I am making a hypothetical solar system with random masses and distances (but still physically and mathematically accurate). I want to calculate the trajectory of a spacecraft, and to do that, I need to know the orbital period. I want to know what the orbital period for half the elliptical orbit is. From perigee to apogee.

According to Kepler's Third Law, $$T = 2\pi\sqrt{a^3\over \mu}$$

Where $T$ equals the Orbital Period, $a$ is the semi-major axis and $\mu$ is the Sun's GM.

In this imaginary solar system case, $\mu =$ $2.4 \times 10^{20}$ and $a =$ $2.0 \times 10^{11}$ metres. This gives a period of 36,275,987.3 seconds. Which gives an orbital period of 18,137,993 seconds for half the orbit.

However, according to this Physics Stack Exchange answer on how to find the transit time for a specific portion of an orbit, gives the formula: $$\tau = \frac{T}{2\pi} \bigg ( E_1 - E_2 - e (\sin E_1 - \sin E_2) \bigg)$$ Where $\tau$ is the transit time, $e$ eccentricity (0.44). Finally, $E_1$ & $E_2$ is the eccentric anomaly of two points. Since I want to find the portion from perigee to apogee, my Eccentric Anomaly is 0 and 180 degrees respectively. I confirmed these two angles with this formula from this source: $$\cos E = \frac{\cos v + e}{1 + e\cos v}$$ Where $v$ = true anomaly, and $e$ = eccentricity (in this case $e= 0.44$). True anomaly is 0 degrees at perigee, and 180 degrees at apogee.

Here is the equation when substituting the values (value of $e$ doesn't really matter because $\sin0^o$ and $\sin180^o$ = 0): $$\tau = \frac{36275987.3}{2\pi}\bigg(0^o-180^o-0.44(\sin0^o - \sin180^o)\bigg)$$ This formula gives the result $-1,039,230,849$ seconds (I guess a little mini side question: Why and should it be negative?).

Question: Why am I getting two different results for the same portion of the orbit; perigee to apogee? Using the first formula, I got $18,173,993$ seconds (about 6 months), but for the second Equation, I got $1,039,230,849$ seconds (about 33 years!). Judging by the results, the former seems correct to me, 6 months. What am I doing wrong with the second equation?


1 Answer 1


A few things come to mind:

1) where you put in the radius $8.76×10^{10}$ I think you meant the eccentricity, which for Earth's orbit is about $0.0167$. Putting in a distance parameter there makes the equation dimensionally inconsistent.

2) Do not forget to convert your angles to radians.

3) To deal with the sign: Measure the angle from the initial perigee/perihelion, and do not reset to zero with each subsequent perigee/perihelion. You do not reset because the angle is actually the total amount of angle traversed from the initial perihelion, including $2\pi$ radians for each completed revolution since. For the second half of the first orbit from zero, it runs from $\pi$ to $2\pi$.

Then it should work.

  • 1
    $\begingroup$ It would for other parts of the orbit. See my edit, we need to verify dimensional consistency too. That's why $e$ appears to me as a dimensionless quantity, which the eccentricity (commonly written $e$) is. $\endgroup$ Mar 28, 2020 at 20:34
  • 1
    $\begingroup$ So how does the angular unit conversion work? That's where the money is for the half-orbit period. $\endgroup$ Mar 28, 2020 at 20:54
  • 2
    $\begingroup$ Yes got it! Thank you very much. Instead of $0^o - 180^o$, it should be in radians. So, $0 rad - \pi$ I'll accept this answer, but can you clarify on point 3. I still get a negative value. I guess I can make the perigee $E_2$, but surely there's a proper mathematical answer as to why it's negative? $\endgroup$
    – Star Man
    Mar 28, 2020 at 21:33
  • 2
    $\begingroup$ "the angle is actually the total amount of angle traversed from the initial perihelion, including 2π radians for each completed revolution since." Even though the "longitude" resets, the total amount of angle does not and that total angle is what you put into the formula. Of course, that makes no difference in the terms where the angle is under a sine or cosine. $\endgroup$ Mar 28, 2020 at 21:42
  • 1
    $\begingroup$ Your formulas are based on time zero at perihelion. Aphelion must be $\pi$ radians from the start (or $\pi$ radians plus some number of revolutions), and the second half going to the next perihelion is the next additional \pi$ radians. $\endgroup$ Mar 29, 2020 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.