Newton's 3rd law and spacewalking

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SUMMARY

The discussion focuses on calculating the distance between an astronaut and a satellite after a push-off during a spacewalk. An 80.0 kg astronaut exerts a 100 N force on a 620 kg satellite for 0.590 seconds, resulting in accelerations of 1.25 m/s² for the astronaut and 0.1613 m/s² for the satellite. After the force application, both objects move at constant velocities, calculated using the momentum change formula F(Δt) = m(Δv). The final distance between them after 1.20 minutes is determined to be 59.5 meters.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly Newton's 3rd law.
  • Familiarity with kinematics equations, specifically xf = xi + vi(t) + 0.5at².
  • Knowledge of momentum and its relationship with force and time, expressed as F(Δt) = m(Δv).
  • Basic proficiency in unit conversions and time calculations in physics problems.
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  • Study the application of Newton's laws in space environments, focusing on force interactions.
  • Learn more about kinematic equations and their applications in various scenarios.
  • Explore momentum conservation principles and their implications in collision and separation events.
  • Investigate the effects of microgravity on motion and acceleration in space missions.
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Physics students, aerospace engineers, and anyone interested in understanding motion dynamics in space environments.

sunbunny
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I'm having troubles with this problem:

An 80.0 kg spacewalking astronaut pushes off a 620kg satellite, exerting a 100N force for the 0.590s it takes him to straighten his arms.How far apart are the astronaut and the satellite after 1.20min ?

I know that i need to somehow set it up to be a kinematics problem and that i need to find the acceleration and how fast the astronaut and the satellite are moving.

I tried to use F=ma where a=100N/80kg and then the same for the orbital

for the astronaut i got the acceleration to equal 1.25m/s^2
for the satellite a=0.1613 m/s^2
and the change in time used was 72-0.59s=71.41s

I then put these into xf=xi +vi(t)+.5at^2 to find the distances

I then added their distances together to get the total distance between them however this way that I did it was wrong.

I got 3187.1m (astronaut) +411(orbit)

any feedbaclk would be great
 
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sunbunny said:
I'm having troubles with this problem:

An 80.0 kg spacewalking astronaut pushes off a 620kg satellite, exerting a 100N force for the 0.590s it takes him to straighten his arms.How far apart are the astronaut and the satellite after 1.20min ?

I know that i need to somehow set it up to be a kinematics problem and that i need to find the acceleration and how fast the astronaut and the satellite are moving.

I tried to use F=ma where a=100N/80kg and then the same for the orbital

for the astronaut i got the acceleration to equal 1.25m/s^2
for the satellite a=0.1613 m/s^2
and the change in time used was 72-0.59s=71.41s

I then put these into xf=xi +vi(t)+.5at^2 to find the distances

I then added their distances together to get the total distance between them however this way that I did it was wrong.

I got 3187.1m (astronaut) +411(orbit)

any feedbaclk would be great
The astronaut and satellite are accelerating only when the force is applied over the 0.59 second period. After that, once contact is gone, they both must move at constant velocity, per Newton's first law.
 
Okay, thank you.

So if they are only accelerating during the 0.59s how would i go about finding the acceleration for them during this time interval?

I was using F=ma but this formula doesn't have time in it. How would you suggest that I go about this?
 
sunbunny said:
Okay, thank you.

So if they are only accelerating during the 0.59s how would i go about finding the acceleration for them during this time interval?

I was using F=ma but this formula doesn't have time in it. How would you suggest that I go about this?
Use F(delta t) =m(delta v)=m(v_f -v_i). That's the same as F=ma.
 
Thank you so much!from the equation you gave me, i found the velocity of the astronaut to be 0.7373m/s and the satellite to be 0.095167m/s. from there, i put these velocities into:

xf=vi(delta t) and then I added the two distances and got the answer
59.5m. Thanks a a lot I really appreciated it!
 
hi i have a similar question to this and when i tried to recreate your workings it wasnt working for me can u please explain to me what you did..thanks :))
 
brunettegurl said:
hi i have a similar question to this and when i tried to recreate your workings it wasnt working for me can u please explain to me what you did..thanks :))

Consider the force applied and the duration.

Applying the force to each mass results in a change in momentum - which gives you the speed of each.

The speeds are in opposite directions so simply add the speeds and determine the distance given the time.
 
so wld i be using the Fdeltat = mdeltav for each of the objects...using this velocity and the second time given in the question figure out the distance and add them together??
 
brunettegurl said:
so wld i be using the Fdeltat = mdeltav for each of the objects...using this velocity and the second time given in the question figure out the distance and add them together??

Basically yes.

Though I think of it more as determining their relative velocity first from the F*Δt on each and then applying the duration of their drift. (The magnitudes of the velocities add since they are in opposite directions,)
 
  • #10
we assume for each object that the vintial is zero right?? so then wld the distance we get be a negative or a positive value??
 

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