# Centripetal issue when considering gravity

Forgive me if my question seems silly, but I am quite baffled. Suppose you have a satellite orbiting a horizontal swing planted into the ground and we want to find the velocity with which the satellite must be moving in order to succeed in taking a fully circular path around the swing (no failure followed by parabolic fall).

Indeed if we solve for $F = m(v^2)/R$, where $F$ can be equated to $g$, we can also solve for the velocity; all nice and good, and the nature of this method makes some intuitive sense but there is a problem (at least for me): the force that the satellite experiences is an inward force towards the center, and therefore the force described in the aforementioned equation is the centripetal force.

However, this conflicts with the method of solving the question, since at the top of the path, the centripetal Force and the gravitational force will be both pointing downwards. Therefore shouldn't the satellite actually be "doubly" compelled to accelerate to the ground? But clearly when we for for $v$ with $g$, we assume that they cancel each other out - hence they point in opposite directions. The idea that the cancel also "explains" the intuition that at the top of the circular path, the satellite will be feeling the most "weightless" it will throughout its revolution.

Additionally from a different reference frame / non centrifugal one, it makes sense that at the top, the most force will be on the satellite to turn its path downward as it is after that instant that the satellite sees its downward acceleration. I am, essentially confused with how to handle centripetal/centrifugal "fictitious" forces in this worked example.

## 1 Answer

I believe your confusion comes from a misunderstanding of the designation of a force as "centripetal".

Any calculation of centripetal force is telling you how much force is needed to make a circular motion take place. This doesn't create the force. There is no guarantee that a force of the calculated size and direction actually exists! You need to go out and look around to see if the force is available from any combination of existing forces.

Suppose you are driving a $1000$ kg car along a flat road, entering a $200$ m radius circle at $20$ m/s.. The centripetal force equation says that the centripetal force needed is:$$F_C=\frac{mv^2}{r}=\frac{1000\times20^2}{200}=2000\text{ Newtons}$$Normally this centripetal force would come from friction between your slightly turned front wheels and the road.

Fine; but your tires are bald, the road is smooth, and someone has dumped litres of motor oil on the ground. The needed centripetal force is still $2000\text{ Newtons}$, but you don't have it, and you won't follow the curve. You'll go straight ahead and wind up in the outside ditch.

So you mount some rockets on the car, thrusting sideways with $1200\text{ Newtons}$ in towards the centre of the circle. This is the only central force present.

Too bad; this force is enough to drive your car at that speed around a circle with a radius given by:$$r=\frac{m v^2}{F}=\frac{1000\times 20^2}{1200}=333.33\text{ m}$$You'll still go off the road.

So you decide to keep the rockets and slow down to $15.5$ m/s. At that speed, the centripetal force needed is:$$F_C=\frac{mv^2}{r}=\frac{1000\times15.5^2}{200}=1201.25\text{ Newtons}$$So you'll drift out a little, but make it around the curve.