How far apart are the astronaut and satellite after 8.67 minutes?

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SUMMARY

The astronaut and satellite separate after 8.67 minutes due to the astronaut's push. The astronaut, weighing 63.8 kg, exerts a force of 156.0 N for 0.870 seconds, resulting in an acceleration of 2.44 m/s². The final velocity of the astronaut is calculated to be 2.12 m/s, which is then used to determine the separation distance of approximately 1102.824 meters after 8.67 minutes. The satellite's motion is negligible compared to the astronaut's, confirming that the astronaut's acceleration is the primary factor in their separation.

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Homework Statement


A 63.8kg spacewalking astronaut pushes off a 678.0kg satellite, exerting a 156.0N force for the 0.870s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 8.67min?

Homework Equations

The Attempt at a Solution


I tried just finding the acceleration of the satellite using Fnet=ma and then using that acceleration to place it into d=1/2at^2 (assuming the satellite was at rest) The answer came up into the thousands so I knew I did something wrong. I have no idea of how to start this question so it'd be great if someone could help out!

Thanks!
 
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alexi_b said:

Homework Statement


A 63.8kg spacewalking astronaut pushes off a 678.0kg satellite, exerting a 156.0N force for the 0.870s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 8.67min?

Homework Equations

The Attempt at a Solution


I tried just finding the acceleration of the satellite using Fnet=ma and then using that acceleration to place it into d=1/2at^2 (assuming the satellite was at rest) The answer came up into the thousands so I knew I did something wrong. I have no idea of how to start this question so it'd be great if someone could help out!

Thanks!
As you said, the satellite's motion is small compared to the one of the astronaut. Most of the separation will be due to the motion of the astronaut, so you should find his/her acceleration. If they want much precision, then you can take into account the displacement of the satellite but it will be small compared to the one of the astronaut.
 
nrqed said:
As you said, the satellite's motion is small compared to the one of the astronaut. Most of the separation will be due to the motion of the astronaut, so you should find his/her acceleration. If they want much precision, then you can take into account the displacement of the satellite but it will be small compared to the one of the astronaut.

But how can i find the acceleration of the astronaut when I barely know anything about them? Would i just be using Fnet= ma, where m= astronauts mass and Fnet= the force they apply on the satellite?
 
alexi_b said:
But how can i find the acceleration of the astronaut when I barely know anything about them? Would i just be using Fnet= ma, where m= astronauts mass and Fnet= the force they apply on the satellite?
Yes. Recall Newton's third law: fi the astronaut pushes on the satellite by 156 N, the force on the astronaut due to the contact with the satellite is also 156 N.
 
nrqed said:
Yes. Recall Newton's third law: fi the astronaut pushes on the satellite by 156 N, the force on the astronaut due to the contact with the satellite is also 156 N.
Okay gotcha! But if i may ask, what does the time the astronaut have in contact with the satellite have to do with the problem?
 
alexi_b said:
Okay gotcha! But if i may ask, what does the time the astronaut have in contact with the satellite have to do with the problem?
The astronaut accelerates only while pushing, so only during this 0.870 s. After he/she loses contact, he/she will continue moving at constant velocity. So you need to find his/her final velocity after being accelerated during 0.870 s. And then you will use that d = vt with t corresponding to the 8.67 min
 
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nrqed said:
The astronaut accelerates only while pushing, so only during this 0.870 s. After he/she loses contact, he/she will continue moving at constant velocity. So you need to find his/her final velocity after being accelerated during 0.870 s. And then you will use that d = vt with t corresponding to the 8.67 min
thanks so much, you've been a great help!
 
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nrqed said:
The astronaut accelerates only while pushing, so only during this 0.870 s. After he/she loses contact, he/she will continue moving at constant velocity. So you need to find his/her final velocity after being accelerated during 0.870 s. And then you will use that d = vt with t corresponding to the 8.67 min
Sorry, just another question came up. Do I assume the astronauts initial velocity is 0? Because other than that I don't see any other way of calculating the astronauts final velocity
 
alexi_b said:
Sorry, just another question came up. Do I assume the astronauts initial velocity is 0? Because other than that I don't see any other way of calculating the astronauts final velocity
Good question. Yes, you can. The point is that initially the astronaut is at rest relative to the satellite. In principle it does not have to be this way and the question should have made it clear, but from experience I can tell you that I am sure they assume this.

By the way, I replied to your other question about two blocks on one another.
 
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nrqed said:
Good question. Yes, you can. The point is that initially the astronaut is at rest relative to the satellite. In principle it does not have to be this way and the question should have made it clear, but from experience I can tell you that I am sure they assume this.

By the way, I replied to your other question about two blocks on one another.
I found the acceleration using the astrounaut to be 2.44 m/s^2
So i just used Vf = Vi + at and solved for Vf which turned out to be 2.12 m/s
This will now become my initial velocity of the satellite.

Using D=vt, d=(2.12m/s)(520.2 s) <-- 8.67 min turned into seconds
and i get 1102.824 m

does this seem correct?
 
  • #11
alexi_b said:
I found the acceleration using the astrounaut to be 2.44 m/s^2
So i just used Vf = Vi + at and solved for Vf which turned out to be 2.12 m/s
This will now become my initial velocity of the satellite.

Using D=vt, d=(2.12m/s)(520.2 s) <-- 8.67 min turned into seconds
and i get 1102.824 m

does this seem correct?
Yes, it is right (You meant "initial velocity of the astronaut").
 
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