I would like to add my own answer based purely on math, which is not as complex as you may think (but I explain each term and everything else so it looks long). We only need a couple equations.
First the Tsiolkovsky Rocket Equation:
$V_f = V_e \ln(\displaystyle \frac{m_i}{m_f})$
(Soon, we'll need this rearranged for $m_f$, which is $m_f = \displaystyle \frac{m_i}{e^\frac{V_f}{V_e}}$)
$V_f$ is the final velocity, or delta v, or change in velocity. If you're going from 0 to 10 km/s, then 10 km/s will be your final velocity.
$V_e$ is the effective exhaust velocity, which is basically how fast the exhaust is propelled out the engine. The faster, the better. $V_e$ is equivalent to the specific impulse (in seconds) multiplied by g (9.80665 m/s/s).
$m_i$ and $m_f$ are the initial and final masses of the whole rocket. These are both total masses.
(The $\ln$ part is a natural logarithm, which is log base e (2.718...). If you don't know what those are, don't worry, just use a calculator. It's noteworthy, though, that this logarithm makes the equation exponential.)
Second, we need to know the tangential velocity of Earth's surface, based on latitude.
Equatorial tangential speed will be $\displaystyle \frac{2\pi*6,371,000}{86,164}$ = 464.58 m/s. (radius of earth = 6371 km, siderial rotational period = 86164 seconds, which is actually 23 hours, 56 minutes, 4 seconds, not 24 hours even.)
For other latitudes, multiply by the cosine of the latitude. At 45 degrees north (or south for that matter), ground tangential speed will be $464.58 * \cos(45°)$ = 328.51 m/s. (be careful your calculator is in degree mode, not radian mode).
One more thing we need to know is the orbital speed our satellite needs. This is typically 7.8 km/s, HOWEVER: Due to air drag, gravity drag, and some obvious vertical acceleration, the typical carrier rocket needs to send its payload to something like 9.7 km/s. This is the figure I will use.
So for prograde orbits, the actual delta v your rocket needs is 9.7 km/s minus the ground speed. Obviously, the larger the ground speed, the less work our rocket needs to do. This is what drives launch sites towards the equator.
Now let's pick some figures for our rocket's final stage. I'll choose some based on the Soyuz' final (3rd) stage.
Let's say this final stage can take a 7-ton payload up to 4 km/s. Its total mass is 30 tons (including the payload), while its empty mass is 9.3 tons (including the payload). This implies that the empty 3rd stage without the payload is always 2.3 tons, and the effective exhaust velocity is 3.415 km/s.
Now with a boost from ground speed at the equator, the final stage only need to take this payload 3.535 km/s. Therefore, we can remove some fuel and add some payload mass. This will keep the total mass constant, because we're just trading fuel for payload.
This is the benefit of ground speed boost: increasing payload mass by reducing needed delta v.
So now it's time for a giant chart:
$$\begin{array}{|c|c|c|c|c|} \hline
\text{site} & \text{latitude} & \text{ground speed} & \text{final stage needed speed} & \text{Final Stage Masses}\\ \hline
\text{Equator} & 0° & \text{465 m/s} & \text{3.535 km/s} & \text{mf = 10.655 t, payload = 8.355 t}\\ \hline
\text{Korou} & 5° & \text{463 m/s} & \text{3.537 km/s} & \text{mf = 10.649 t, payload = 8.349 t}\\ \hline
\text{US Virgin Islands} & 10° & \text{458 m/s} & \text{3.542 km/s} & \text{mf = 10.633 t, payload = 8.333 t}\\ \hline
\text{Cape Canaveral} & 28° & \text{411 m/s} & \text{3.589 km/s} & \text{mf = 10.488 t, payload = 8.188 t}\\ \hline
\text{Tanegashima} & 30° & \text{403 m/s} & \text{3.597 km/s} & \text{mf = 10.464 t, payload = 8.164 t}\\ \hline
\text{Baikonur} & 46° & \text{323 m/s} & \text{3.677 km/s} & \text{mf = 10.221 t, payload = 7.921 t}\\ \hline
\text{Plesetsk} & 62° & \text{218 m/s} & \text{3.782 km/s} & \text{mf = 9.912 t, payload = 7.612 t}\\ \hline
\text{North Pole} & 90° & \text{0 m/s} & \text{4.000 km/s} & \text{mf = 9.300 t, payload = 7.000 t}\\ \hline
\end{array}$$
So comparing your two cases, Korou and Cape Canaveral, we see that we can get up to an additional 161 kg of payload. It may not seem like a lot, but keep in mind, we are paying tens of thousands of dollars for every pound we put into space.
By the same token, a reduction of 52 m/s may not seem like a lot either, but the exponential nature of the Tsiolkovsky Rocket Equation means that this is nothing to sneeze at.
In case you're wondering about the very small differences at first, but the larger differences later, see this graph, because the cosine function is exponential too. 45 degrees is halfway between pole and equator (50 percent), but take the cosine of it and you get something closer to 71 percent.
But guess what. It's not over:
This only holds for launches due east. If you're at 46 degrees north and launch due east, your sputnik ends up in an orbit inclined 46 degrees. But Baikonur launches in a 51.6 degree orbit, and used to launch in a 63 degree inclined orbit! This is to keep the sputnik over Russia, not China, for the early phase in case the rocket fails and the sputnik (or Soyuz with people aboard) comes back down. For these other inclinations, you'll need to do a vector subtraction in only one dimension to find the rocket needed speed.
Spy sputniks typically go into polar orbit (about a 90 degree inclination). So large ground speed is actually a bad thing. This actually drives launch sites towards the poles, if it's a launch site intended for spy sputniks, like the Plesetsk Cosmodrome.
There's an alternate way to apply benefits: keeping the payload mass constant, but sending it to a higher orbit.
If you're launching into a retrograde orbit (opposite of the way the Earth turns), then being closer to the equator is a disadvantage. Israel faces this problem as the only safe direction to launch their orbital rockets is to the west. I believe they orbit things in an approximately 120 degree inclination.
Latitude is not the only factor in choosing sites. Other factors include weather, security (USSR chose a central inland area for this), and transportability (Cape Canaveral is transportable by sea barge, so they can ship huge rockets without a problem).
Don't assume that the ratio of masses for any two examples in the chart will be the same ratio for the same latitudes of a different rocket. You really need to look at the rocket's final stage, see what it's Ve is and other parameters. Then use those numbers in the equations and make your own chart.
Earth isn't a perfect sphere. Equatorial radius is actually a bit more than 6371 km. There may also be a mountain or plateau somewhere that would give a decent boost by extra radius (which yields extra ground speed) and extra height above mean sea level (which yields less vertical acceleration needed). Of course, remote and inhospitable areas incur their own expenses if you're really thinking about building there, but these sort of boosts once spawned the concept of balloon-launched rockets, and still today there is talk of rockets launched from airplanes as well as those exotic launch systems like the Launch Loop which can run up a mountainside for the same "Newton's Cannonball" physical boost.
Alright. There's my answer based purely on math. If you dare, you can plug in your own example and figure out the exact benefits for your specific case.