Calculating Maximum Deceleration of Re-entry Capsule

AI Thread Summary
The discussion revolves around calculating the maximum deceleration of a re-entry capsule returning to Earth, given a re-entry velocity of 7.4 km/s and a flight-path angle of 10 degrees. Participants are attempting to use the formula a max = V^2 re-entry b sin y/ 2e, but are encountering discrepancies in their calculations. One user consistently arrives at an incorrect high value of 9.298x10^10, while another suggests that the atmospheric scale height and the base of the natural logarithm are being misapplied. Clarifications indicate that the BC number pertains to a subsequent part of the problem and that proper division and multiplication by e are crucial for accurate results. The conversation highlights the importance of careful calculation and understanding of the formula components.
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Homework Statement




Contact lenses being manufactured in space are returned in a re-entry capsule to Earth for distribution and sale. If the re-entry velocity is 7.4km/s and the re-entry flight-path angle is 10 degrees, determine the maximum deceleration it will experience and at what altitude? The capsule’s BC is 1000 kg/m^2

Homework Equations



a max = V^2 re-entry b sin y/ 2e


The Attempt at a Solution



a max = 7400m/s^2 (0.000139m^-1 sin(10)) / 2(2.7182)

the answer should be 24.813 , I keep getting a very high number of 9.298x10^10 ... why am I so off?

where .000139m^-1 is the atmospheric scale height and 2.7182 is the base of the natural logarithm.
 
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When I run the same numbers I get 243.122 m/s^2

Where does the BC number figure into the problem?
 
the BC number is actually for the second part of the question. You got closer to me. I tried my calcs again and still getting the same off-base number. I've plugged it into the calculator just like this.

7400^2 (0.000139m^-1 sin(10)) / 2(2.7182) and get 9.298x10^10
 
You apparently are dividing by 0.000139m^-1.
 
D H said:
You apparently are dividing by 0.000139m^-1.

...and multiplying by e.
 
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