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If a satellite is in a stable circular orbit and goes about 41% faster (escape velocity) then it leaves its host forever. I get that. However, what if it speeds up by less than 41%?

Intuitively, it would seem to make the satellite move farther away from the host and thus enter a higher (more distant) orbit.

However, according to my understanding, a stable orbit requires the satellite to move more slowly the farther away it is from the host. For example, the earth moves more slowly around than the sun than Venus because it is farther away from the sun than Venus.

So, if a satellite speeds up then the stable orbit would be closer to the host, not farther away. What am I missing here?

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  • $\begingroup$ As to the question in the title, without any further specification the outcome depends on whether the acceleration is up or down. $\endgroup$
    – my2cts
    Aug 29, 2020 at 10:47
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    $\begingroup$ Have you ever played Kerbal Space Program? It is a wonderful game that features realistic orbital mechanics $\endgroup$ Aug 29, 2020 at 12:34
  • $\begingroup$ @my2cts The direction in which the satellite speeds up is of no importance. $\endgroup$ Aug 29, 2020 at 16:37
  • $\begingroup$ @descheleschilder it does not matter if it is up or down by the same amount, but the vector direction of the impulse certainly matters. $\endgroup$
    – ProfRob
    Aug 29, 2020 at 18:28
  • $\begingroup$ @RobJeffries Well to a certain extent. If you change the direction (and magnitude, i.e. the speed) the satellite **could hit the host, but only for a small range of angles for which it will crash on the host. BANG!!! I'm not sure if this range depends on the radius of the host (assuming it's a spherical planet). $\endgroup$ Aug 29, 2020 at 22:39

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The parameters of stationary orbits depend on the energy and orbital momentum. For circular orbits, we have a simple relationship for orbital velocity and orbit radius $v^2 =\frac {GM}{r}$. It follows that Venus moves faster than Earth, and Mercury moves faster than Venus. However, for elliptical orbits, the speed does not depend only on the radius. Maneuvering with the increasing speed at a given point of the orbit leads to a change in the shape of the orbit. For example, a circular orbit becomes an ellipse, parabola, or hyperbola. Figure 1 shows examples of maneuvers with a transition to an elliptical orbit (the speed increases by 30%) and a parabolic orbit (the speed increases by 41%).

fig1

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  • $\begingroup$ You increased the speed in a particular direction though? $\endgroup$
    – ProfRob
    Aug 29, 2020 at 18:35
  • $\begingroup$ @RobJeffries For circular orbit we have nonzero tangential component of speed only, so it increases by 30% or 41%. $\endgroup$ Aug 29, 2020 at 21:57
  • $\begingroup$ You have assumed that the increase in speed is provided by adding to the tangential velocity. The trajectory on the left does not look like that if there is a radial component added. $\endgroup$
    – ProfRob
    Aug 29, 2020 at 22:09
  • $\begingroup$ @RobJeffries Yes, it looks like tangential component been increased by 30%. $\endgroup$ Aug 30, 2020 at 10:33
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If the satellite will reach a velocity somewhat smaller than the escape velocity then the satellite (coming out of the circular motion), the satellite will get both closer and farther away.
It will follow an elliptic trajectory, the earth being one of the ellipse's two focus points. This means the satellite will pass both the aphelion (the point that closest to earth, smaller than the radius of the circle) as the perihelion (the point which is the furthest distance removed from the earth, bigger than the radius of the circle). See the picture below, where the $F$'s are the two foci (the earth stays virtually put on one of these).
If the satellite will get out of the circular orbit with a velocity higher than the escape velocity it will escape us forever. Depending on the speed, the corresponding trajectory will be a parabola or a hyperbola. See the second picture where the ellipse is a the intersection of a flat plane with a cone (as are the parabola and the hyperbola).

enter image description here fig.1 enter image description here fig. 2

For a certain range of angles of the increase in speed the satellite will crash on the host. The majority of angles though will not make this happen. The range of angles for which a collision does appear depends, or not, on the mass of the host. I'm not sure if this range of angles is scale-independent. You have to calculate it. It's Sunday and I'm not in the mood to calculate...

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  • $\begingroup$ Why the F**K a downvote? $\endgroup$ Aug 29, 2020 at 3:20
  • $\begingroup$ What do you mean by "the satellite will get both closer and farther away". Compared to the original orbit? $\endgroup$
    – ProfRob
    Aug 29, 2020 at 8:22
  • $\begingroup$ @RobJeffries That the satellite will move both through the aphelion and the perihelium. $\endgroup$ Aug 29, 2020 at 10:58
  • $\begingroup$ @RobJeffries Thanks... $\endgroup$ Aug 29, 2020 at 11:21
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    $\begingroup$ You don't read anybody else's answers? Even the ones you "edit". $\endgroup$
    – ProfRob
    Aug 29, 2020 at 18:23
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What you are missing is that as the satellite moves farther away from the Earth, it slows down because of work done by the gravitational field.

Since you have given the satellite more total energy (kinetic plus potential) and because the total energy of a bound orbit is $-GMm/2a$, where $a$ is the semi-major axis, then to increase the energy (make it less negative), $a$ must increase.

For a tangential $\Delta v$, the resulting orbit is an ellipse with a perigee at the point where the velocity was added to the satellite. This is because both energy and angular momentum are conserved - when the satellite returns to the same radius it must have the same tangential velocity (to conserve angular momentum), but then cannot have an extra radial velocity component because that would change the total energy. This is a Hohmann transfer orbit.

On the other hand, a radial impulse adds energy but does not change the angular momentum. The result is an elliptical orbit with a larger $a$, but with a perigee closer to the Earth. The satellite moves outwards first but then falls back. When the satellite returns to the same radius then the tangential part of its velocity must be the same as before the impulse, but to conserve energy there must be an inward radial velocity taking it closer to the Earth. i.e. a more eccentric ellipse is produced than if the same impulse is given tangentially.

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An orbit of a mass $m$ in the Newton potential of an immobile (for simplicity) mass M can be characterised by two constants of the motion energy, $E=mv^2/2-GMm/r$ and angular momentum L. E fixes v as function of r and L subsequently its direction. For a circular orbit v and r are constant. The increase r for a circular orbit you first need to boost the speed. The orbit becomes elliptical and r varies in time. At the desired r you then give a second boost to set v to the value and direction belonging to a circular orbital of that r. So you increased the speed (at least) twice but the kinetic in the end decreases. The extra kinetic energy was converted into potential energy.

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Since the stable velocity v for a given orbital radius $R$ is given by $v = \sqrt\frac{GM}{R}$, I would assume that the satellite spirals outwards at an accelerating rate. Since its velocity is greater than the stable velocity for that orbit, it will start to increase the radius of that orbit. As the radius increases, the magnitude of the gravitational force decreases, and thus the rate of spiraling increases (therefore, an accelerating spiral). I'm not sure myself either because this would also imply that the satellite will eventually leave the host, but this is my best guess.

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    $\begingroup$ "best guesses" should be comments, not answers. $\endgroup$
    – garyp
    Apr 20, 2019 at 18:22

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