1
$\begingroup$

Let's assume a spaceship is at rest and that it increases the acceleration from 0 to g m/s in 1 second and that it then continues accelerating at 1g for an indefinite length of time.

  1. This means that after 1 second the spaceship would be traveling at roughly 35km/h (9.81 x 3600 / 1000) and would be traveling at roughly 70km/h after 2 seconds, right?
  2. As long as the spaceship accelerates at a constant 1g a person in the spaceship would feel artificial 1g (thrust) gravity pushing them towards the thruster, but this artificial gravity would stop as soon as the acceleration stopped, right?
  3. As there is no friction or air resistance in space (and assuming the spaceship's course is far enough from gravitational fields), the spaceship would feel no forces acting on it once the thrust stopped and would hence keep moving at a constant speed indefinitely, right?
$\endgroup$
4
  • $\begingroup$ Thanks, I improved the explanation based on your comment $\endgroup$ Apr 10 at 5:49
  • $\begingroup$ The speed at t=1 would be smaller because the acceleration isn't 1 g for the whole second. If the acceleration in that first second increases linearly, then the speed at t=1 is half of what it'd be from a constant 1 g acceleration, so ~ 17.652 km/h. $\endgroup$
    – PM 2Ring
    Apr 10 at 6:06
  • $\begingroup$ @PM2Ring He actually means the acceleration is $g$ m/s/s for one second. In other words, it goes from v=0 to v=9.8 m/s in one second. $\endgroup$
    – joseph h
    Apr 10 at 6:31
  • $\begingroup$ @joseph Perhaps that is what Adrien means, but that isn't what they wrote. $\endgroup$
    – PM 2Ring
    Apr 10 at 6:54
0
$\begingroup$
  1. If you mean accelerate to a velocity $9.8m/s$ then yes. Remember $$v = a\times t$$ if $a=g=9.8 \ ms^{-2}$ is the acceleration and the object starts from rest. Your conversion to $km/h$ looks correct. So after two seconds the rocket is travelling with a velocity $70 \ km \ hr^{-1}$

  2. Yes. They would feel an inertial force equal to $-mg$ if the ship is accelerating at $+g \ ms^{-2}$. So when the acceleration stops, this inertial force is no longer acting, since there is no forward force acting, the person's body is no longer resisting a change. Remember Newton's first law An object at rest will stay at rest and an object in motion will stay in motion with a constant velocity unless acted upon by an unbalanced force

  3. Yes. If there are no forces acting on it, then it will continue with the velocity it had at the instant the thrust was turned off. Newton's second law states that if the net force on an object $$F_N=\sum_i^n F_i$$ where $F_i$ is each force, and if all the $F_i$ sum to zero, then $$F_N=ma_N=m\frac{dv}{dt}=0$$ or $$v=\text{constant}$$

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$ Apr 10 at 6:46
0
$\begingroup$

To point 1

If the acceleration is varying over the first second then you can't simply use the final acceleration to find the speed after that second.

  • With a constant acceleration of $9.8\;\mathrm{m/s^2}$, meaning $9.8$ metres-per-second added every second, then sure, you will from rest reach a speed of $9.8\;\mathrm{m/s}$ ($35\;\mathrm{km/hr}$) after one second and $19.6\;\mathrm{m/s}$ ($70\;\mathrm{km/hr}$) after two seconds.

  • If the acceleration is not at $9.8\;\mathrm{m/s^2}$ for the full first second, though, then you won't reach a speed of $9.8\;\mathrm{m/s}$ during that first second. Rather, assuming the acceleration to increase linearly from $0\;\mathrm{m/s}$ to $9.8\;\mathrm{m/s}$, you only gain half the speed, so $4.9\;\mathrm{m/s}$ ($17\;\mathrm{km/hr}$) after one second and then $14.7\;\mathrm{m/s}$ ($52\;\mathrm{km/hr}$) after two seconds.

The gained speed is the area under the acceleration graph. By assuming a linearly increasing acceleration, this area is exactly half of what it would have been if the acceleration had started out at full value. If the acceleration increase is not linear, it might be a bit harder and you might need a modified version of a kinematic equation. To solve it.

To points 1 and 3

Yes and yes.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks @Steeven for the clarification $\endgroup$ Apr 11 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.