Numerically, to what extent is the orbital time period of ISS affected during the time when it boosts back higher to gain the lost altitude, using Soyuz thrusters/engines?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ By coincidence just a few hours ago I estimated a change of 1.2 seconds for a 1 km change in a 437 km LEO in this answer! $\endgroup$– uhohApr 12, 2018 at 9:35
-
2$\begingroup$ The change of 1.2 seconds for a 1 km change seems to be a good aproximation for heights between 350 and 450 km. $\endgroup$– UweApr 12, 2018 at 9:39
-
1$\begingroup$ @Uwe it's reassuring to see everyone is getting the same thing! ;-) $\endgroup$– uhohApr 12, 2018 at 9:40
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
Assuming a circular orbit, I calculated some times and speeds for different heights using an online calculator from this page:
Height │ Orbital period │ Speed [km] │ [h:min:sec] │ [m/s] ───────┼────────────────┼───────── 350 │ 1:31:23 │ 7701.7 360 │ 1:31:35 │ 7696.0 370 │ 1:31:48 │ 7690.3 380 │ 1:32:00 │ 7684.6 390 │ 1:32:12 │ 7678.9 400 │ 1:32:24 │ 7673.2 410 │ 1:32:37 │ 7667.6 420 │ 1:32:49 │ 7661.9 430 │ 1:33:01 │ 7656.3 440 │ 1:33:14 │ 7650.7 450 │ 1:33:36 │ 7645.0
For a height difference of 10 km, the orbits period change is only 12 to 13 seconds.
For an elliptical orbit with 403/406 km height, the period is 1:32:30, the same period as for a circular orbit with 404.5 km height.
-
$\begingroup$ @leftaroundabout: Thanks for the edit, the table looks much better now. $\endgroup$– UweApr 12, 2018 at 16:17
$\begingroup$
$\endgroup$
1
The orbit f the ISS is almost perfectly circular - at the moment it sits at 403/406 km, with an eccentricity of 0.0002209.
This means I will just assume the orbit is circular and ignore eccentricity. This makes the answer a bit less precise, but I will also use Keplers law and drag etc, which introduce errors.
If we check the height of the ISS, we get the following graph:
As you can see, the differences are minmal - the lowest point was an average of 404.1km, the highest was 405.5km. Those are averages, though.
We know that
$ T=2\pi {\sqrt {\frac {a^{3}}{\mu }}} $
Know that the Earth isn't exactly round and the gravitational field is no point mass, so this is merely an approximation. Using 404.1km, we get 92.644 minutes, and using 405.5km we get 92.673 - a difference of 0.029 minutes, or 1.74 seconds for that timeframe between the highest and lowest orbit. Individual changes are smaller and thus affect the period less.
Historically, the differences have been larger. The ISS has been at 415km height for some time, and during shuttle use it has been at around 345 - 355km, and was then boosted to the current 400km orbit.
$\begingroup$
$\endgroup$
5
A vague question.
Though lets do atleast the order calculation. From kepler's third law $$ a^3 = K T^2 $$ Where $a$ is semi-major axis, $T$ is the time period and $K$ is proportionality constant.
Given $1km$ adjustment done in semi major axis(raising maneuver)
Apogee: 408 km Perigee: 401.1 km
$ a = 7209.1km $
Then change in $ T_{now} = 1.0002 T_{prev} $
Assuming period for 100 minutes. Change is in order of 0.02 minutes or 1.2 seconds.
So atleast a order of seconds can be expected change in time period of the orbit.
-
$\begingroup$ The period of an orbit with Apogee: 408 km Perigee: 401.1 km is 92 minutes and 30 seconds, not 100 minutes. $\endgroup$– UweApr 12, 2018 at 13:53
-
$\begingroup$ How is the question vague? It is very specific. $\endgroup$ Apr 12, 2018 at 14:22
-
$\begingroup$ I was just calculating the order @uhoh. $\endgroup$ Apr 12, 2018 at 14:25
-
$\begingroup$ It’s vague because, how much orbit is lost is not specified. So all one can do is rough calculation based on assumed altitude loss $\endgroup$ Apr 12, 2018 at 14:27
-
$\begingroup$ @Prakhar You can extrapolate that easily from historic data $\endgroup$ Apr 13, 2018 at 10:49