmetrically, and ends up at 1 AU on the far side. In order to get that result, you need to impose some kind of regulator that modifies the $1/r^2$ attraction at short distances, which (as you correctly note) causes the ellipse trick to no longer work.

• This isn't a serious problem. Just let the sun have finite radius, and let the elliptical orbit be such that the dropped comes within two radii of the sun at perigee, and then let the size of the sun go to zero. Both the elliptical orbit and the path of the object dropped straight into the sun are well-defined and they converge to the same time, even though the behavior is different in the way I originally described. – Mark Eichenlaub Sep 10 '18 at 20:21
• @MarkEichenlaub I don't understand your response. The ellipse trick gives the exact answer in the artificial limit where the sun is a true point particle, in which case the dropped object does not pass through it. It does not give the correct answer in the (physically realistic) situation where the Sun has finite radius, in which case the object does pass through it. But it gives the approximately correct answer for the trajectory until the particle enters the Sun, at which point it stops being even qualitatively correct, because the point approximation fails completely inside the Sun. – tparker Sep 10 '18 at 21:03
• The two solutions (physically realistic and point-idealized/ellipse trick) agree very closely as long as they're in the appropriate regime (outside the Sun). Once the particle enters the Sun they diverge completely, so there's no reason to assume the ellipse trick trajectory ("bouncing off the Sun") is even qualitatively accurate. – tparker Sep 10 '18 at 21:06
• @MarkEichenlaub Does this answer your question? – tparker Sep 15 '18 at 22:10

There is an approximate answer to this. A mass released from 1AU with no orbital velocity will drop to the sun. It will experience an acceleration towards the sun that increases $a~\propto~1/r^2$ and the Earth orbiting the sun has an acceleration projected along the direction this object drops along approximately equal to this, at first. So this object will reach the sun within $3$ months.

Now let's do this more exactly. For the object falling towards the sun in a radial direction we will have the total energy $$E~=~\frac{m}{2}\dot r^2~=~\frac{GMm}{r}.$$ We set the total energy to be the gravitational energy at the start of the evolution where $\dot r~=~0$ and so $$-\frac{GMm}{R}~=~\frac{m}{2}\dot r^2~=~\frac{GMm}{r}.$$ I want to isolate $dt$ and integrate, and so this turns into $$\int dt~=~\frac{1}{\sqrt{2GM}}\int_R^{r'}\frac{dr}{\sqrt{1/r~-~1/R}},$$ for $r'$ the radius of the sun and $R$ the orbital radius. This evaluates as $$T~=~\frac{Rr}{\sqrt{2GM}}\left[\sqrt{\frac{1}{r}~-~\frac{1}{R}}~+~\frac{i}{2}R^{3/2}ln\left(iR^{3/2}~+~2Rr\sqrt{\frac{1}{r}~-~\frac{1}{R}}~-~2i\sqrt{R}r\right)\right]\Big|_R^{r'},$$ which is fairly complicated

Now to work this we know the Earth orbits at $1.5\times 10^{8}$km and the sun has a radius of about $7.\times 10^5$ km and so we can write this as $$T~\simeq~\frac{1}{\sqrt{2GM}}R^{3/2}\left(\frac{2}{3}~-~\frac{r}{5R}\right)\Big|_R^{r'}$$ The orbital period for a circular orbit is $T~=~2\pi\sqrt{GM}R^{3/2}$ It is not hard to see that the approximation is about $.74$ of a quarter of an orbital period.