5
$\begingroup$

I am attempting to quantify how aggressively returning spacecraft reenter the atmosphere. There is likely a trade-off between aerobraking more gradually and splashing down accurately. I'm hoping that the deorbit trajectory perigee information will help to clarify what spacecraft typically do. If you can cite your sources, that would be very much appreciated. But if you happen to know from personal experience, please just share your knowledge.

Similar info on spacecraft returning from other places would be helpful too.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ See space.stackexchange.com/questions/12011/… Shuttle changed its perigee to ~30 nm $\endgroup$
    – Organic Marble
    Commented May 17 at 22:36
  • $\begingroup$ Thanks! In that question and its answers I found this old article which was helpful. I'm surprised to learn how much they lowered the perigee. My initial assumption was that they would want to aim higher because that would be easier on the overall delta-v budget and the thermal reentry system. $\endgroup$
    – phil1008
    Commented May 17 at 23:09
  • 1
    $\begingroup$ There are a lot of misunderstandings about shuttle. $\endgroup$
    – Organic Marble
    Commented May 17 at 23:39
  • 2
    $\begingroup$ A common misconception is that a shallower entry is somehow generally easier on the TPS. It's actually almost the opposite: although entering at a shallower angle does lower the peak heat flux,it increases the total heating a lot (and the latter of these two parameters is often the more important one). However,you can't enter too steeply either as that is mostly restricted by peak deceleration (if you have fragile human cargo on board,or wings you don't want ripped off). $\endgroup$
    – TooTea
    Commented May 18 at 13:36
  • 2
    $\begingroup$ Shuttle entries were constrained by many things, but a too-steep entry pushed the outer TPS temperature limit, a too-shallow entry pushed the structural temperature limit. $\endgroup$
    – Organic Marble
    Commented May 19 at 12:12

1 Answer 1

Reset to default
5
$\begingroup$

When a spacecraft departs from the International Space Station (ISS) and performs its deorbit burn, it transitions from a circular orbit to an elliptical orbit.

Now, based on your high reputation, I assume you already know how this all works, but for those of us who don't I'm going to include the process below:

Deorbit Burn: During the deorbit burn, the spacecraft’s engines fire to reduce its velocity. This change in velocity (often denoted as ΔV or delta-V) is crucial for reentry. The spacecraft aims to lower its altitude significantly, allowing it to reenter Earth’s atmosphere.

Elliptical Orbit: After the deorbit burn, the spacecraft enters an elliptical orbit. The perigee (closest point to Earth) occurs at the lowest altitude, while the apogee (farthest point from Earth) occurs at the highest altitude. The altitude at the perigee depends on the specific mission and spacecraft design.

Typical Altitudes: The exact altitude at the perigee varies based on mission requirements, safety considerations, and reentry profiles. Here are some examples: ISS Deorbit: When the ISS is deorbited, the perigee altitude is typically lowered to around 50 kilometers (31 miles) above Earth’s surface. Source, I paraphrased

Orion MPCV (Example Calculation): Let’s consider an example using the Orion Multi-Purpose Crew Vehicle (MPCV). Suppose the Orion MPCV needs to change its altitude from 343.5 kilometers to 96.5 kilometers at perigee. We can calculate the required burn time using the following steps: Determine the change in altitude: ΔAltitude = Original Perigee - New Perigee = 343.5 km - 96.5 km = 247 km. Use the conversion factor: 0.379 m/s² per kilometer. Calculate the required delta-V: ΔV = ΔAltitude × 0.379 = 247 km × 0.379 m/s² = 93.613 m/s. Apply Newton’s Second Law: F = ma, where F is force (thrust), m is mass, and a is acceleration. Solve for acceleration: a = F / m.

Rearrange the acceleration equation to find the time required for the specific velocity change: t = ΔV / a. Plug in the values: t = 93.613 m/s / (53,000 N / 25,848 kg) ≈ 3.34 seconds.

These values can vary based on mission specifics, spacecraft design, and operational constraints. Engineers carefully plan and execute deorbit burns to ensure safe reentry and landing.

Improve this answer
$\endgroup$
5

Not the answer you're looking for? Browse other questions tagged or ask your own question.