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An artificial satellite is moving around the surface of earth. If the magnitude of the gravitational constant starts decreasing at a constant rate, then what would the effect on the path of the satellite be?

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$\let\om=\omega$ If $G$ varies very slowly, we can use the adiabatic invariants theorem. In present problem angular momentum is an adiabatic invariant. Therefore we have $$L = \mu r^2 \om = \mathrm{const.}$$ where $\mu$ is reduced mass: $$\mu = {M\,m \over M + m}.$$ Then $\om=k/r^2$, with $k$ some constant. Third Kepler's law says $$\om^2 r^3 = G\,(M + m)$$ and substituting for $\om$ $${k^2 \over r} = G\,(M + m)$$ i.e. $$r \propto {1 \over G}.$$

Conclusion: if $G$ decreases, $r$ increases.

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    $\begingroup$ What would be the justification for why $L$ is an adiabatic invariant? $\endgroup$ – N. Steinle Nov 24 '18 at 21:23
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The Satelite Motion can be represented in polar coordinate $r(t)$ and $\varphi(t)$

The equations of motion are:

${\frac {d^{2}}{d{t}^{2}}}r \left( t \right) -r \left( t \right) \left( {\frac {d}{dt}}\varphi \left( t \right) \right) ^{2}=-\mu\, \left( r \left( t \right) \right) ^{-2} $

$r \left( t \right) {\frac {d^{2}}{d{t}^{2}}}\varphi \left( t \right) +2\, \left( {\frac {d}{dt}}r \left( t \right) \right) {\frac {d}{dt}} \varphi \left( t \right) =0 $

$\mu$ is proportional to $G$ and $r$ is the satellite radius

We simulate the satellite path ($x=r\,\cos(\varphi)\,,y=r\,\sin(\varphi)$) with various values of $\mu$

Satellite Path

Result: If $\mu\propto G$ decreases, the satellite path increases

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It will spiral outward.

The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.

So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.

Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.

So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.

Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.

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  • $\begingroup$ Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it. $\endgroup$ – Elio Fabri Nov 24 '18 at 17:09
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Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.

If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.

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To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$

$x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,

$y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,

$x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.

The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.

fig1

If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, t\ge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.

fig2

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