Suppose no planet exist in the solar system, only the sun.

Then place a mass at a distance $r$ from the sun.

At what velocity does the mass 'hit' the sun?

Suppose both mass and sun are points in space and movements happen in a line. I know the relativistic consequences of this, just simplify things a bit...

My question rises from calculating the integral $v = a(t)dt$.

I know the acceleration by Newton's law, given by $G\frac{M}{r^2}$, but it depend on the distance and I'm not integrating over distance, but over time.

In the end, $G*M$ are constant, hence I am getting integral of $\frac{1}{r^2} dt$, and I'm stuck.

I'm not looking for answers involving mechanical conservation of energy, simply answers involving this integration method.

Also, how much time does this mass take to hit the sun?

I wrote out the integral of $\int G\frac{M}{r^2}t, dt$, but cannot go further with my actual knowledge.

  • $\begingroup$ What's so bad about using energy conservation? The Work-Energy Theorem is essentially the integration you want to do, except it's generalized to all the possible force laws that will give sensible results. Any answer you get that does direct integration is just a downgraded version of the theorem's proof. $\endgroup$ Commented Dec 8, 2017 at 18:58

2 Answers 2


So we can look at this in terms of differential equation: $$\frac{d^2r}{dt^2}=-\frac{G(M+m)}{r^2}$$where $M$ is the mass of the sun and $m$ is the mass of the object that is going into the sun. $($I have written it like this in case the mass of the object $m$ is large enough to create a noticeable acceleration of the sun$)$. Multiplying both sides by $dr$ we get: $$\int_{r_0}^r\frac{dv}{dt}dr=\int_{r_0}^r -\frac{G(M+m)}{r^2}dr$$ $$\int_0^v v\ dv=\int_{r_0}^r -\frac{G(M+m)}{r^2}dr$$ $$\frac{1}{2}v^2=\frac{G(M+m)}{r}-\frac{G(M+m)}{r_0}$$ $$v=\sqrt{\frac{2G(M+m)}{r}-\frac{2G(M+m)}{r_0}}$$ Where $r_0$ is the initial radius and $r$ is the final radius $($both measured from the centre of the sun$)$. Now getting onto your second question we should continue from where we left off with:$$v=\frac{dr}{dt}=\sqrt{\frac{2G(M+m)}{r}-\frac{2G(M+m)}{r_0}}$$ $$\frac{dr}{dt}=\sqrt{\frac{2G(M+m)r_0-2G(M+m)r}{rr_0}}$$ And rearranging for $dt$ we get: $$dt=\frac{\sqrt{rr_0}dr}{\sqrt{2G(M+m)r_0-2G(M+m)r}}$$ $$dt=\sqrt{\frac{r_0}{2G(M+m)}}\sqrt{\frac{r}{r_0-r}}dr$$ Substituting in for $$u=\sqrt{\frac{r}{r_0-r}}$$ and rearranging we get $$dr=\frac{2r_0u}{(1+u^2)^2}du$$ and so $$dt=\sqrt{\frac{r_0}{2G(M+m)}}\frac{2r_0u^2}{(1+u^2)^2}du$$ Now, as $r\to r_0,\ u\to\infty$ and as $r\to r_f,\ u\to u_f$ where $r_f$ is the radius of the sun and $$u_f=\sqrt{\frac{r_f}{r_0-r_f}}$$ $$dt=\frac{2r_0^{3/2}}{\sqrt{2G(M+m)}}\int_{u_f}^{\infty}\frac{u^2}{(1+u^2)^2}du$$Solving this for $t$ gets us: $$t=\frac{2r_0^{3/2}}{\sqrt{G(M+m)}}\bigg(\frac{\pi}{2}-\bigg(\tan^{-1}(u_f)-\frac{u_f}{1+u_f^2}\bigg)\bigg)$$

Hope this helps :). If you need me to expand on any points I can just leave a comment.

  • 1
    $\begingroup$ I agree that one of the integration limits is $\infty$ ($r \to r_0$), but the other limit is at the surface of the sun, not the center. The answer might not be much different, practically, but $r \ne 0$ at the surface, so the other limit for $u$ is not exactly 0. $\endgroup$
    – Bill N
    Commented Dec 8, 2017 at 19:45
  • $\begingroup$ Ahh I completely forgot! I did this ages ago for two objects with negligible radius. I'll go and edit that, thanks. $\endgroup$ Commented Dec 8, 2017 at 19:46
  • $\begingroup$ It's also interesting that the square of the time to fall in is proportional to the cube of the starting distance. $\endgroup$
    – Bill N
    Commented Dec 8, 2017 at 19:46
  • $\begingroup$ Thanks indeed for the answer. but why is that you have taken the sum of the masses instead of the product in the acceleration formula? $\endgroup$ Commented Dec 8, 2017 at 19:49
  • $\begingroup$ @f126ck The gravitational acceleration on the mass from the sun is $\frac{GM}{r^2}$. However, the mass also exerts a force on the sun, giving an acceleration of $\frac{Gm}{r^2}$. In case the mass is large enough for this acceleration to be notable I have taken $a=\frac{GM}{r^2}+\frac{Gm}{r^2}=\frac{G(M+m)}{r^2}$ $\endgroup$ Commented Dec 8, 2017 at 19:57

I tried to do a quick code of the above equations and got this graph of t and r_0. Setting the r_0 as the sun earth distance. And the masses of that of the sun and the earth. The peak reaches around 7^26 seconds which is about 2 months. Also feel free to correct me if im wrong! enter image description here


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