# Newton's second law from different reference frames

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?

• Why are you using dots "." where you should have commas ","? 1.000 is 1 Newton, not 1,000 Newtons. What is 0,01? – joseph h Jan 28 at 2:45
• @Drjh it's a European convention – electronpusher Jan 28 at 2:46
• 100 kg is a meteoroid, not an asteroid. or is that100,000? That's on the fence. – JEB Jan 28 at 2:50
• @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. – Philip Jan 28 at 2:50
• Sorry about that. Different regional convention. Edited for clarity. – polytheneman Jan 28 at 3:41

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.

• 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. – Umaxo Jan 28 at 6:17
• @Umaxo exactly.. they are just mathematical tool.. – Ankit Jan 28 at 6:19
• if you agree, what do you mean by "Not much." In you answer? – Umaxo Jan 28 at 6:20
• @Umaxo because Newton's law is still valid with some corrections in it . That's why I wrote not much. – Ankit Jan 28 at 6:22

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.