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consider a ballistic missile in reentry trajectory with initial velocity of 1000 m/s. as missile enters the atmosphere, air drag decelereting its velocity so missile's acceleration becomes negative. in this situation how many g's i mean acceleration due to gravity missile felt? positive? negative or zero g? is there any eqution for that?

another question is when air drag is equal to acceleration of gravity is it means that g is one?

sorry about my english

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    $\begingroup$ Good question, but in my opinion it belongs to Physics Stack Exchange. $\endgroup$ Commented Jan 5, 2023 at 23:39
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    $\begingroup$ I’m voting to close this question because it's about physics, not aviation. $\endgroup$ Commented Jan 6, 2023 at 11:11

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Just to compare apples to apples, Force = mass × acceleration = mass × $delta$ Velocity/time

Drag is also a force. This makes it easy, because forces can be added and subtracted.

Before entering the atmosphere, there is no drag, so downward force is mass × 9.8 m/s$^2$. G force felt is 0. Free fall. Velocity is increasing.

As drag starts, Net downward force is (mass × gravity) - drag force, still positive, but less. G force is greater than 0 but less than 1. The object continues to speed up, at a lower rate.

Downward acceleration stops when (mass × gravity) - drag force = 0 This is terminal (constant) velocity. G force felt = 1 towards earth.

If you have ever watched a Falcon 9 launch, terminal velocity decreases as the booster falls into thicker atmosphere. Now acceleration is negative, and G force is greater than 1. The rocket is slowing down.

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    $\begingroup$ Minor correction: acceleration is not v/t, it is ∆v/∆t. Both are very different. $\endgroup$ Commented Jan 5, 2023 at 23:43
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    $\begingroup$ I guess that's what we're trying to show as net force. Positive delta v is less as drag kicks in, eventually going to zero, then negative as the rocket slows down in thicker atmosphere. $\endgroup$ Commented Jan 5, 2023 at 23:44
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    $\begingroup$ If you accelerate from 1000 m/s to 1002 m/s in one second, what is the acceleration? "v" (average v) is 1001 m/s, ∆v (change in v) is 2 m/s. By using v like you did, we get an acceleration of 1001 m/s², by using ∆v like I suggest, the acceleration we get is 2 m/s². Clearly, the latter is the actual acceleration. I get your point, and for your case, the more appropriate term would be dv/dt (instantaneous acceleration) instead of ∆v/∆t. But v/t is wrong because it doesn't represent acceleration, as shown in my little calculation above. $\endgroup$ Commented Jan 6, 2023 at 0:02
  • $\begingroup$ @AdityaSharma please feel free to edit in the deltas. I agree this is a more accurate way of describing acceleration. $\endgroup$ Commented Jan 6, 2023 at 0:35
  • $\begingroup$ It won't let me edit, it says "edits must be atleast 6 characters long" 😅, you can yourself just add a delta before velocity and time. $\endgroup$ Commented Jan 6, 2023 at 0:44

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