# Why can't the Chinese rocket Long March 5B keep revolving around the Earth without re-entering?

The out-of-control part of the Chinese rocket Long March 5B is expected to re-enter the atmosphere of the earth and crash on the earth in a few hours. (Today is May 8th, 2021) At the moment, it is revolving around the earth in an elliptical orbit and completes a revolution in 90 minutes. Since April 29th, 2021 it has gone around the earth more than 100 times and the size of the orbit is getting smaller and smaller.

Why can't this rocket keep revolving around the earth until it is completely burnt in the atmosphere or hits the skyscrapers (in a path parallel to the earth surface and not perpendicular to it, so to speak)? What laws of physics would cause its re-entry into the atmosphere? It is losing altitude but what prevents it from losing altitude gradually and continuously (without a sudden change in its path)? It is already experiencing the gravitational pull of the earth but what is supposed to cause its 'sudden' downfall towards the earth and crashing on the earth?

• I think you will find this useful: phet.colorado.edu/sims/html/gravity-and-orbits/latest/… – Natru May 8 at 17:07
• Mashup, a very simple answer is, you may not realize the rocket remains are not powered at all. It's that simple .. it is slowly falling. As you know, things like communications satellites can just "keep going around" - but they are much much further away. – Fattie May 9 at 13:51

You are describing something called orbital decay, which is well-understood. Here are the basics, now including more detail:

With increasing height, the earth's atmosphere gradually fades out and becomes vacuum. Objects in low earth orbit will therefore be subject to a tiny amount of frictional braking from the tiny amount of gas molecules they encounter. This makes them slow down a tiny bit in response, and that causes them to settle into an orbit that is a tiny bit closer to the earth.

The closer an orbit is to the earth, the faster the orbiting object has to move to remain in orbit, so as the friction robs the object of kinetic energy it sinks into a lower orbit and speeds up, encounters denser air, more friction, more drag, lower orbit, speeds up, etc., etc.

This process feeds on itself until the object is no longer going fast enough to stay in orbit at all and its path begins to tilt downward as it falls. Friction with the air, and the formation of a super hot shock wave just in front of it, heat it up and it softens, melts, and (usually) disintegrates- and the smallest pieces are burnt up before they make it all the way through the atmosphere. The biggest pieces can sometimes make it all the way down to hit the surface of the earth.

• It's a bit stronger than that. Friction does not cause the orbital speed to decrease - it makes the orbit speed up. This is because locally friction slows the satellite down, but this just makes it go into a lower orbit, which converts potential energy into kinetic energy and makes the speed increase. (ie, friction does decrease the total energy, but not the kinetic energy.) This is what really powers the exponential feedback loop that makes the deorbiting event so sudden. – Emilio Pisanty May 8 at 20:45
• The edit is a massive improvement. As a similar observation: "until the object is no longer going fast enough to stay in orbit at all" - that's true (the object is too slow to orbit at that altitude) but it's also going faster than at any time before that. Orbital mechanics is just weird that way =). – Emilio Pisanty May 8 at 22:28
• I think you'sre wrong about objects falling straight down at tremendous speed. Objects reenter almost horizontally, heat from air friction which slows them. If they don't break into pieces (breakup is likely for anything not specifically designed for reentry) this continues until they either hit the ground with significant horizontal velocity, or slow until the horizontal velocity is near zero. At that point, they drop straight down at terminal velocity, which isn't all that fast. – jamesqf May 9 at 2:47
• For instance, here's a picture of a falling meteor in its "dark flight" phase, captured by a Norwegian skydiver: universetoday.com/110963/… And of course looking at reentry profiles of space capsules will show the same behavior. – jamesqf May 9 at 2:51
• @Vikki-formerlySean though likely, the analysis does tell it would also be consistent with a meteor — it picks the pebble explanation because it is much, much more likely, and Occam's razor. But in the context of this comment here, what matters is: had it been an actual meteor, it would have looked exactly like this. Which is consistent with jamesqf's point. – spectras May 9 at 20:36

The force of gravity causes it to re-enter the earth since it did not have the essential conditions to be in a orbit, like satellites. In fact, for each orbit with a certain radius ($$r$$), there is a certain speed ($$v$$) so that the object remains in the orbit, $$v = \sqrt {\frac{{GM}}{r}},$$ where $$M$$ is the earth mass and $$G$$ is the gravitational constant. See this figure from WikiPedia:

Any deviation from this speed will disrupt the orbital motion and the object will go out of orbit (the object goes into an orbit with a smaller/bigger radius or it could even fall into the Earth or it could even escape from the gravitational influence of Earth). Therefore, there must be a precise balance between velocity and orbital radius. The Chinese rocket did not have such conditions because it was out of control. It has finally fallen to Earth over the Indian Ocean..

When it enters the Earth's atmosphere, the rocket's speed in both horizontal and vertical directions is reduced due to air resistance force (collision with air molecules). On the other hand, due to the gravity of the earth, a downward force is always applied to the rocket. As a result, the rocket's horizontal velocity is constantly decreasing while its vertical (downward) velocity is constantly increasing. In summary:

Before entering the Earth's atmosphere: $${v_x} = {\rm{constant}}, \, \,{v_y} = gt + {v_{y_0}}$$

After entering the Earth's atmosphere:

$${v_x} = {\rm{is\,decreasing}}, \, \,{v^\prime_y} = gt + {v^\prime_{y_0}}$$

(The downward direction is considered positive.)

The constant decreasing in horizontal velocity causes the rocket to travel less and less horizontally over time. This means its trajectory becomes more and more vertical. This situation is illustrated in the following figure

Note that the scales of this painting are not realistic.

• BTW, $v=\sqrt{\frac{{GM}}{r}}$ is only valid for perfectly circular orbits. For an elliptical orbit, we can use the vis viva equation, $v^2=GM\left(\frac2r-\frac1a\right)$, where $a$ is the mean radius of the orbit (that is, the mean of the minimum & maximum radii). Of course, both equations ignore atmospheric drag. – PM 2Ring May 9 at 9:43
• @PM2Ring, Thanks for your comment. I know, a circular orbit is for simplicity and, of course, it is valid for earth's satellites with a good approximation. In that WikiPedia link in my answer, different types of orbital motions have been summarized. BTW, the general conclusion, which is important here, is correct for both cases. – SG8 May 9 at 12:10