I come across this interesting problem comet-x-earth. It was an exam problem asking the time that a comet will be spent inside the Earth's orbital. I make an illustration for the problem:


The comet is in a parabolic orbit (i.e. its total energy = 0). It comes inside the Earth's orbit. The nearest position of the comet's orbit is $\beta R$, where $\beta <1$ and $R$ the radius of earth orbit (assume circular). The problem asked for the time that the comet travels from point $A$ to point $B$.

The link had a nice answer starting with the polar equation of a parabolic, determined the parameter of equation at the nearest point, found out the intersection position $A$ and $B$; then wrote down the integral.

Out of curiosity, I like to ask the question: is there a way of solving this problem without referring to the equation of the comet's orbit?

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    $\begingroup$ I really like this question. If I get a moment I will try to write a brief Python script to simulate some scenarios. Thanks for posting. $\endgroup$ Commented Jan 26, 2021 at 20:42
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    $\begingroup$ There's an even simpler solution that uses only Kepler's second law. Let the total area swept out by the comet between points A and B be $A$, and let the total time be $T$. Then the constant rate at which area is swept out is $dA / dt = A / T$, but we also know that $dA/dt = (\beta R) v / 2$ when the comet is at the vertex of the parabola, where $v$ is its speed there. $\endgroup$
    – knzhou
    Commented Mar 11, 2021 at 6:27
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    $\begingroup$ Thus, by evaluating $A$ using some geometry, the problem can be solved without even needing any calculus. $\endgroup$
    – knzhou
    Commented Mar 11, 2021 at 6:28
  • $\begingroup$ I suspect this is not a simpler way, It will route back to the same integral. $\endgroup$
    – ytlu
    Commented Mar 11, 2021 at 6:33
  • $\begingroup$ Or, you have to employ the equation of the trajectory, which comes from a more complicate integral. $\endgroup$
    – ytlu
    Commented Mar 11, 2021 at 7:02

2 Answers 2


Is there a way of solving this problem without referring to the equation of the comets?

Yes. You do not need to use either the polar form or the Cartesian form of the comet's parabolic trajectory. You can work with the differential equations satisfied by the trajectory, rather than with their solution.

Write the differential equations for the conservation of energy and the conservation of angular momentum, in polar coordinates. They will involve $\dot r$, $\dot\theta$, and $r$. Eliminate $\dot\theta$ to get a relationship between $\dot r$ and $r$.

There will be an unknown constant, the specific angular momentum of the comet. (This is its angular momentum divided by its mass.) Evaluate it by imposing the conditions that at closest approach, $r=\beta R$ and $\dot r=0$. (Do this in both of the original differential equations.)

Separate the variables in the equation relating $\dot r$ and $r$ to get a differential relationship of the form $dt=f(r)dr$. Integrate the right side from $r=\beta $R to $r=R$. (This integral has a different form from the integral hint in the original question.) Double this time to get the time from A to B.

At the end, you will need to recognize a particular combination of $G$, $M$ (the mass of the Sun), and $R$ as the period of the Earth's orbit, namely one year.

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    $\begingroup$ The original question was closed as homework-like -- it was on an exam -- but its answers were not deleted. Therefore I felt that it was OK to provide another answer. However, I didn't want to provide another complete solution to a homework-like problem, so I only sketched this approach. Moderators should consider deleting all three answers to this exam problem. $\endgroup$
    – G. Smith
    Commented Jan 25, 2021 at 20:35

I try to work it out following the clues from G.Smith for a complete record.

in 2-d polar coordinate, the velocity $\vec{v}$

$$ \vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} = v_r \hat{r} + v_{\theta} \hat{\theta} $$

The radial component $ v_r = \dot{r} = \frac{dr}{dt}$. The time $T$ from point $A$ to point $B$ can be write as the integral:

$$ T = 2 \int_{\beta R}^R \frac{dr}{v_r}. \tag{1} $$ We then will try to relate $v_r$ as function of $r$ in order to do this integral.

Conservation Laws

Conservation of total energy $$ E(t) = -\frac{G M m}{r(t)} + \frac{1}{2} m ( v_r^2 + v_{\theta}^2 ) = 0 \tag{2} $$

At the nearest point, $\vec{r} = v_{\theta 0} \hat{\theta}$, $v_{r0} = 0$ at $r_o = \beta R$. We find the speed at the nearest point.

$$ v_{\theta 0} = \sqrt{ \frac{2GM}{\beta R} } \tag{3} $$

Conservation of total angular momentum

$$ L_z = m r v_\theta = m r_o v_{\theta 0} = m \sqrt{2 G M\beta R } \tag{4} $$ We obtain a $v_\theta$ as a function of $r$ from Eq.(4) $$ v_\theta (r) = \sqrt{2 G M\beta R } \frac{1}{r} \tag{5} $$

Function $v_r(r)$

Substitute Eq.(5) into Eq.(2) to find the $v_r$ as function of $r$:

$$ v_r = \sqrt{ \frac{2GM}{R} } \sqrt{ \frac{R}{r} - \beta \frac{R^2}{r^2} }. $$

Place this relation into the Eq.(1) to complete the integral form:

$$ T = 2 \sqrt{\frac{\beta R}{2 G M}} \int_{\beta R}^{R} \frac{dr}{\sqrt{ \frac{\beta R}{r} - \beta^2 \frac{R^2}{r^2} } } $$


Let $\xi = \frac{r}{\beta R}$, the integral becomes

$$ T =2\beta R \sqrt{\frac{\beta R}{2 G M}} \int_1^{1/\beta} \frac{\xi d\xi}{\sqrt{\xi - 1} } = \sqrt{\frac{2 \beta^3 R^3}{ G M}} \int_0^{1/\beta - 1} (\sqrt{\eta} + \frac{1}{\sqrt{\eta}} ) d\eta. $$ In the fianl expression, we substitue $\eta = \xi -1$.

$$ T = \sqrt{\frac{2 \beta^3 R^3}{ G M}} \{ \frac{2}{3} (\frac{1}{\beta} - 1)^{3/2} + 2 (\frac{1}{\beta}-1)^{1/2} \} $$ Replace $R$ by the period of Earth $T_0 = 1 year$ using relation $T_0^2 = \frac{4\pi^2}{GM} R^3$, we finally have: $$ T = \frac{T_0}{ 3 \pi } (1 + 2 \beta) \sqrt{2(1-\beta)}. $$


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