I may or may not waste more time on this but who cares? You won't take the slightest bit of notice!
Well, if your spacecraft orbits Earth at 7 635 m/s leaving the ISS at 450 000 m altitude it needs a lot braking to start re-entry at 120 000 m/altitude.
It won't come into it here but no unpowered satellite can orbit fro more than a fraction of an orbit below 180 km - the air resistance at orbital speed is too high,
I see that you haven't bothered to read even this
Returning from Space: Re-entrySuppose the spacecraft is orbiting at an altitude of 450 km with an orbital velocity of 7640 m/s and the reentry altitude is at an altitude of 120 km with the required entry angle.
All that needs be done is to reduce the velocity enough to enter an elliptical orbit (another Hohmann transfer orbit) with the perigee enough below that 120 km to get the right entry angle.
I'm not going to attempt to calculate exactly what that perigee need be - I'll "guess" and put that perigee at 80 km.
The velocity needs to be reduced by only 107 m/s to 7533 m/s a the apogee for a perigee of 80 km.
If there were no air drag, after half an orbit, in 45 minutes the spacecraft would be down to the 80 km and travelling at 7965 m/s but air drag starts before then.
Sometime before reaching the reentry interface any unwanted fuel and modules not needed (eg the Soyuz equipment module) are disposed of to burn un on there own.
But the point is it takes very little reduction in velocity to start the deorbit manoeuvre.
This is not an ideal diagram but it might illustrate this process:
Not that the spacecraft is orbiting clockwise in it.
The "landing burn" would only be needed if the atmosphere were not dense enough for a complete atmospheric reentry (eg on the Moon, minor planets, asteroids and possibly on Mars). On Earth an atmospheric reentry can slow the craft to where parachutes can be deployed - this is even done on Mars with two stage parachutes.
Coming back from the Moon you velocity at 120 000 m altitude is >11 000 m/s when starting re-entry ... and it is pretty fast.
Sure, you are travelling very fast!
As soon as you contact the Earth atmosphere, you start to heat up and ... burn up.
Sure, you start to heat up!
But a properly designed and oriented heat shield will cause most of the kinetic energy to be expended in heating the air to plasma temperatures and not the reentering module.
There is nothing left of you.
Incorrect! You simply say these things with no evidence to back your claim - present the evidence or run away!
If you have a "properly designed and oriented heat shield" most of the energy ends up heating the atmosphere.
But there have been a number of reentry mishaps - including this:
Soyuz Hard Landing: Equipment Module Failed to Separate – OfficialAnd an earlier more serious explosive bolt problem (simulated as Russia did not release much):
Soyuz 4 & 5 - Docking, Spacewalks and Nearly Burning Up In The Atmosphere by Scott Manley