Frequently Asked Questions
- Why do satellites fall from orbit?
- Why does space hardware come apart during reentry?
- Are there examples of objects that have survived reentry?
- How much material from a satellite will survive reentry?
- What is a "debris footprint?"
- Can we predict where debris will land?
- Can we control where debris will land?
- Has anyone been hit by falling debris?
- How fast will debris be moving when it lands?
- What is the overall risk from reentry debris?
- Why do reentering bodies experience extreme heating?
Spacecraft Reentry FAQ:
Why do reentering bodies experience extreme heating?
One way to examine reentry heating is to consider conservation of energy as expressed by the first law of thermodynamics. The first law states that the total energy change of a system is equal to the sum of heat and work interactions between the system and its surroundings. Let's apply this to reentry of the space shuttle. After the engines are fired and turned off, the shuttle glides into the atmosphere, and so no work is being performed. For this case, the first law says:
| E,f - E,i | = | Q | |
| where, | E,f - E,i | = | change in shuttle energy |
| = | final energy - initial energy | ||
| Q | = | heat interaction with atmosphere |
Next, we need to account for the shuttle's energy. This is in two main forms: potential energy (PE) and kinetic energy (KE). PE is energy available due to the shuttle's position in the planet's gravitational field. Change in PE for the shuttle is the product of its weight and its altitude change. This change turns out to be small when compared to the change in KE, so we will ignore it. KE is the energy the shuttle has due to the motion of its mass, and is equal to one-half the mass multiplied by the square of the velocity (V):
KE = 0.5 x mass x V x V
Using this, we can now write the first law for the shuttle as:
KE,f - KE,i = Q
When the shuttle is near landing, the velocity is very low compared to its orbital velocity. This means the final kinetic energy (KE,f) is small when compared to the initial kinetic energy (KE,i). So, ignoring KE,f and rearranging terms gives:
KE,i = -Q
This says the heat that must be dissipated during reentry is equal to the initial kinetic energy of the shuttle. Let's calculate this heat:
| Q = KE,i | = | 0.5(mass)(V x V) |
| = | 0.5(68000 kg)(7900 m/s)^2 | |
| = | 2,120,000,000,000 J | |
| = | 2,010,000,000 BTU |
This is a tremendous amount of heat. It would bring over 6 million kg of water to boil. Not all of this heat goes into the shuttle, however. Most of this heat goes into the air around the shuttle, and is lost in the vehicle's wake. Only about 1/1000 of the heat goes into the shuttle, but this is still a great deal. This is the reason the shuttle needs a thermal protection system, which shields the underlying structure from the extreme heating.
In the absence of a thermal protection system, the structural elements of the shuttle would begin to melt and break away, ultimately destroying the vehicle. This effect commonly occurs whenever a satellite reenters the atmosphere. Satellites are usually built without reentry thermal protection systems. As the reentry heating builds up, structural elements begin melting, and the satellite breaks apart, leaving a long trail of burning debris behind.
Further questions? Contact us at cords@aero.org .