I just posted Why is the centrifugal force radially outwards?, but that may not be 9th grade level.
To avoid semantics and corner cases, let's restrict this discussion to a single mass undergoing uniform planar circular motion around a point called The Origin.
A centripetal force is any and all forces that point inward to the origin. If it's Earth going around the Sun, then it's the Sun's gravitational pull. If it's David's rock in his sling, then it's the tension force in the rope. On a loop-de-loop at Six Flags, it's a combination of gravity and the contact force of the rails on the wheels.
A centripetal force can be in an inertial frame, or a non-inertial frame. All that matters is that:
$$ f = m\frac{v^2} r $$
where $r$ is the distance to the origin (again, we're 2D in a plane to avoid semantics around "axis" vs "origin". etc), and $v$ is the tangential velocity (again, I am tempted to say $r$ is the radius of curvature and $v$ is that tangential velocity--that would apply to an arbitrary path--and this is why I want to stick to uniform circular motion in a plane).
In any and all inertial frames: the centrifugal force is zero. One of the biggest lessons in classical physics is that nature and physical law does not care about our coordinate systems. A foundational law is Newton's:
$$ \vec F = m\vec a $$
and it holds in inertial frames with no extra work.
In a rotating frame, there is a problem with Newton's Law: it's violated.
Here on Earth, we have the precession of a Foucault pendulum, for example.
For a simpler example (always start simple and work your way up), consider a 2D reference frame centered on the Sun, with the Earth at $R$. (pls ignore Barycenter for now--the Earth revolves around the center of a stationary Sun).
In this frame:
Earth moves at
$$ v = \frac{2\pi\,{\rm A.U.}}{\rm year} $$
in a circle of radius:
$$ R = 1{\rm A.U.} $$
The Sun's gravity applies a centripetal force, such that the acceleration of the Earth is:
$$ a_{\oplus} = -\frac{v_{\oplus}^2}{R} = 4\pi^2
\,{\rm\frac{A.U.}{\rm year}^2}
$$
That minus sign indicates the acceleration is radially inward.
That requires a force:
$$ f = M_{\oplus}a_{\oplus} $$
and it is from gravity:
$$ f_g = -GM_{\odot}\frac{M_{\oplus}}{R^2} $$
The minus sign indicates the force is attractive.
You can verify:
$$ f_g = M_{\oplus}a_{\oplus} $$
and Newton's Laws work.
Now let's go to a rotating frame. Same origin, but we're going to rotate the frame at:
$$ \Omega =2\pi\,{\rm year}^{-1}$$
(Since it's a 2D coordinate system, we don't need a $\hat z$ modifying it).
In the coordinate system, the Earth sits at $R$ with velocity:
$$ v'_{\oplus} = 0$$
with acceleration:
$$ a'_{\oplus} = 0 $$.
Well, now there is a problem. The Sun's gravity still works, so that:
$$ f_g \ne M_{\oplus}a_{\oplus} $$
Newton's law isn't working.
We can restore physics by introducing a fictitious centrifugal force:
$$ f_c = +M_{\oplus}\Omega^2 R = M_{\oplus}\frac{4\pi^2}{1 {\rm A.U.}} $$
The plus sign indicates that this new force is radially outward.
Now the sum of all external forces is:
$$ F_{tot} = f_g + f_c = 0 = f_{cent}$$
and $F=ma$ holds: physics doesn't care about your coordinates.
Now this is the simplest example. Had I chosen:
$$\Omega \ne 2\pi\,{\rm year}^{-1} $$
then Earth would have a non-zero velocity, breaking
$$ \Omega R = v $$
and Newton would be out of balance. This is remedied by introducing a velocity dependent Coriolis Force:
$$ \vec f_C = -2\vec \Omega \times \vec v' $$
Note that the angular velocity is that of the coordinate system, not the Earth's in its orbit; moreover, $v'$ is not the inertial velocity of Earth, it's the velocity of the Earth in the rotating coordinate system.
As an example, suppose I chose $\Omega$ such that Earth appeared to revolve backwards one per year.
Let's call the yearly revolution $\Omega_0$, so I start with:
$$\Omega = 2\Omega_0 $$
Then we have gravity:
$$ f_g = -M_{\oplus}\frac{v^2} R =
-M_{\oplus}\Omega_0^2 R$$
and the centrifugal force
$$ f_c = +M_{\oplus}\Omega^2 R = +4M_{\oplus}\Omega_0^2 R$$
We are way out of balance now, but now we add the Coriolis force with:
$$ v' = (\Omega_0-\Omega) R = -\Omega_0 R $$
so:
$$ f_C = -2\Omega v' = -4\Omega_0^2 R $$
The total$^1$ of all those forces is the centripetal force:
$$ f_{cent} = f_g + f_c + f_C = f_g = M_{\oplus}a_{\oplus} $$
which keeps the Earth on track to revolve once per year.
Newton is restored!
Note: the centripetal force is the sum of all of the radial forces, so that's not just gravity. It includes the inertial forces, too.
I prefer the term "inertial" over "fictitious" for the following reason(s):
If you start with Special Relativity (a 4D spacetime), and include acceleration: gravity, $f_g$, is also an inertial force caused by having a coordinate system that does not follow a geodesic. Moreover, the centrifugal and Coriolis force appear as gravitation. Of course, the math is much more difficult, but in the end it works because:
PHYSICS DOES NOT CARE ABOUT YOUR COORDINATE SYSTEM.
[1] If you want to say only inward forces are centripetal, that is a matter of choice. In that case: $f_g + f_C$ cancels $f_c$ and has enough left over to keep Earth in orbit. See also: https://en.wikipedia.org/wiki/Eötvös_effect, for yet another modification.
