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An astronaut with a mass of $100 kg$ pulls on a rope attached to a $100,000 kg$ asteroid with a strength of $1,000 N$. The rope pulls back on the astronaut with the same force. The astronaut accelerates towards the asteroid at $10 m/s²$ ($a = \frac{F}m = \frac{1,000}{100}$), while the asteroid accelerates towards the astronaut at $0.01 m/s²$ ($a = \frac{F}m = \frac{1,000}{100,000}$).

Now, from the astronaut's reference frame he's pulling on the $100,000 kg$ asteroid with a strength of $1,000 N$ and as a result, the asteroid is accelerating towards him at $10 m/s²$, when, from Newton's second law, it should be accelerating at $a = \frac{F}m = \frac{1,000}{100,000} = 0.01 m/s²$.

Am I correct in concluding that Newton's second law breaks down in the astronaut's reference frame? Would I also be correct in concluding that this is because the astronaut is not in an inertial reference frame?

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    $\begingroup$ Why are you using dots "." where you should have commas ","? 1.000 is 1 Newton, not 1,000 Newtons. What is 0,01? $\endgroup$ – joseph h Jan 28 at 2:45
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    $\begingroup$ @Drjh it's a European convention $\endgroup$ – electronpusher Jan 28 at 2:46
  • $\begingroup$ 100 kg is a meteoroid, not an asteroid. or is that100,000? That's on the fence. $\endgroup$ – JEB Jan 28 at 2:50
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    $\begingroup$ @Drjh In France, for example, the decimal point and the comma have their roles reversed, as in this question. However I think of the benefit of many readers here, polytheneman, it might help to change them around. $\endgroup$ – Philip Jan 28 at 2:50
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    $\begingroup$ Sorry about that. Different regional convention. Edited for clarity. $\endgroup$ – polytheneman Jan 28 at 3:41
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Am I correct in concluding that Newton's second law breaks down in the astronaut's reference frame? Would I also be correct in concluding that this is because the astronaut is not in an inertial reference frame?

Not much. But yes, you got confused because the astronaut is a non - inertial observer.

And to compensate for this problem in Newton's law from a non - inertial frame, the term Pseudo force was introduced whose magnitude is equal to

$F_{pseudo}= (Mass_{being \; observed})(acceleration_{of\; observer})$

Do you must count this force too when using Newton's law from a non inertial frame.

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    $\begingroup$ Depends on the precise formulation of the law. If we used definition form wikipedia which is quite common the rate of change of momentum of a body over time is directly proportional to the force applied, and occurs in the same direction as the applied force, then the law is not applicable for non-inertial frames. Pseudoforces are certainly not applied forces. They are just mathematical trick to retain the formula F=ma, but conceptually are very different things. $\endgroup$ – Umaxo Jan 28 at 6:17
  • $\begingroup$ @Umaxo exactly.. they are just mathematical tool.. $\endgroup$ – Ankit Jan 28 at 6:19
  • $\begingroup$ if you agree, what do you mean by "Not much." In you answer? $\endgroup$ – Umaxo Jan 28 at 6:20
  • $\begingroup$ @Umaxo because Newton's law is still valid with some corrections in it . That's why I wrote not much. $\endgroup$ – Ankit Jan 28 at 6:22
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The second law is still valid, but you're right that the acceleration will differ between the inertial frame (neutral observer) and the astronauts frame. F = ma is still obeyed by all parties.

Whether the astronaut sees himself as accelerating toward the rock or the rock accelerating toward him is a matter of preference.

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