Problem, Newton's 3rd Law Astronaut/Satellite

1. Mar 25, 2007

DantheMan10

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations

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

3. 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

2. Mar 25, 2007

hage567

3. Mar 25, 2007

DantheMan10

which is... i've read my book, haven't seen that in there or implemented, it may be called something different

4. Mar 25, 2007

hage567

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.

$$I=\Delta{p}=F\Delta{t}$$

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. Mar 25, 2007

denverdoc

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. Mar 25, 2007

DantheMan10

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. Mar 25, 2007

denverdoc

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: Mar 25, 2007
8. Mar 25, 2007

hage567

denverdoc, using the impulse theorem, I get the same answer you do.

9. Mar 25, 2007

DantheMan10

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. Mar 25, 2007

denverdoc

Hage for my own edification, can you tell me how you did it.

11. Mar 26, 2007

hage567

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?