The answers to Where does the extra kinetic energy come from in a gravitational slingshot? state that in a gravitational slingshot the object being accelerated "steal" speed from the planet (or moon). Does that mean that an excessive number of g-slingshot could stop a planet?
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$\begingroup$ Note that if you're talking actual slingshots by actual devices sent from earth, then the answer is different from if you're hypothesising some kind of science-fiction scenario where we can use (perhaps over time) masses adding up to totals comparable to the mass of planets. Random impacts from space-junk (meteors) no doubt adds up to far more impulse on any planet that what we could send in real life. $\endgroup$– Steve JessopCommented Sep 9, 2014 at 11:49
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$\begingroup$ Related: physics.stackexchange.com/q/38542/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Sep 9, 2014 at 13:02
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2 Answers
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No, from experimentation point of view (check Emilio's comments). Yes, from the "practical"-theoretical point of view. No, from the rigorous theoretical point of view. (edited paragraph).
Let's assume a planet with mass $m_p$, and a object in which will make a slingshot, of mass $m$. Planet has speed $v_p$. Mass $m$ has speed $v$. After slingshot, mass $m$ has speed $2v_p + v$.
Linear momentum must conserve. So, initially $p_i$ and after slingshot $p_f$. $$ p_i = mv + m_p v_p = m(v + 2v_p) + m_p v_p' = p_f $$
Where $v_p'$ is the final velocity of the planet. We can isolate it: $$ v_p' = \frac{mv + m_p v_p - m(v + 2v_p)}{m_p} = \frac{m_p v_p - 2mv_p}{m_p} = v_p - \frac{2m}{m_p}v_p $$
Therefore, the variation of planet speed $\Delta v_p = v_p' - v_p$ is: $$ \Delta v_p = -\frac{2m}{m_p}v_p $$
Now we can throw up $n$ times a mass $m$ object to peform slingshot, which means, $n$ slingshots. If $m_p >> m$ (which is of course true since you won't slingshot a planet in another planet) it is valid the approximation such that $\Delta v_p \approx dv_p$ and then we integrate over $n$ slingshots. $$ \frac{dv_p}{v_p} = -\frac{2m}{m_p}dn \quad\Longrightarrow\quad \int \frac{dv_p}{v_p}dn = -\int \frac{2m}{m_p}dn $$
$$ \ln v_p = -\frac{2m}{m_p}n + C $$
We can find out the integral constant such that be in function of $v_0$, where $v_0$ is the initial speed of the planet. Then we get a function $v_p(n)$, which means, velocity of the planet is dependent from the amount $n$ of slingshots performed.
We end up with: $$ v_p(n) = v_0 \exp\left(-\frac{2m}{m_p}n\right) $$
Where $n$ is the number of slingshots. So, you can see each slingshot the planet speed drops exponentially, and therefore, rigorously never reaches zero. But, it will be close enough to zero after a lot slingshots. A nice observation, from here we can notice that after a slingshot, the planet speed is independent from the initial speed of mass $m$ object. Only depends on mass $m$ of the object.
answered Sep 9, 2014 at 12:39
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1$\begingroup$ On the other hand, planet masses are in the range of $10^{23}$ to $10^{27}$ kg, and even the ISS is on the order of $10^4$ kg. This makes the exponential term be, optimistically, on the order of $$\exp\left(-n/10^{19}\right).$$ It would require a quintillion slingshots to make a dent on the planet's speed, so when you say "Yes, from the practical point of view", it is still a pretty theoretical point of view. $\endgroup$ Commented Sep 9, 2014 at 13:03
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$\begingroup$ @EmilioPisanty, haha, true. But algiogia is not interested on this. Question: "Does that mean that an excessive number of g-slingshot could stop a planet?" So, short answer, in practice, yes =) $\endgroup$ Commented Sep 9, 2014 at 13:07
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1$\begingroup$ I disagree. In theory, yes: you do enough slingshots until it grazes the sun and then Roche lobes and atmospheric drag will do the rest. However, you cannot get "in practice" statements by ignoring the fact that planet masses are astronomically big. Calculate, for example, the time interval between slingshots you would need to be done before the Sun turns into a red giant. $\endgroup$ Commented Sep 9, 2014 at 13:15
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$\begingroup$ I think I understood now. By practice I meant the exponential is pretty close to zero after large enough $n$, so, indeed still theory instead of experimentation. Maybe I should add quotes. Thanks =D. $\endgroup$ Commented Sep 9, 2014 at 13:21
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$\begingroup$ You also have to remember that while you slowdown a planet through gravity slingshots the planet would get a smaller semi-major axis and actually on average get an higher velocity. $\endgroup$ Commented Sep 9, 2014 at 14:29
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In theory yes. In practice it would require an enormous amount of momentum - you would have to steal the planet's speed/momentum/kinetic energy with by slingshotting objects that have very high velocities and/or a very large total mass. Because momentum is conserved, you could stop the planet with mass $M$ and orbital speed $v$ using a mass $M/4$ of small objects, all moving at $v$ in the opposite direction, all perfectly aimed. ($v$ is 30 kilometers per second for Earth.) These three factors (total mass, speed, and aim) can compensate for one another, in that if you have less of one you can make up for it with the others.
Actually, there is a name for this process: dynamical friction. If you have some large object passing through a field of randomly moving bodies, its velocity will approach the average velocity of the smaller bodies over time.
One example of this is two galaxies merging, each with a supermassive (millions of times the mass of the Sun) black hole at its center. In general the merger won't be exactly head-on, and the black holes' initial trajectories would have them miss each other by a large margin, but the black holes will both be dragged toward the center of the merged system by dynamical friction.