U-turn in Deep Space

Question

While writing a physically realistic game ( "Asteroid Defender" ) a physical question came up whether Diag.1 or Diag.2 or Diag.3, correctly depicts reality.

In deep space (away from other celestial bodies), a perfectly spherical asteroid of mass m moves in a straight line with velocity $\overrightarrow{V0}$ relative to point C (red dot). Its motion is constant and uniform since no forces are acting on it.

The asteroid has uniform density so its Center of Mass (CoM) coincides with its geometric center. The asteroid is rigid and does not deform when touched or pushed. The asteroid does NOT spin about its CoM. The pale green rectangles appearing on the asteroid visualize the lack of asteroid's spin. This is depicted at times t-1 and t0 on the diagrams.

At time t1 a maneuverable spacetug (space-pusher for European readers) applies a force $\overrightarrow{F1}$ to the surface of the asteroid at point P1 (small yellow dot) via a rigid and flat pushplate, which is mounted in front of the spacetug (thick blue line). This force vector lies on a line connecting point `P1' and the CoM, thus it is incapable of causing the asteroid to spin about its CoM.

As time progresses, the spacetug continuously varies the direction of the applied force in such manner as to cause the asteroid to traverse a semicircular path (U-turn) of radius r centered around point C. The magnitude of this force remains constant throughout the U-turn - only its direction changes continuously.
At all times, the applied force vectors lie on lines connecting the CoM with the points at which the pushplate touches the surface of the asteroid (e.g.: P1 at t1, P2 at t2, P3 at t3, P4 at t4, P5 at t5). The pushplate does NOT slip on the surface of the asteroid and does not spin it about its CoM - the pushplate only pushes the asteroid. This is depicted on the diagrams at times from t1 to t5.

Once the asteroid completes the 180 degrees of the U-turn, the spacetug disengages and allows the asteroid to move away in a straight line at the velocity $\overrightarrow{-V0}$ which is parallel but opposite to the initial approach. The kinetic energy of the asteroid before and after the U-turn is the same. The asteroid does not spin about its CoM as it departs. This is depicted at times t6 and t7 on the Diagrams.

QUESTION: Which Diagram correctly depicts reality in this scenario?

Please justify why one diagram correctly depicts reality and the remaining ones - do not.

Diag. 1, depicts the lines (P1_CoM, ... P5_CoM) connecting asteroid's CoM and the points at which the pushplate touches the asteroid's surface (P1 at t1, ... P5 at t5), as always passing through the center of the U-turn ( point C ). The vectors ( $\overrightarrow{F1}$, ... $\overrightarrow{F5}$ ) lie on these lines. Zoom for more details. Diag. 2 and Diag.3 depict the lines (P1_CoM, ... P5_CoM) connecting asteroid's CoM and the points at which the pushplate touches the asteroid's surface (P1 at t1, ... P5 at t5), as passing through points (Q1, ... Q5), respectively, which do NOT coincide with point C.
In other words: the lines (P1_Q1, ...P5_Q5) on which the force vectors lie ( $\overrightarrow{F1}$, ... $\overrightarrow{F5}$ ), pass a certain distance x away from the point C.
Zoom for more details. Zoom for more details.

The red dashed line P0_Q0 is just a helper line that passes through the CoM at t1 and through the CoM at t5 and through point C. This line cannot be seen without zooming in.

-------------- EDIT ----------------
A question arose in the comments to Kamil's answer, whether it is possible to have a sum of two vectors $\overrightarrow{A}$ + $\overrightarrow{B}$ such that the magnitude of this sum is the same as the magnitude of the vector $\overrightarrow{A}$ alone?
The answer is "Yes", but that is possible only when the angle between these two vectors is >90º and <270º. See the formal proof here: https://i.stack.imgur.com/WK9nJ.jpg

Another EDIT: In response to the objction raised by Luke Pritchett in the comments below, I am linking an answer relevant to his objection: Asteroid Spin Prevention while Pushing

Answer 1

The get a semi-circular trajectory the transverse acceleration must be non-zero and constant. It is quite, simple. If the asteroid is moving with speed $v$, and a constant transverse acceleration of $a=a_T$ is applied, then the asteroid is going curve with a radius of curvature equals to $r = v^2/a_T$. The sweep rate is going to be $\omega = a_T/v$. The exit velocity is $v$, as there is zero longitudinal acceleration to speed up or slow down the asteroid.

This corresponds to Diagram 1.

Diagrams 2 and 3 are incorrect because the asteroid is not going to traverse a semi-circular path. Both are subsets of the general problem, where the line of action has a moment arm $d$ from the instant center of rotation (point C). For Diagram 2, $d>0$ and for Diagram 3 $d<0$. Of course, Diagram 1 is $d=0$.

Considering the lead angle $\theta$ formed by $d$ across $r$ (the radius of curvature) the acceleration $a$ is decomposed into two components

$$ \matrix{ a_T = a \cos \theta & a_L = a \sin \theta } \;\tag{1}$$

The trigonometry of the problem is such that $d = r \sin \theta$

The equations of motion are:

$$ \matrix{ \dot{v} = a \sin \theta & \frac{v^2}{r} = a \cos \theta} \; \tag{2} $$

The solution of the above at every instant is

$$ \boxed{ r = \sqrt{d^2 + \left( \frac{v^2}{a} \right)^2 } \\ \dot{v} = \frac{a^2\;d}{ \sqrt{v^4 + a^2 d^2} } }$$

which means that the radius depends on the speed, and the speed keeps changing in a non-linear fashion depending on the sign of $d$. Thus the path curvature changes with time making the asteroid trace a spiral shape.

Answer 2

At any moment the force component in the line of (tangent to) the momentary velocity changes the magnitude of the velocity (i.e. speed), but not the direction; the force component perpendicular (normal) to the line of the momentary velocity changes the direction of the velocity, but not its magnitude.

In diag. 1 the force is always perpendicular to the line of the momentary velocity, so the speed remains $V_0$.

In diag. 2 there's always a force component against the velocity; this reduces the speed, so it cannot be $V_0$ at the end of the maneuver.

In diag. 3 there's always a force component adding to the speed, so it cannot be $V_0$ at the end of the maneuver.

In either case the asteroid can move along the semicircle, but 2 and 3 require the spacetug to gradually change the magnitude of the perpendicular component of the force, not only the direction. This is because the perpendicular component that would keep a mass $m$ on the given circular trajectory with the radius $r$ depends on the speed $v$:

$$ F_p=\frac { m v^2 } r$$

I think it may be possible to keep the magnitude of the force constant in cases 2 and 3. Non-constant perpendicular component would require a non-constant tangent component, so the overall magnitude could stay constant. Still the non-zero tangent component would reduce (diag. 2) or increase (diag. 3) the speed over time.

From the three diagrams only the first one can give you $- \overrightarrow {V_0}$.

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Note U-turn in space is a waste of fuel. If the spacetug just applied force to the left, it could eventually stop the asteroid and then accelerate it to $- \overrightarrow {V_0}$. Planes in the atmosphere perform U-turns along semicircles because it's very easy to get normal forces from aerodynamics; plus they need to maintain speed, so they don't stall. In space, unless you need a specific trajectory, just push to the left long enough to change $\overrightarrow {V_0}$ to $- \overrightarrow {V_0}$.

Answer 3

An object with a center of mass that orbits a point in a circular path at radius $r$ has position vector $$\vec{x}(t) = r(\cos \theta(t), \sin\theta (t))$$ and hence must experience net force $$\vec{F}_{net} = mr\dot{\theta}^2 (-\cos\theta,-\sin\theta) + mr\ddot{\theta}(-\sin\theta,\cos\theta)$$ which has magnitude $$|\vec{F}_{net}| = mr\sqrt{\dot{\theta}^4+\ddot{\theta}^2}$$

For the magnitude of the force to be constant we must have $$ \vec{F}\cdot\dot{\vec{F}} = 0$$ $$\Rightarrow \dot{\omega}(2\omega^3+\ddot{\omega})=0$$ where $\omega = \dot{\theta}$ is the angular speed. There are two solutions: $\dot{\omega} = 0$ and $2\omega^3 + \ddot{\omega} = 0$. The second solution does not work because if $\omega >0$ then $\ddot{\omega} <0$, but that would mean the object could not come out of the semi-circular path at the same speed it started. This means that the object must travel the semi-circle at a constant speed, with $\dot{\omega} = 0$.

Looking at the equation for the net force we see that if $\ddot{\theta} = 0$, the force always points to the center of the circle. And finally, if the object is to not spin as it orbits the force must also point to the center of mass of the object. So if the object travels at a constant speed your Diagram 1 is the only correct answer.

Answer 4

To reverse the direction of the craft without orbital assist, the most fuel efficient way would be to fire thrusters exactly opposite to the direction traveled, until the craft comes to a complete stop and then starts moving back. the diagrams shown would rotate the craft but not efficiently reverse it's course. diagram one could reverse its coarse if the thrusters were fired continuously at t3 until the craft came to a complete stop and then came to desired opposite velocity. Merely rotating a projectile will not reverse its coarse. To efficiently rotate a craft you only need one off center burn to start it spinning and then one equal and opposite burn to stop its spinning at the desired point.