# What is the proper vector form for the Gravity Formula?

We are familiar with the equation for calculating the velocity of an object in orbit given the central mass (or the mass given the orbital velocity): $$\frac{v^2}{r} = \frac{MG}{r^2}$$ I'm trying to understand this as a net of forces in vector form (assuming a unit mass in these formulas): $$\vec{\frac{v^2}{r}} = \vec{\frac{MG}{r^2}}$$ The centripetal force is directed towards the center (opposite the normal vector of the surface), but so is the gravity. The scalar form (assuming positive force is directed radially outward) would be: $$-\frac{v^2}{r} = -\frac{MG}{r^2}$$ So how does this balance? The net force is supposed to be zero, but if you rearrange, you get: $$-\frac{v^2}{r} +\frac{MG}{r^2} \ne 0$$ How do you arrange these components to get a net force of zero (assuming there's no such thing as centrifugal force)?

• How do you know on your last line that the sum is not 0?
– user97166
Nov 2 '15 at 1:52
• Because moving $\frac{MG}{r^2}$ on the same side of the equation as the centripetal force would be turning it into anit-gravity. When setting this up as a balance of forces equation, I'm starting with the fact that I know that the centripetal and gravitational forces are both pointing in the same (negative) direction. How do you get two negative forces to balance to zero?
– user32023
Nov 2 '15 at 1:58
• Yeah, but as it turns out $\frac{v^2}{r}=\frac{MG}{r^2}$, so the sum is indeed $0$. No antigravity here. The centripetal force is not an added force, the centripetal force is provided by gravity.
– user97166
Nov 2 '15 at 2:02
• Donald, this is a really important idea if you want to work problems involving orbits, and it is so simple it is easy to miss. There isn't a separate centripetal force: it always comes from some combination of physical forces present in the problem. Nov 2 '15 at 2:09

After a comment it becomes clear that there is deep misunderstanding here and the question title has nothing to do with the actual problem.

Let's get something basic clear: there is no "the centripetal force". That is no force out there that magically decides to come into being when an object moves in a circle. Rather "centripetal" is a label for those real force that point toward the center.

A centripetal force is one that points toward the center and causes acceleration transverse to the direction of motion. Given that the motion has velocity $v$ and radius $r$ you can know that the size of the acceleration is $v^2/r$ and the strength of the force is $mv^2/r$.

In this case that centripetal force is the force of gravity. That is the only force at work here. And because it points to the center we can also label it "centripetal" and know that it's magnitude must be $mv^2/r$ in order for the orbit to be circular.

As a general rule one identifies the centripetal force in any particular situation as the sum of force components pointing toward the center of rotation minus the sum of force components pointing away from the center: $$F_c = m\frac{v^2}{r} = \sum_i F_{i,in} - \sum_i F_{i,out} .$$

When you ask how to arrange things to get zero force you are asking the wrong questions: there is and should be a net force acting on the orbiting body because it is undergoing continuous acceleration.

A couple of the usual incantations are

$$F_g = -G\frac{M}{r^3}\vec{r} \,,$$

and

$$F_g = -G\frac{M}{r^2}\hat{r} \,.$$

In both cases the direction is taken as the negative of the radius vector to the point of consideration, giving it the correct (inward) direction. The choice to use the unit vector or not seems to be taken on a personal basis rather than on the basis of a well articulated reason.

• Thanks, but I don't see how that helps with balancing the forces. I agree that gravity is directed inward (in the negative direction), but so is the centripetal force.
– user32023
Nov 2 '15 at 1:45
• @DonaldRoyAirey The forces are not balanced because the object is accelerating around the central mass $M$. dmckee has given an excellent explanation while answering your question. If you need vector notation consider $$-G\frac{m_1m_2}{r_{12}^2}\hat{r}_{12}$$ where the unit vector points from $m_1$ toward $m_2$. Nov 2 '15 at 3:11

This is a very common "gotcha" for novices, and something that is not covered well at all in textbooks, in my opinion. The problem is one of semantics, not of physics. The term centripetal force does not name a physical force the way gravitational force and electrostatic force do. The words centripetal force stand for the net result of real forces in the case where the net force points to some center.

For example, on a ferris wheel, there are two forces on a person: gravity (due to the earth) and the normal force (due to the chair). The vector sum of these points toward the axis, so we call the sum the centripetal force. There is no additional force that we would call the centripetal force.

In your case there is one force on the object, gravity. There is not an additional centripetal force.

• This. The language we use "the centripetal force" is a huge tripping point. Especially as it is usually introduced shortly after a chapter in which gravity, friction and "the normal force" all make their appearance. It's a wonder that every last student isn't confused on the matter. Nov 2 '15 at 2:30

A pedagogical note: Because students regularly try to look for the centripetal force and get confused (as noted by other answers here), I try to emphasize that it is the acceleration which is centripetal rather than the force (even though the sum is center directed). Because the acceleration is always the result of a sum, I have found that fewer students in my classes are confused by this statement for circular motion:

$$\frac{1}{m}\sum_i (F_r)_i=-a_r=-\frac{v^2}{r}.$$

The minus sign arises if you define the positive radial direction is outward from the center of the orbit.

Do the sum of forces, find the acceleration, and lastly decide whether the acceleration (or one of its components) is centripetal. Then you won't be looking for a centripetal force.

• Hmmm ... nice. Just this semester I've started toying with the notion of treating known circular motion in a manner parallel to our treatment of equilibrium. "If you know an object is in equilibrium then you its acceleration is zero and therefore so is the net force; if you know an object is in circular motion, then you know it's acceleration is $v^2/r$ toward the center and therefore that the net force is $mv^2/r$ in the same direction. Nov 2 '15 at 5:28

Centripetal Force isn't a real force. It is a construct for representing the sum of different forces that cause circular motion. In the case of an orbit, the gravitational force is the centripetal force, which is what you stated in the first equation. They are NOT opposites, as you asked in the question. They do not become arranged to have a net force of zero. A net force of zero would mean that the object would not be in circular motion. An object in circular motion always has a net force (Centripetal Force) directed towards the center of rotation. In the case of an orbit, the gravitational force is the only force acting on the object, and is the centripetal force. If you spun an object attached to a string around in a circle, the tension in the string would be the centripetal force.

• By your own reasoning the gravitational force is not a real force. That is incorrect.
– Gert
Nov 2 '15 at 2:02
• @Gert you have misunderstood the argument here. "Centripetal" is and always has been a label applied to a set of real forces. The real force here is gravity (ignoring the complications of GR for the moment), and there is no separate centripetal force. Thery are the same. Nov 2 '15 at 2:05
• @dmckee: I'll bear the semantics in mind but it sounds much like a distinction without a difference to me.
– Gert
Nov 2 '15 at 2:13
• This is not a semantic distinction, it is fundamental. Gravity is a force. "Centripetal" isn't a force, it is a label for a sum of force components. What goes in the sum varies from problem to problem. Here is is gravity, for a car navigating a traffic-circle it is friction, for a stuntman riding around a vertical loop-dee-loop it is the normal force on his tires plus $mg\cos\theta$ (with theta measured from vertical up). Nov 2 '15 at 2:19

You are a bit confused by the meaning of the centripetal and centrifugal force. First, in a reference frame at rest relative to the center of mass, the gravity IS the centripetal force (just because it points towards the center). My impression is that you assumed that there where two forces, gravity and the centrifugal force. Second, in a reference frame in which the rotating mass it at rest, there must be two forces that balance each other. What is this second force? it is called a pseudoforce, that appears when you are not in an inertial frame (the rotating frame is not inertial). In the case of a rotating frame of reference reference the pseudo force is a centrifugal force, opposite in direction to gravity. That is why in this reference frame the mass is at rest.