I did some research and I think I can answer my question myself now, after all. I hope you find it interesting.
As it turns out, the technology of the Kepler space telescope would indeed allow detection of all Solar system planets except Mercury and probably Mars, i.e. all of them are big enough to be seen by it from a distance of about 2,000 ly. However, that doesn't mean that we would also be able to detect (=scientifically confirm) all of them, even if we would look at the planets' plane head on.
The reason for this is the long time we would need to gather the data. Kepler observes transits of planets in front of their sun. To get a planet confirmed there's more than just one transit observation needed. So if we look at Jupiter (large enough to be detected), with an orbital period of twelve years it would take 36 years of data to confirm its existence by this measure/method. Even harder: Saturn, Uranus, Neptune with orbital periods of 29, 84, and 165y, resp. However, Kepler would be able to detect and confirm Earth and Venus with about 3 years of data. That Kepler has found so many large planets so far has actually nothing to do with their size but with their relatively short orbital periods.
So the answer is: Yes, if we would look at the Solar system with the Kepler space telescope from 2,000 ly away and from an angle to see the planets transit the Sun, we would have been able to confirm Venus and Earth with the transit observation method. Maybe we would have seen one or the other transit by one of the four gas giants but not enough of them. However it would still be possible to use the data to try to confirm them with alternative methods (such as radial velocity observations) from other, even groundbased, telescopes.
That said, as the Kepler telescope has now stopped gathering further data and has been operating for about four and a half years it's quite possible that we find some Earth and Venus sized planets if their orbital periods are not much longer than 1, maybe 2 years. And while the transit method might not lend itself to easily detect even gas giants if they are too far out, they might be identified with other methods once everyone zooms in on the most interesting planetary systems.
why do the mean losv and dispersion look this way?
The radial velocity profile tells you that the stars in the disk are rotating in a coherent fashion. You can calculate the rotational velocity by correcting for the inclination: $v = {v_{rad}\over \sin(\theta)}$. If you plot the velocity profile as a function of the distance $r$ form the center of the galaxy, you will obtain the famous flat rotation curve often used as proof for the existence of dark matter.
Indeed, from the velocity profile you can infer the density profile as follows. In order for a star to be in a circular orbit, its velocity must satisfy
$${v(r)^2 \over r} = -{d \Phi(r) \over d r}$$
Poisson equation in a spherical potential (the DM halo is approximately spherical and it contributes the most to the potential) is:
$${1 \over r^2} {d \over dr}(r^2 {d\Phi \over dr}) = -4\pi G \rho$$
Substituting the first equation in the second one obtains
$$4\pi G \rho = {1 \over r^2} {d \over dr}(r v^2) = {1 \over r^2}(v^2 + 2rv {dv \over dr} )$$
$$\rho(r) = {v(r)^2 \over 4 \pi G r^2} (1 + 2 {r \over v(r)}{dv(r) \over dr})$$
If $v(r)$ is constant (flat profile) then $\rho(r) \propto r^{-2}$, also called singular isothermal sphere profile and this should roughly match the radial density profile of your DM halo.
In the bulge, things are different. Stars are not rotating coherently, each start follows its own orbit. This leads to a low average radial velocity (because the random velocities average to zero) and to a high radial velocity dispersion $\sigma_{rad}$. Assuming that the velocity distribution is isotropic, you can calculate the total velocity dispersion as $\sigma = \sqrt 3 \sigma_{rad}$.
From it, you can estimate the total mass of the bulge using the virial theorem.
$$2K = -U \implies M\sigma^2 = \alpha {G M^2 \over R}$$
$$M = {R \sigma^2 \over G \alpha}$$
Where $R$ can be the radius of the bulge and $\alpha$ is a geometrical factor of the order of $1$, that depends on the specific density profile of the bulge.
Additionally, you could use the $M-\sigma$ relation to estimate the mass of the central black hole, but in your case you don't mention any black hole, so I think this may not be relevant
Best Answer
I'm going to work with gravitational potential, a scalar, rather than the gravitational field, a vector.
The deepest gravitational potential well in the solar system is that of the Sun. The next deepest well is that of Jupiter, but it is only 1% as deep. Earth's is 3000 times shallower.
Suppose all the planets happened to line up in a row, but with their correct spacing based on the semimajor axis of their orbit. Then the gravitational potential along that line would look like this:
The horizontal axis is in astronomical units. The vertical axis is gravitational potential, normalized so that the potential at the center of the Sun is -1. That is 50 times further down than the graph goes.
The little dip of a few pixels at 1 AU is Earth's well, superposed on the Sun's well. You can also see an even smaller dip to the left around 0.7 AU which is Venus. Mercury and Mars's wells are too small to see. The four large dips are Jupiter, Saturn, Uranus, and Neptune.
The planets have such small radii compared with their distances from the Sun that their gravitational wells are basically just spikes when you look at the whole solar system out to 32 AU.
We can zoom in to the vicinity of Earth to look at its well in more detail:
At this scale, the Sun's well appears flat.
Here is a greyscale density plot of the gravitational potential in the ecliptic plane showing the outer planets Jupiter, Saturn, Uranus, and Neptune, in their relative positions today:
The Sun is at the center but its well has not been included because it makes it impossible to see the planets' wells. (They are too shallow by comparison, and there aren't enough shades of gray.) The planets' wells look big and blurry here because I've taken the logarithm of the potential to make the wells less spiky. Otherwise, you'd just see a pixel or two for them. The other planets' wells don't show up because their wells are so shallow.