Technically this answer doesn't specify what slows it down, but it does explain how it does.
I've never done anything with orbital mechanics before today, but I got bored and read a few articles, linked below and made a scale model of the Parker solar probe's final orbit in my program I wrote from scratch!
What you see here is the arguments of the orbit of the parker solar probe:
0.043 AU is the distance from the sun it will end up at, this is the first oblong orbit that you see.
0.738 AU is the final distance it will be travelling after the 6 fly-bys of venus it makes.
- With each flyby of venus it lowers the lowest point of the orbit.
- When it starts at earth, it's going to be at approximately 1 AU away from the sun!
- The animation shows the difference between the starting orbit (Earth) and it's final one.
Note: AU is arbitrary units in the graph, don't pay attention to the scaling for AU.
If you look at the points on the graph, every single individual point is a specific unit of time.
The only important thing to know is that the time it takes to go between any two dots is the same.
- Notice how the dots are super far apart closer to the sun!
- This means that it takes a the same amount of time to traverse more distance.
- This is also known simply as "going faster".
At it's final orbit, closest to the sun, it will be going 700,000 km/h (wikipedia).
- The Earth revolves around the sun at about ~107200 km/h (wikipedia).
- The parker solar probe will be lowering one end of its orbit to a measly 7.3% of earth's lowest.
- This is done by using Venus' gravity to repeatedly slow (yes, slow!) the highest orbital point.
- By slowing the highest point, you stay there for a longer time and a smaller amount at the lowest.
- However, you travel the same amount of distance on both sides of the orbit!
This is called eccentricity (cue the graph to the bottom left, and the outer ellipse):
- "Eccentric" is the word to describe "non-circular" orbits, or ones with a low point.
- Earth is about .01673 eccentric, that's basically nothing, we're a perfect circle.
- Parker solar probe will be approximately .846 eccentric by my calculations (probably off).
Now, lastly, lets look at that velocity graph.
- Notice how the axis of the graph shrinks dramatically when we go from "Parker" to "Earth".
- The maximum and minimum vector velocities shrink with a less "eccentric" orbit.
- With a perfectly circular orbit, you're not changing speed at all (simplified greatly).
- Look at how evenly the dots are spaced, all line segments are equal!
- This means they're equal in time (length) and speed (distance between two dots over time)!
PS: Anybody please correct anything glaringly wrong with this answer, I want to learn too.