U-turn in Deep Space

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://imgur.com/LELihq9

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

• Isn’t there a stack about games and games design? – user207455 Jul 13 at 14:47
• Yes, there is, but this is a physics question. The application does not determine its nature. – George Robinson Jul 13 at 14:58
• "pushplate does NOT slip on the surface of the asteroid and does not spin it about its CoM - the pushplate only pushes the asteroid." I'm pretty sure this is inconsistent. In order for the push to not spin the asteroid the force has to always point through to CoM to point c. This means the point of application has to rotate also be on the line from CoM to point c, which rotates around. To make this happen either the plate has to slide along the outside of the asteroid or the asteroid has to spin around at the same rate it rotates around c. – Luke Pritchett Jul 13 at 15:52
• @Luke: There is a third solution. The pushplate can relatively "roll" on the surface of the asteroid...and this is exactly what the Diagrams show. Also, I cannot agree with you that: "...In order for the pushplate to not spin the asteroid the force has to always point through to CoM to point C" - pointing to the point Cis not necessary to prevent spin, only pointing to CoM is necessary for spin prevention. I can open up another question about it if it will help. – George Robinson Jul 13 at 17:07
• The force must point to c in order for the asteroid to move in a circle around c at a constant speed. The force must point to the CoM for the planet not to spin. Therefore, the force must point through both. I did not consider rolling though. That makes Diagram 1 correct. – Luke Pritchett Jul 14 at 0:02

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 $$- \overrightarrow {V_0}$$.

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 $$- \overrightarrow {V_0}$$.

• @GeorgeRobinson If both vectors are perpendicular to the velocity, they lay on a plane perpendicular to the velocity, so their sum lays on the same plane, so the sum is also perpendicular to the velocity. – Kamil Maciorowski Jul 13 at 16:41
• Which vectors are you referring to? If we talk about vec addition, let's talk about adding velocity vectors to velocity vectors, or force vectors to force vectors. It is obvious that force causes acceleration (a=F/m), but acceleration is not the same as velocity...albeit it is the rate of change of it. You wrote: "...changes the direction of the velocity, but not its magnitude." - why not the magnitude, as well? When I throw a rock horizontally, the perpendicular gravitational attraction causes a parabolic trajectory, in which both the magnitude and direction of the rock's velocity changes. – George Robinson Jul 13 at 16:57
• @GeorgeRobinson so where is the gravity you mention for the rock - you state you are in deep space so the rock will travel in a straight line. – user207455 Jul 13 at 17:56
• @Mike: The rock example was just an illustration that an addition of another perpendicular vector changes the direction and magnitude of the vector sum. In my Original Posting there is no gravity. However, the different, albeit related, parabolic scenario can be accomplished with a thruster in lieu of gravity (and act just like a horizontal rock throw) . See: imgur.com/9dwtEPu – George Robinson Jul 14 at 10:08
• NOTE: How in this example, the appearance of acceleration in the Y direction causes velocities V1, ...V8 to be added to the perpendicular V0, which changes BOTH the magnitude and direction of the resultant vector sum VT. All without gravity. See: imgur.com/9dwtEPu – George Robinson Jul 14 at 10:13

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.

• How do you reconcile your answer with the proof at imgur.com/LELihq9 ? In Diag.1, the initial F1 is perpendicular to V0 so it can only cause perpendicular velocity components to be added to V0. According to the proof, the MAGNITUDE of the vector sum of V0 + another velocity vector, cannot be equal to the magnitude of V0 when the angle between the summands is 90deg. If you want to see what happens when another velocity vector is added to V0 at 90deg, then see this different, albeit related, scenario at: imgur.com/9dwtEPu – George Robinson Jul 14 at 6:53
• The problem with that proof is that it doesn't use infinitesimal changes in velocity. If we want to know whether $v$ has constant magnitude at all times we need to take the derivative of $|\vec{v}| = \sqrt{\vec{v}\cdot\vec{v}}$. The derivative of that is $\frac{d|\vec{v}|}{dt}=\frac{\vec{a}\cdot\vec{v}}{|\vec{v}|}$, so the speed only remains constant when $\vec{a}\cdot\vec{v} = 0$. Again, we are not adding vectors perpendicular to $v$, we are adding infinitesimal vectors perpendicular to $v$. There is no – Luke Pritchett Jul 14 at 13:26
• I thought that rules of vector addition are not affected by the size of its summands. Anyway, how can adding infinitely small vectors, which are PERPENDICULAR to V0, decrease (...and eventually even reverse) it to (-V0) ? – George Robinson Jul 14 at 13:41
• Because you add a new infinitesimal at each instant. By adding up many infinitesimals at each instant you can get any change you want over a period of time. You have to consider it this way because you have a constantly changing force, and hence you have to consider infinitesimal time changes. – Luke Pritchett Jul 14 at 13:46
• Consider two velocity vectors $\vec{v}_{A,B}$ with the same magnitude and angle $\theta$ between them. The angle $\Delta v = \vec{v}_B-\vec{v}_A$ makes with $\vec{v}_A$ is determined by $\Delta v\cdot \vec{v}_A = \vec{v}_B\cdot \vec{v}_{A} - |\vec{v}_A|^2 = |\vec{v}_A|^2(\cos\theta - 1)$. As we consider velocity vectors that are very close to each other on the circular path we have $\theta\rightarrow 0$ and so $\cos\theta-1\rightarrow 0$ and $\Delta \vec{v}\cdot \vec{v}_A \rightarrow 0$. – Luke Pritchett Jul 14 at 14:18

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.

• How do you reconcile the addition of a velocity v_a (caused by the acceleration a) perpendicular to the initial velocity v (existing before this addition) resulting in a vector sum v_sum, which according to this proof must have a magnitude greater than the initial velocity v` ? – George Robinson Jul 26 at 23:57
• @GeorgeRobinson - it is well known that a perpendicular acceleration only changes the direction of velocity and not its magnitude. – ja72 Jul 27 at 18:14
• Why? Over any non-zero time interval, doesn't an acceleration always change the magnitude of the velocity component laying on the same line as the acceleration vector (or force vector) ? ...as in this example – George Robinson Jul 28 at 13:27
• @GeorgeRobinson there is no tangent component ever of acceleration and thus the speed doesn't change. The comments aren't a great place to prove this. Ask a new question if you don't believe me, specific to perpendicular acceleration. – ja72 Jul 28 at 14:39
• I'd like to create a new question about perpendicular acceleration. Could you you propose the title of this question so I do not twist your words? – George Robinson Jul 28 at 16:49

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.

• Thank you for the answer, but the rocket fuel efficiency is irrelevant to this problem. The semicircular trajectory is mandatory. – George Robinson Jul 13 at 16:14