How Long Does It Take for a Space Tug and Asteroid to Meet When Pulled Together?

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In summary, the problem discusses astronauts using a space tug and a 5770-kg asteroid to mine asteroids. The tug pulls on the asteroid with a force of 510 N, causing both objects to accelerate towards each other. The question asks for the time it takes for the two objects to meet. To solve this problem, the accelerations of both objects must be found and added together. The formula X=Vot+0.5at^2 is used to solve for time, with the initial velocity (Vo) being zero and the displacement (X) being 550m. It is important to not assume one acceleration to be negative, as the objects are moving towards each other and their accelerations must be added.
  • #1
v12alp
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Hi everyone, this problem is truly pain in the a**. I understand that Newton's third law applies here, but whatever I have tried nothing worked out. I know that the force applied is the same in magnitude, but opposite in the direction and accelerations of these two masses are different...but no luck. Can anyone please help me out?



At a time when mining asteroids has become feasible, astronauts have connected a line between their 3090-kg space tug and a 5770-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 510 N. Initially the tug and the asteroid are at rest, 550 m apart. How much time does it take for the ship and the asteroid to meet?

 
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  • #2
For starters can you find the accelerations of each of the two objects?
 
  • #3
Hi, yes, as a matter of fact I did. I assumed down direction to be negative so F=ma, a=F/m for the first the tug a (tug)=510N/m For the asteroid a=-510N/m (asteroid). From the formula Xf=Xo+Vot+1/2at^2 it became clear that Vo=0 m/s they started from rest and hence Xf-Xo=1/2at^2 or simply 550m=1/2at^2. Now, what the hell is acceleration here? Of what object? At this point I have no clue!
 
  • #4
try forming an equation for each object that describes its position, and then use simultaneous equations
 
  • #5
it did not work, I'm not sure about one thing while tug is pulling asteroid, is tug moving? Ok I'm completely lost, I need someone's help badly, please!
 
  • #6
Are you assuming they are talking about some kind of winch? If the tug is pulling the asteroid, of course the tug is moving! Earlier you said "accelerations of these two masses are different". No, they are not! As long as the tow line is taut (and assuming it can't stretch), the two masses (tug and asteroid) have exactly the same motion.
 
  • #7
HallsofIvy said:
Are you assuming they are talking about some kind of winch? If the tug is pulling the asteroid, of course the tug is moving! Earlier you said "accelerations of these two masses are different". No, they are not! As long as the tow line is taut (and assuming it can't stretch), the two masses (tug and asteroid) have exactly the same motion.

ok if they have the same motions they must have the same accelerations which means they will never meet? I thought about winch, but there is nothing in the problem shows it. Can you please tell me if Xf=Xo+Vot+1/2at^2 is the correct equation to use for it because honestly I have no clue what to do right now and the problem is due pretty soon ((
 
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  • #8
HallsofIvy I re-read you message again and I have a question about your point: when you said that they have the same accelerations which is reasonable assuming taut's stretching is trivial, how in the world they suppose to meet? I mean we can neglect gravitational forces between then since the setting of the problem is in space. Damn that leaves no room for thought.
 
  • #9
May well be wrong, but the way I interpret the question is that the motor is fired long enough to establish tension in the line, then cut out, so that the force that established the tension is no longer operative. At that point, the tension will exert a brief acceleration on both objects toward each other. Any velocity achieved during the motor firing can be ignored.
 
  • #10
denverdoc said:
May well be wrong, but the way I interpret the question is that the motor is fired long enough to establish tension in the line, then cut out, so that the force that established the tension is no longer operative. At that point, the tension will exert a brief acceleration on both objects toward each other. Any velocity achieved during the motor firing can be ignored.

There is not enough info to calculate that...moreover there is no gravity present from what I understand. Considering if two objects are moving, especially one is pulling the other: there are to scenarios, either one is stationary and using some mechanism to pull the other towards itself (winch) or one has greater velocity. I have no damn clue
 
  • #11
v12alp said:
There is not enough info to calculate that...moreover there is no gravity present from what I understand. Considering if two objects are moving, especially one is pulling the other: there are to scenarios, either one is stationary and using some mechanism to pull the other towards itself (winch) or one has greater velocity. I have no damn clue

I think sometimes questions are worded deliberately vaguely to make you think thru all the scenarios, do it both ways, as I suggest and with a winch. See what you come up with.
 
  • #12
denverdoc said:
I think sometimes questions are worded deliberately vaguely to make you think thru all the scenarios, do it both ways, as I suggest and with a winch. See what you come up with.

I have done it and computer says that all answers are wrong...now I'm trying to write two separate equations for displacement and make then equal and solve for time we'll see...
 
  • #13
When the tug pulls on the asteriod, both objects move (accelerate) toward each other. So, find the acceleration for both objects, add them, and plug the value into the formula, X= Vot + .5a(t^2). The Vo here is zero, so just X = 550m, and you are solving for t.
**Don't assume one of the accelerations to be negative; the objects are moving towards each other, so relative to each other, you have to add them.**
 

Related to How Long Does It Take for a Space Tug and Asteroid to Meet When Pulled Together?

1. What is the "Complicated Asteroid problem"?

The "Complicated Asteroid problem" refers to the challenge of accurately predicting the trajectories of asteroids and their potential impact on Earth.

2. Why is the "Complicated Asteroid problem" important?

The "Complicated Asteroid problem" is important because it helps us understand and potentially prevent catastrophic asteroid impacts on Earth. This information is crucial for the safety and survival of our planet and its inhabitants.

3. How do scientists predict the trajectories of asteroids?

Scientists use a variety of methods to predict the trajectories of asteroids, including observing their positions and movements through telescopes and using computer simulations to model their orbits.

4. What makes the "Complicated Asteroid problem" difficult to solve?

The "Complicated Asteroid problem" is difficult to solve because it involves many complex factors, such as the gravitational pull of other objects in our solar system, the irregular shapes and compositions of asteroids, and the potential for asteroids to break apart or change course due to collisions or other forces.

5. What are some potential solutions to the "Complicated Asteroid problem"?

Potential solutions to the "Complicated Asteroid problem" include developing advanced detection and tracking technologies, launching missions to study and potentially deflect or redirect asteroids, and implementing early warning systems and emergency response plans in case of a potential impact.

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