I was doing simulations in GMAT and I could observe that if I increase the mass of the satellite, the satellite falls slower ... and if I reduce the mass of the satellite, then it falls faster (I only changed the mass, the drag area, the initial altitude, etc, everything stayed the same). What is this all about?
$\begingroup$
$\endgroup$
9
-
2$\begingroup$ en.wikipedia.org/wiki/Ballistic_coefficient $\endgroup$– Organic MarbleCommented Aug 21, 2020 at 23:14
-
$\begingroup$ It can be helpful to wait 24 hours before accepting an answer, in this case would suggest considering if you want to move your accepted tick to uhoh's answer that delved deeper into the math. $\endgroup$– GremlinWrangerCommented Aug 22, 2020 at 9:58
-
1$\begingroup$ Suppose you have a water jug that you use 1 litre a day. Instead of using 4 litre jugs, one day you use a 12 litre jug. You notice it lasts much longer. Why? Drag=usage rate, capacity=mass. $\endgroup$– Harper - Reinstate MonicaCommented Aug 22, 2020 at 13:39
-
$\begingroup$ @PeterCordes I can't edit! Another edit is awaiting approval. Can you? $\endgroup$– Deschele SchilderCommented Aug 23, 2020 at 8:25
-
$\begingroup$ @PeterCordes Great! For sure it will be accepted! Thanks. $\endgroup$– Deschele SchilderCommented Aug 23, 2020 at 8:31
|
Show 4 more comments
3 Answers
$\begingroup$
$\endgroup$
4
That's a great software-based experiment!
What is this about?
It's about drag and Newton's 2nd law of motion!
$$F = \frac{dp}{dt} = ma$$
but in the context of orbital mechanics.
We can re-arrange Newton's law as $a = F_D/m$ where $F_D$ is the drag force, and the drag equation is
$$F_D = \frac{1}{2} \rho v^2 C_D A$$
where $\rho$ is the density at that altitude, $v$ is the speed, $C_D$ is the drag coefficient of the spacecraft's shape (usually between 0.5 and 1.0) and $A$ is the cross-sectional area of the spacecraft viewed from the direction of motion.
Put the two together and you get
$$a = \frac{1}{m}\frac{dp}{dt} = \frac{1}{2m} \rho v^2 C_D A$$
note: If you re-run your simulation in GEO where there is essentially zero atmospheric drag because $\rho$ out there is extremely close to zero, you'll see that you can drop the mass even further without detecting much loss of altitude. But with very very low mass, if you have photon pressure turned on in your program then sunlight could start to do strange things to your orbit, because you've effectively built a solar sail.
Other effects like gravitational perturbations from the Sun and the Moon will always be present, but those accelerations are independent of mass of the spacecraft because it cancels out (i.e. drop a feather and a hammer on the moon and they accelerate the same).
If this were a linear problem, you'd stop here and say that the deceleration was inversely proportional to the mass.
But in orbit things work counterintuitively. While you'd think that you lose linear momentum $dp/dt$ you actually speed up as you lose altitude and drop deeper into Earth's gravitational well.
You can read further at: https://www.grc.nasa.gov/WWW/K-12/rocket/newton2r.html
-
1
-
1$\begingroup$ @RossPresser done, thanks! That one is hard-wired in my brain (along with several others). It's a constant battle... :-) $\endgroup$– uhohCommented Aug 22, 2020 at 22:26
-
1$\begingroup$ This relation between drag and mass is taken to a relative (for satellites) extreme in GOCE: en.wikipedia.org/wiki/… which I think needed to be close to Earth to accurately sense the changes in gravity, but "streamlined" (have a low frontal surface area) to not decelerate too much in order to reach a mission duration of at least 20 months. Some called it "the Ferarri of Space" phys.org/news/2013-09-ferrari-space-fall-earth.html $\endgroup$– a.t.Commented Aug 23, 2020 at 6:57
-
$\begingroup$ @a.t. I'm intrigued, so I've just asked Exactly how "Ferrari-like" were (was) GOCE? Was its drag coefficient as low as the car's? $\endgroup$– uhohCommented Aug 23, 2020 at 10:59
$\begingroup$
$\endgroup$
1
From Force=Mass*Acceleration for a given starting drag force (from the constant drag area, altitude and velocity) increasing the mass will reduce the acceleration (deceleration in this case) slowing the satellite and causing the perihelion to lower.
So being heavier does not directly make the satellite fall slower, but does change how quickly atmospheric drag reduces the orbit.
answered Aug 21, 2020 at 23:16
-
$\begingroup$ would be improved by incorporating Organic Marble's comment $\endgroup$– user20636Commented Aug 22, 2020 at 9:39
$\begingroup$
$\endgroup$
This is not always the case. Suppose the satellite with the higher mass has, say, the form of a sphere, and the lighter one that of a long massive spear with a sharp point gradually extending to the small diameter of the spear, which has a higher mass at the front than at the back. On entering the atmosphere, the spherical satellite will experience more friction than the spear-shaped satellite.
The spear-shaped satellite will, due to the friction forces, adjust its orientation, so it is heading to earth vertically, thereby experiencing less drag force. Therefore it will arrive on earth before the lighter satellite.
Of course, it also depends on the ratio of both masses, but if the mass of the spear-shaped is not too much lower than the spherical one, it will surely arrive on earth faster. Which means falling faster.
