# Why does trajectory in the space curved by gravity, depend on the speed?

I am sorry about the probably naiive nature of this question (I am a software eng, not a physics student):

I (think I) understand the popular curved "trampoline" model of 2-dimensional space, curved by a mass in the center ("bottom") of the trampoline. A free-falling particle will orbit the center, and if the velocity of the particle is larger, the orbit will be farther from the center. This somehow follows from the particle following the "straight" line, where the straight line is defined, that locally, if you take two points on the line, the line is the shortest path between these two points. So if a particle on a larger orbit, suddenly performs "an engine burn" in the direction opposite of traveling, to lose some speed, it will then fall to a lower orbit.

But why will it fall? The velocity vector has the same direction as before, only it is shorter - but the straight line is the same as before, so it should travel along the same (larger) orbit, only slower...

What am I confusing here?

It's easier to think about things like this if you use a 2D (1D-space + time) analogue. When the object changes its 1D spatial speed, the direction of the overall 2D vector changes. Notably, the overall magnitude of the spacetime velocity is always the same and equal to the speed of light (modulo a $$+/-$$ sign that depends on an arbitrary convention) $$\vec{v}\cdot\vec{v} = c^2.$$
• That's not quite right either. Both the space and the time are curved. For example the spacetime metric for a point mass (Schwarzschild metric) has curvature in the radial and time directions ($\hat{r}$ and $\hat{t}$ in spherical coordinates). Feb 14 at 18:57