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Problem, Newton's 3rd Law Astronaut/Satellite

  • #1

Homework Statement


An 92.0kg spacewalking astronaut pushes off a 625 kg satellite, exerting a 90.0 N force for the 0.460 seconds it takes him to straighten his arms.


Question: How far apart are the astronaut and the satellite after 1.50 mins ?

Homework Equations



Newton's 3rd and 2nd law, and kinematics i guess


The Attempt at a Solution



I did F=ma to find the accel of the man and the sat, F = 90N for both of them because of Newton's 3rd law.

I then took that into the kinematics but I also took 90 (1.5mins) - the .46seconds and used that to find position with vi = 0 and accel = to what I found earlier. From there I took the same kinematic eq. pluged in what I got previous for the inital distance, turned accel to velocity and accel to 0 since now the guy and sat were no longer accelerating. I did this for both bodies and got approx 193m which is wrong.



Please tell me what I am doing wrong and/or how to implement the laws as needed. I already looked at a similar problem on here but all that was said is use the laws. But that person only offered the problem no attempt at the solution, so I am hoping my attempt will get me more noteworthy help. If you need me to I can scan my work, it however is not going to be pretty
 

Answers and Replies

  • #2
hage567
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I'd start with the impulse-momentum theorem.
 
  • #3
which is... i've read my book, haven't seen that in there or implemented, it may be called something different
 
  • #4
hage567
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There's no mention of it in your book? I'm not sure what else it would be called. It is based on Newton's second law.

[tex] I=\Delta{p}=F\Delta{t} [/tex]

where I is the impulse, delta(p) is the change in momentum, F is the force, and delta(t) is the time interval the over which the force acts.

Think conservation of momentum.
 
  • #5
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I was under the impression that the impulse-momentum theorum, can be derived from N-2,3 and kinematics, in other words, the approach of the OP should work; eg

Vsat=a*t=90/Msat and Vast=a*t=-90/92 so Vsat=90/625*t and Vast=-90/92*t so the relative velocity apart from each other is
90(1/625+1/92)*t= 1.12*0.46=0.52m/s; 90 seconds later, 46.46m apart. Not sure why this is so different than the 193 offered as none of the work is shown.
 
  • #6
ok i'm going to go through this and try to understand it and I went through again and there is no impulse formula in my book...hmm.. but i do remember it now from HS phys. Thank you for the help so far guys
 
  • #7
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ok i'm going to go through this and try to understand it and I went through again and there is no impulse formula in my book...hmm.. but i do remember it now from HS phys. Thank you for the help so far guys
Dan,

did you reconcile your results with mine? Just curious. If you take the impulse/delta P approach you end up with
Force*time=M1V1+M2V2 for the two objects.
 
Last edited:
  • #8
hage567
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denverdoc, using the impulse theorem, I get the same answer you do.
 
  • #9
I have yet to go completely over it, I have been working on Statics and Aerospace programming for the evening, I should get to it sometime tonight or tomm morning.

The only problem with this Force*time=M1V1+M2V2, is I do not have their velocities
 
  • #10
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denverdoc, using the impulse theorem, I get the same answer you do.
Hage for my own edification, can you tell me how you did it.
 
  • #11
hage567
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denverdoc, I'll send you a PM.

Dan, I don't understand why you are subtracting the 0.46s from the 90s in your original post (assuming I'm understanding what you did). Can you show more of your calculation?
 

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