On May 13, 2013, China launched a rocket on a suborbital trajectory to high altitude. China announced that the launch was part of a project to study space weather and that the probe carried out an experiment at high altitude. A report in China News (translation) stated that the launch reached an altitude of about 10,000 km.
Some U.S. reports, however, interpreted the launch as the test of a rocket that could carry an anti-satellite (ASAT) weapon to very high altitudes. A Pentagon official said they believed the rocket went “nearly to geosynchronous Earth orbit” (GEO, or 36,000 km altitude) but did not place any objects in orbit.
It does not appear possible to determine the true purpose of the flight without more information. China may, of course, have conducted a space experiment on a booster that could also be used to lift an ASAT weapon into space.
However, by looking at the early trajectory of the rocket and its reentry point, it is possible to learn something about the altitude the launch reached.
Apogee of the May 13 Launch
The rocket was launched from the Xichang Satellite Launch Center in western China. The direction of the launch early in flight is known from the locations of the launch site and the ground area announced before the launch where debris (a faring or an empty stage) from the launcher was expected to fall to Earth. This zone, called a NOTAM, or Notice to Airmen, is a warning to airplane traffic about falling debris (Figure 1).
The Pentagon report of the launch stated that parts of the rocket reentered the atmosphere over the Indian Ocean, which is southwest of the launch site. Since the rocket was launched to the southeast, the duration of its flight must have been long enough that the rotation of the Earth moved the Earth’s surface far enough east to place the Indian Ocean under the reentry point. That allows an estimate of the duration of the rocket flight, and therefore of the maximum altitude that its payload reached.
The Xichang launch center is at 28 degrees north latitude. The rocket was said to have reentered at a latitude somewhat south of the equator (private communication, August 2013). For the first part of this analysis, I assume it reentered over the equator in the Indian Ocean.
To understand the geometry of the flight, it is useful to first consider where the launch would land if the Earth was not rotating (while still taking into account that part of the initial velocity of the launcher that comes from the Earth’s rotational motion). The ground track on the nonrotating Earth follows the initial direction of the launch shown in Figure 1.
The rocket’s track stretches southeast from the Xichang launch site (Figure 2) and must travel far enough south to reach the equator. If it continues along its initial direction, the rocket would land at the point labeled A in Figure 2, at a ground distance of 7,200 km from the launch site.
The fact that the Earth is rotating means that the impact point will effectively move westward by a distance Ve·t during the rocket’s flight, where Ve is the rotational speed of the Earth’s surface (about 0.46 km/s at these latitudes) and t is the flight time of the rocket from launch until reentry.
In order to be safely away from land, the reentry point must be roughly at the point B in Figure 2, or at a point west of B. The Earth’s rotation must therefore carry the point B to the point A during the flight time of the rocket. The ground distance between points A and B in Figure 2 is 8,500 km, which means that the flight time of the rocket must be longer than 8,500 km/(0.46 km/s), or 5.1 hours.
Now consider two trajectories from Xichang to point A, one with an apogee of 10,000 km and one with an apogee of 30,000 km.
To calculate these trajectories, I used a computer model of a rocket based on the Chinese CZ-3 space launcher. However, since the total flight time is much longer than the time the rocket booster is burning, the results are insensitive to the details of the boost phase.
Computer modeling shows that the trajectory with an apogee of 10,000 km has a flight time of 2.0 hours, and the trajectory with an apogee of 30,000 km has a flight time of 6.7 hours.
This implies that the launch must have gone significantly higher than 10,000 km, since the trajectory to 10,000 km gives too short a flight time for the rocket to reenter over the Indian Ocean. A launch that reentered at point B with a flight time of roughly 5 hours would mean the launch must have reached at least 24,000 km. A trajectory with an apogee at GEO would have a flight time of about 7.9 hours, during which time the Earth would rotate 13,000 km, which would put the reentry point near the coast of Africa.
The conclusion is that if reentry occurred over the equator in the Indian Ocean, and if the launch followed the trajectory shown in Figure 2, the apogee of the rocket must have been in the range of about 24,000 to 35,000 km.
A Maneuvering Upper Stage?
This estimate would change, however, if the upper stage(s) of the rocket maneuvered, so that only the first two stages followed the southeasterly trajectory shown in Figure 1. The NOTAM zone and circles in Figure 1 indicate where debris was reported to have fallen. The circle at the southeastern end of the red line in Figure 1 (which is about 1,000 km from the launch site) appears to be roughly consistent with where the empty second-stage booster may have fallen. This would imply that the rocket did not begin to maneuver away from this path until after the second stage burned out. The second stage would have burned out at a ground range of only about 100 km from the launch site (due to the steep angle of the trajectory), so maneuvering would show up as a bending of the ground track starting at a point close to the launch site.
The upper stage(s) may then have used part of their thrust to rotate the trajectory of the rocket so that it was heading in a more southerly direction. A maneuver of this kind, however, requires a lot of fuel, which means less is available to increase the rocket’s speed. We estimate that even a large maneuver of delta-V = 1 km/s would only rotate the horizontal component of the velocity by about 30 degrees (see technical note). That would result in the trajectory shown in Figure 3 (again assuming a non-rotating Earth), which would land on the equator at point C, a ground distance of 5,300 km from the launch site.
The distance from C to B is roughly 6,000 km, which would require the rocket flight time to be 6000 km/(0.46 km/s), or 3.6 hours, for it to land in the Indian Ocean. A trajectory with that flight time would correspond to an apogee of about 18,000 km. If the rocket had a longer flight time and reached a higher altitude it could reenter at a point west of B in the Indian Ocean. In particular, the Earth could rotate by about 10,000 km during the rocket’s flight and the rocket could still reenter over the Indian Ocean. That corresponds to a flight time of 6.0 hours, or an apogee of about 27,000 km. The U.S. statement that the apogee was near GEO would seem to rule out this case.
A More Southern Route?
Another possibility is that the rocket reentered somewhat south of the equator. If reentry occurred at about 15 degrees south latitude, then reentry could occur some 2,500 km east of the reentry point considered above (see point B in Figure 4), near the coast of Australia.
However, as the rocket’s trajectory travels south of the equator, it moves the (non-rotating Earth) impact point A east by essentially the same distance, so that the distance from A to B is about 8,000 km. The rotational speed of the Earth at this latitude is 0.44 km/s, so the minimum flight time of the rocket must be about 5.1 hours, which is the same as the first case above. In this case the distance from the launch site to impact point is 10,600 km, so that the apogee of the trajectory would have been at least 23,000 km. If the impact point were west of B, the apogee could easily have been as high as GEO.
A similar analysis applies if the rocket reentered at 15 degrees south latitude but maneuvered to bend the trajectory southward, as considered in the second example above. Analysis shows that the minimum flight time for a maneuver using delta-V = 1 km/s would be 3.2 hrs, so that the apogee of the trajectory would be at least 16,000 km. If the reentry point were west of B, the apogee could have approached GEO.
There is unfortunately not much publicly available information about the March 2013 launch. If the information used in this analysis is correct, it suggests that the Chinese launch reached, or could have reached, altitudes much greater than 10,000 km. If the rocket did not maneuver horizontally, its flight time suggests it may have reached altitudes of 23,000 to 36,000 km, depending on where in the Indian Ocean it reentered.
If the rocket maneuvered as considered above, it could lower the apogee of the trajectory. But even if it maneuvered very significantly, it appears difficult to get the apogee as low as 10,000 km. Moreover, if the maneuvering fuel had been used instead to increase the speed of the rocket it would have been able to reach a much higher altitude. Either way, this analysis seems to imply that the rocket was capable of reaching altitudes well above 10,000 km, even if it did not actually do so in this launch.
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