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If a body starts to fall freely from Mars orbit towards the Sun, then how can we calculate the time to touch the Sun' surface?

I have tried the third equation of motion that says R=(1/2 dg)*dt^2. Then what must be the limits? If needed, you can add a comment for the work I did and want it as a pic.

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  • $\begingroup$ yes definitely add a pic of the work you did! This is always encouraged in Stack Exchange. $\endgroup$ – uhoh Feb 28 '19 at 7:29
  • $\begingroup$ If you can format your work using MathJax/LaTeX, that would be even better. $\endgroup$ – Nathan Tuggy Feb 28 '19 at 7:31
  • $\begingroup$ I see you've asked two questions (1, 2) in Physics SE and they weren't well received with comments explaining that you should show what work you have done. So yes, definitely, add your work right from the start. By the way, if you search Physics SE for "The Time That 2 Masses Will Collide Due To Newtonian Gravity" will lead you to several good answers. $\endgroup$ – uhoh Feb 28 '19 at 8:13
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A quick-and-dirty way to get an approximate answer is to note that, as the initial orbital velocity of the body approaches zero, its orbit will approach a (degenerate) ellipse with a minor axis of zero and a major axis equal to the body's initial distance from the center of the Sun (i.e. the radius of Mars' orbit).

In particular, this means that the orbit will have a semi-major axis equal to half the initial distance of the body from the Sun, i.e. half of the semi-major axis of the (approximate circular) orbit of Mars itself.

Since we know that the period of an elliptical orbit is proportional to its semi-major axis raised to the power of $3/2$, this means that reducing the semi-major axis of an orbit by one half will reduce the orbital period down to $1/2^{3/2} = 1/\sqrt8 = \sqrt2/4 \approx 0.35$ times the original period.

As the time it takes for an orbiting body to fall from the apoapsis of its orbit down to the periapsis is exactly half of its orbital period, this means that your initially stationary body will fall to the Sun in a time that is $\sqrt2/8 \approx 0.18$ times the orbital period of a body in a circular orbit at the same initial altitude (such as, in this case, Mars).

We can then look up the (sidereal) orbital period of Mars and multiply it by $\sqrt2/8 \approx 0.18$ to get the solution to your question.

Ps. This method does make a couple of approximations worth noting. One is implicit in the question itself, which merely specifies the starting altitude of the body as "Mars orbit", without further specifying whether that means aphelion, perihelion or somewhere in between. In effect, this method assumes that the starting altitude is "somewhere in between", specifically at the semi-major axis of the orbit of Mars.

The other notable approximation made is that we're effectively assuming the radius of the Sun to be negligible. As the Sun's radius is less than 0.3% of the semi-major axis of Mars, and as the falling body will in any case spend the majority of its fall time in the outer parts of its orbit where it's moving slowest, the error introduced by this approximation will be even less than that. Compared to the roughly ±14% uncertainty introduced by the poorly specified starting altitude due to the eccentricity of Mars' orbit, that is indeed negligible.

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  • $\begingroup$ I'm utterly confused by your math here. Why are you reducing the semi-major axis by 1/2? $\endgroup$ – Loren Pechtel Mar 2 '19 at 16:11
  • $\begingroup$ @LorenPechtel: Because a circular(ish) orbit at distance $d$ from the Sun has a semi-major axis of $d$, whereas an orbit falling from distance $d$ straight into the Sun has a semi-major axis of $\frac12d$. Does that help? If not, please describe the nature of your confusion in more detail. $\endgroup$ – Ilmari Karonen Mar 2 '19 at 18:14
  • $\begingroup$ why do you think that is the case? $\endgroup$ – JCRM Mar 2 '19 at 19:03
  • $\begingroup$ @JCRM: Because that's how it's defined? The semi-major axis is half the major axis of the orbit, i.e. half the distance from the perihelion to the aphelion (which are always located on opposite sides of the orbit, because that's how Keplerian orbits work). For a roughly circular orbit, like that of Mars, the perihelion and the aphelion are at roughly the same distance from the Sun. For a body free-falling from that distance into the Sun, the aphelion is still at that same distance, but the perihelion is inside the Sun. $\endgroup$ – Ilmari Karonen Mar 2 '19 at 20:15
  • $\begingroup$ So you're saying when the velocity is zero the orbital energy becomes $\epsilon = -\frac{\mu}{r}$ and because the semi major axis is $a = -\frac{\mu}{2\epsilon}$ it follows that it is $\frac{r}{2}$ $\endgroup$ – JCRM Mar 2 '19 at 20:31

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