MHB Stopping a Spaceship: Calculating the Required Force from Backthrusters

[Dustinsfl]

So the question reads:

A spaceship is accelerating at 1000m/s^2. How much force is required from the backthrusters to completely stop the spaceship?

In space, nothing truly stops. You may not be accelerating but you are definitely orbiting something (planet, moon, sun, etc). We can put something at a stable Lagrange point, but the points are in an non inertial reference frame so even then the seemingly stationary object isn't stopped.

Is this question answerable?

 

[Ackbach]

So the question reads:

A spaceship is accelerating at 1000m/s^2. How much force is required from the backthrusters to completely stop the spaceship?

In space, nothing truly stops. You may not be accelerating but you are definitely orbiting something (planet, moon, sun, etc).

Actually, not necessarily. You could be in outer space, in a micro-gravity environment with virtually no gravitational force on you. You could also, theoretically, come to a stop in the reference frame of the fixed stars.

We can put something at a stable Lagrange point, but the points are in an non inertial reference frame so even then the seemingly stationary object isn't stopped.

Is this question answerable?

As is, I should think not. I think you'd need the know the mass of the spaceship. Are you given that?

Another thing that puzzles me: what force is causing the $1000$ m/s² acceleration? Is this the force that the backthrusters must counter?

 

[Nono713]

So the question reads:

A spaceship is accelerating at 1000m/s^2. How much force is required from the backthrusters to completely stop the spaceship?

This question makes absolutely no sense without being given the spaceship's mass and its current velocity or at least the time it's been accelerating. As it is the spaceship could be going at near the speed of light or at a few metres per second (in some frame of reference) and the answers would be widely different.

With the current data all you can say is that an acceleration of 1000 m/s^2 in the opposite direction will cause the spaceship to stop accelerating (in whatever frame of reference you are using). But it will still be moving. And never mind that you can't actually compute acceleration from force without being given the mass of the spaceship.

In space, nothing truly stops. You may not be accelerating but you are definitely orbiting something (planet, moon, sun, etc). We can put something at a stable Lagrange point, but the points are in an non inertial reference frame so even then the seemingly stationary object isn't stopped.

Well, "nothing truly stops" is hardly a useful definition of motion. Evidently two objects which aren't moving in the exact same direction at the exact same speed appear to be moving from one another's reference frame. Otherwise, they appear immobile relative to one another. Barring relativity, if I'm moving from your perspective, then you're moving from mine. You don't have to be in space for this to be the case, when I walk the entire planet is in fact receding beneath my feet, and so on. So I'm not sure what the statement is trying to say.

Also, if you are orbiting something then you are definitely accelerating in some frame of reference (but not in others). It's all relative.

Is this question answerable?

No, since it is meaningless.

 

[chisigma]

One of the most critical problems in Apollo Missions on the 'sixty years' was the landing of the space shuttle on lunar ground. Of course the use of a parachute is not applicable and the only way to contrast the gravity and stop the shuttle is the appropriate use of a posterior rocket. If the initial speed $v_{0}$ at time $t_{0}$ and the 'stop time' T are given, then it is possible to program the thrust of the rocket as function of time to be $S_{1}$ for $t_{0} < t < t_{0} + T$ in order to have v = 0 at $t = t_{0} + T$ and $S_{2} < S_{1}$ for $t > t_{0} + T$ in order to mantain $v=0$ for $t > t_{0} + T$ ... Kind regards$\chi$ $\sigma$

 

[Nono713]

One of the most critical problems in Apollo Missions on the 'sixty years' was the landing of the space shuttle on lunar ground. Of course the use of a parachute is not applicable and the only way to contrast the gravity and stop the shuttle is the appropriate use of a posterior rocket. If the initial speed $v_{0}$ at time $t_{0}$ and the 'stop time' T are given, then it is possible to program the thrust of the rocket as function of time to be $S_{1}$ for $t_{0} < t < t_{0} + T$ in order to have v = 0 at $t = t_{0} + T$ and $S_{2} < S_{1}$ for $t > t_{0} + T$ in order to mantain $v=0$ for $t > t_{0} + T$ ... Kind regards$\chi$ $\sigma$

I assume you meant the LEM (lunar excursion module) ;D the space shuttle never left low Earth orbit and did not have near enough delta-v to even reach the moon, let alone land on it and get back home. Good analysis, though in practice this analytical solution is improved over time using a PID controller to account for gravitational anomalies, rough terrain, rotation, and so on...

To think all this was done on a computer infinitely less powerful than your average cellphone these days... impressive.