How Long Will It Take for a Spaceship and Asteroid to Meet?

[ShamTheCandle]

Homework Statement

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500 kg space tug and 6200 kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 450 m apart. How much time does it take for the ship and the asteroid to meet?

Homework Equations

$$F=ma$$

$$d=v_0t+\frac{1}{2}at^2$$

The Attempt at a Solution

I attempted to solve this problem by dividing 490 N by the mass of asteroid, and after finding acceleration, I used kinematics equation to solve for t. (of course, knowing that initial velocity is zero). But I doubt it is correct, because I didn't even use mass of space tug. And is it necessary to use it? This question is even-numbered question with double star (which implies very hard) in my textbook and no answer at the back. Thanks for helping me.

 

[cepheid]

This problem seems pretty weird to me. If the spaceship's engines are providing a thrust of 490 N, then the spaceship is going to accelerate as well, not just the asteroid. In fact, the spaceship's acceleration will be larger than the asteroid's, because it is less massive. Therefore, the spaceship will want to pull away from the asteroid, which can only happen if the cable stretches (and eventually snaps) and the answer is that the ship and the asteroid don't meet. If the cable is rigid (can't stretch or break), then it's more like a rod and the whole 9700 kg system accelerates as one unit. The answer is still that the spaceship and asteroid don't meet. Am I missing something?

 

[ShamTheCandle]

Thank you Cepheid for your response. But there is one thing that I don't understand:

Therefore, the spaceship will want to pull away from the asteroid...

Can you explain this?
In my opinion, the spaceship will come closer to asteroid.

 

[rl.bhat]

The spaceship exerts the force on the asteroid by pulling the line. As a reaction asteroid pulls the spaceship with equal force through the connecting line. Find the acceleration of spaceship and the asteroid. Since they are in the opposite direction, find the relative acceleration.
Using kinematic equation find t.As you have mentioned, the initial relative velocity of spaceship and asteroid is zero.

 

[cepheid]

As a reaction asteroid pulls the spaceship with equal force through the connecting line.

Right, I totally missed that part. So, does that mean the net force on the spaceship is zero?

EDIT: But the net force on the asteroid is not zero. It's being tugged on. If it starts to accelerate forward, the rope will lose tension, and the spaceship's engines will no longer be counteracted...I don't know why this problem is confusing me so much.

 

[rl.bhat]

Right, I totally missed that part. So, does that mean the net force on the spaceship is zero?

No.
Spaceship experiences only one force that is the reaction of asteroid. So it accelerates towards the asteroid.
Why don't you consider that the engine in the spaceship is winding the rope attached to asteroid, just like we do during fishing, and pulling the asteroid towards it?

 

[cepheid]

What about the thrust from the engines? Sorry if I'm being dense...

 

[cepheid]

No.
Why don't you consider that the engine in the spaceship is winding the rope attached to asteroid, just like we do during fishing, and pulling the asteroid towards it?

That never occurred to me. I assumed that the back of the spaceship was attached to the asteroid by a rope of fixed length and that the astronauts were firing rockets mounted to the ship (also facing towards the rear). I guess my scenario doesn't make much sense though. Yours makes more sense, although I have to say that it is not obvious to me from the wording of the problem that that is what was meant. Oh well, sorry for confusing the issue.

 

[rl.bhat]

That never occurred to me. I assumed that the back of the spaceship was attached to the asteroid by a rope of fixed length and that the astronauts were firing rockets mounted to the ship (also facing towards the rear). I guess my scenario doesn't make much sense though. Yours makes more sense, although I have to say that it is not obvious to me from the wording of the problem that that is what was meant. Oh well, sorry for confusing the issue.

It is alright.