# Physics with Calculus/Mechanics/Gravity

It may seem obvious that gravity is both the force that makes objects fall and makes planets orbit, but the fact that gravity explains both of these motions only goes back to Isaac Newton in the 17th century. He showed that the acceleration of a falling object and the acceleration of the moon in its orbit can both be explained by an attractive force acting between any two objects given by

${\displaystyle F=-{\frac {GMm}{r^{2}}}}$

where ${\displaystyle M}$ and ${\displaystyle m}$ are the masses of the two objects, ${\displaystyle r}$ is the distance between them, and G is a fundamental constant, later measured to be equal to ${\displaystyle 6.67\times 10^{-11}Nm^{2}/kg^{2}}$. The minus sign indicates that gravity is attractive.

One interesting feature of gravity is that the gravitational mass, ${\displaystyle m}$, that appears above is quantitatively identical to the inertial mass that appears in ${\displaystyle F=ma}$. Albert Einstein noted this coincidence and it led him to the most successful current theory of gravity known as General Relativity. Although gravity may seem simple, at a fundamental level we still do not yet understand it. General Relativity is incompatible with the quantum theory of elementary particles known as the standard model. Many physicists have attempted to build a quantum theory of gravity, but neither a coherent theory nor experimentally verifiable predictions have yet been obtained. Newton's theory of gravity which we study in this course remains as a powerful tool for predicting the motion of objects with sizes between galaxies and atoms, which includes most objects of everyday relevance on earth. In Newton's theory, gravity is simply a property of matter that causes each mass to attract every other mass with the force given above.

Gravity is a conservative force, which means that the work done moving a small mass from one point to another depends only on the end points, and not the path taken. I will demonstrate this with a single point mass. With this result, it is almost obvious that the force from any arrangement of mass will also be conservative. To see this, note that the gravitational force is the sum of the parts and that work done by one force is the sum of the work done by its parts. Thus, when at test particle is moved through an arrangement of particles exerting gravitational force, if the work done by each of them depends only on the end points, then the sum depends only on the endpoints, which is what we wanted to show. Now, consider a single particle of mass M at the origin, and a test particle of mass m moving from A to B. To simplify calculations, assume that the path connecting A and B lies in the xy plane, although the argument will easily but rather tediously extended to 3 dimensions. Now, it is easier to work in polar coordinates.

When the particle moves a small distance ds, the work done is done only by the force acting in the r direction and thus only movement in the r direction matters. Since this holds exactly for small distances, it holds for large distances and the work done can only depend on how the particle moves in r. That is, we may neglect ${\displaystyle \theta }$. Now, if the particle moves a small distance dr one direction then a little while later, at the same radius, in the other direction, the work cancels out because both the force and displacement are the same. That means that work is independent of the path and gravity is conservative!

Since gravity is conservative, you can see that we can define a function of position such that the work done from moving from a to b is just this function at a minus the function at b. Such a function is called the potential energy -- the same potential energy you already know! If instead we remove it one step, and divide by the mass m, then we have a function that is independent of the particle in question and we call it the potential. Now, it would be useful to see some properties of potential.

First of all, the defining quality of potential, ${\displaystyle \phi (\mathbf {r} )}$ is that if you move a particle of mass m from a point a to a point b along any path, then the work done is

${\displaystyle m\phi (a)-m\phi (b)}$.

Now, to find the other important property of potential, consider moving a particle of mass m a little bit, ${\displaystyle d\mathbf {r} =(dx,dy,dz)}$ in a force field F. The work done by going in the x direction is

${\displaystyle -m(\phi (x+dx,y,z)-\phi (x,y,z))=F_{x}dx}$

with similar expressions for y and z.

Dividing by dx and letting dx go to zero, we have,

${\displaystyle F_{x}=m{\frac {\partial \phi }{\partial x}}}$ which just means take the derivative of ${\displaystyle \phi }$ treating y and z as constants.

If we do the same thing for y and z, we have

${\displaystyle F=m({\frac {\partial \phi }{\partial x}},{\frac {\partial \phi }{\partial y}},{\frac {\partial \phi }{\partial z}})=\nabla \phi }$.

The upside down triangle is just shorthand for the first expression.

This is amazing -- all the information about the force, a vector, is contained in a scalar. Three components for the price of one. The reason we can do this is because we know that the force is conservative, as you can see, that greatly restricts the number of possible fields. In the case of one dimension, you can see that the force is the derivative of the potential energy. In fact, the gradient (the name for taking the partial derivatives with respect to each variable and putting them into a vector) is one generalization of a derivative to many dimensions.

It is now easy to see one final important property of potential, that it is unique up to an additive constant. If you added a function depending on any of the variables, then at least one of the partial derivatives would not be zero and you would get a different force. But if you add any constant, you get the same force because the derivative of a constant is zero.