How much greater are the tidal forces from heavenly bodies during apparent retrograde motion, compared to normal?

Earth's tides are mostly affected by The Moon and somewhat by The Sun. I read somewhere that the average tidal forces exerted by Jupiter are about a one-thousandth that of The Moon.

But, here's my question: These tidal forces will be at their most extreme when Jupiter (& Mars, etc.) is in Apparent Retrograde Motion, presumably particularly when there is the greatest apparent acceleration (incl. change in direction) as seen from Earth, which I picture in my mind like Jupiter's hand reaching out and cracking a whip in earth's direction. This also occurs when Earth is closest to these planets, so the gravitational effects will be greater for that reason alone.

Will the changes in gravitational forces at these most extreme times cause significant changes to tides, storm events, Earth's rotation, orbit or satellite orbits, etc.?

The gravitational acceleration of course goes like $GM/R^2$ and therefore any other body in the Solar System pulls a little stronger on the leading edge of the Earth than on the trailing edge of the Earth. Essentially all of the rock on the earth just gets pulled "on average" like the center of the Earth, but the fluids on the earth actually feel this force, it is not resisted by those internal forces holding things down to the ground. This difference between the forces at the center and the forces at the surface therefore causes a tendency to produce tides and there are in fact 2 of them, one when the object pulls the water up over the ground and another when the object pulls the ground out from under the water, so you get high tide at both the side of Earth facing the object and the side facing away, with low tide at points looking 90 degrees away.

However $R$ is in general vast in comparison to the radius of the Earth $r$ and therefore this difference leads directly to the tidal acceleration, which is just a Taylor expansion (standard calculus trick for small differences) of the above gravitational acceleration: $$\frac{GM}{(R + r)^2} \approx \frac{GM}{R^2} - \frac {2GM r}{R^3}.$$ In other words, the tidal force is weaker by a factor of $2r/R$ where $r$ is the radius of Earth and $R$ is the distance to the thing whose gravity is acting on Earth.

Note for example that the Sun's net gravitational acceleration (net = on the center) on us is ~6000 μm/s2 by the $GM/R^2$ formula, whereas the Moon's gravitational acceleration on us is ~30 μm/s2, a factor of 200 smaller. However because R is so large, tidally (tidal = subtracting the edge from the center) these numbers flip: the Moon is 0.00257 AU away compared to the Sun being 1 AU away, 1/0.00257 is 389, or roughly 400: so when we look at the situation tidally the 400 and the 200 cancel, and the Moon's tidal forces are about two times stronger than the Sun's; the Moon has a tidal force of about 1.1 μm/s2 whereas the Sun's tidal force is about 0.51 μm/s2. That's why the Moon determines the tides and the Sun just makes them stronger (at full moon or new moon) or weaker (at half-moons).

Now Jupiter is on average at 4.2 AU from Earth making its net gravitational force only 0.32 μm/s2. So in addition to Jupiter having a weaker force on us in general, where its net force is less than the Sun's tidal force, it's also even further away than the Sun!

Its tidal force is in fact 0.0000032 μm/s2. So you heard wrong, or more likely you heard right but misinterpreted: Jupiter's net gravitational force is less than the Moon's net gravitational force by a factor of a thousand; but Jupiter's tidal force is less than the Moon's tidal force by a factor of 3 million.

So now you want to talk about the variation in that. Bless your heart! In September of this year (2016) Jupiter got closer than it's going to get until 2022, coming to a distance of 368 million miles which is 3.96 AU.

So it's normally at ~4.2 and this year it came closest at ~4.0, the difference is $s = -0.2\text{ AU}$ and that s/R = -0.2/4.2 is about a -5% difference. Do you know what's coming next? Another differential expansion!

So now we know the tidal force is $-2GMr/R^3$ and we expand once more in $R$ to find $+6GMrs/R^4$, meaning that the tidal force grows to only $(1 + 3|s|/R)$ of its size when this happens.

In other words it might get 15% larger, maybe even 20%, but it'll still be roughly 3 million times smaller than the tidal forces of the Moon and the Sun.