*Notes to a video lecture on UNIZOR.COM*

__Laws of Newton - Problem 1__

*Problem A*

Prove that gravitational force of a point mass is

**conservative**.

That is, prove that the work performed by the force of gravity of a point mass onto an object moving along a trajectory from point

*A*to point

*B*depends only on positions of these endpoints and independent of the trajectory an object moves along.

*Proof*

As was proven in an earlier lecture

*Field, Potential*of this chapter of the course, to prove that a force is conservative, it is sufficient to show that the force is a negative gradient of some scalar function called

**potential**.

According to the Newton's Law of Gravitation, the vector of the force of gravity produced by a point mass

*M*and attracting a point mass

*m*positioned on a distance

*r*from mass

*M*is directed along the line connecting them towards a mass

*M*, and its magnitude equals to

*F = G·M·m/r²*

Let's define Cartesian coordinates with a center at a point mass

*M*.

Coordinates of point mass

*M*is (

*0,0,0*).

Vector

*represents a position of a point mass*

**r =**||x,y,z||*m*in this system.

Now we can express the force as a vector in this system using the fact that vector

*represents a unit vector directed from mass*

**r**/r*M*to mass

*m*.

**F =**−(G·M·m)·(**r**/r³)In this formula we have added to the magnitude of force a multiplier

*that represents the unit radial vector directed from*

**r**/r*M*to

*m*and the minus sign to change the direction of the vector towards mass

*M*because the gravity attracts.

Now we will define a scalar function, the gradient of which equals to the vector of gravitational force.

Consider for now only a variable part of the vector of force

*.*

**r**/r³Its representation in coordinate form is

*||x/r³,y/r³,z/r³||*, where

*r=(x²+y²+z²)*.

^{½}Let's define a function

*R(x,y,z) = R(r) =*

= 1/r=1/(x²+y²+z²).

= 1/r=1/(x²+y²+z²)

^{½}Its gradient

∇

*R = ||∂R/∂x,∂R/∂x,∂R/∂z||*

can be explicitly calculated as

*∂R/∂x = −½(x²+y²+z²)*

= −x/(x²+y²+z²)³ = −x/r³

^{−3/2}·2x == −x/(x²+y²+z²)³ = −x/r³

Analogously,

*∂R/∂y = −½(x²+y²+z²)*

= −y/(x²+y²+z²)³ = −y/r³

^{−3/2}·2y == −y/(x²+y²+z²)³ = −y/r³

*∂R/∂z = −½(x²+y²+z²)*

= −z/(x²+y²+z²)³ = −z/r³

^{−3/2}·2z == −z/(x²+y²+z²)³ = −z/r³

Comparing this with an expression for the force of gravity, we see that the difference between vector of force

**F =**−(G·M·m)·(**r**/r³)and the gradient of defined above function

*R(x)*

∇

*R=||−x/r³,−x/r³,−x/r³||=−*

**r**/r³is only a constant multiplier.

Hence,

*∇*

**F**= (G·M·m)·*R*

Therefore, scalar function

*U(r) = −(G·M·m)·R(r) =*

= −(G·M·m)/r

= −(G·M·m)/r

where

*r=(x²+y²+z²)*

^{½}has the required property, its negative gradient equals to a vector of the gravitational force.

From this follows that the force of gravity is

**conservative**.

*Note 1*

This function depends not only on the field properies (mass

*M*of the source of the field and distance

*r*from it), but also linearly depends on a property of another object (mass

*m*).

To make a concept of

**potential**a property of the field only, the function

*U(r)*is called a

**field potential**when mass

*m*is a unit of mass, in which case

*U(r) = −G·M/r*

*Note 2*

We can prove that gravity is a conservative force directly by following the same logic we used to prove that, if the force can be represented by a gradient of a potential, the work performed by this force is independent of the trajectory.

Consider now any two points in space

*A*and

*B*and some trajectory that point mass

*m*takes to move from

*A*to

*B*.

The work done by any force on an object moving along some path consists of all small amounts of work the force performs on any small piece of trajectory and, by definition, equals to an infinite sum (that is, integral along a path) of infinitesimal increments *that is, differentials) of work, each of which is a scalar product of the vector of force

*and infinitesimal vector of the increment of position along a trajectory*

**F***d*

**r***W*

_{[AB]}= ∫

_{[AB]}

*d*∫

**W**=_{[AB]}

**F·**d**r**To calculate the scalar product

*, we can express both in coordinate form*

**F·**d**r**

= −(G·M·m/r³)·

**F**= −(G·M·m/r³)·||x,y,z|| == −(G·M·m/r³)·

**r***d*

**r =**||dx,dy,dz||*- a scalar*

·(x·dx+y·dy+z·dz)

**F·**d**r**= −(G·M·m/r³)··(x·dx+y·dy+z·dz)

Notice that

*x·dx + y·dy + z·dz =*

d(x²/2 + y²/2 + z²/2) =

= (1/2)d(r²) = r·dr

d(x²/2 + y²/2 + z²/2) =

= (1/2)d(r²) = r·dr

Therefore,

= −(G·M·m/r²)·dr =

= −(G·M·m)·d(1/r) =

= d(−G·M·m/r)

**F·**d**r**= −(G·M·m/r³)·r·dr == −(G·M·m/r²)·dr =

= −(G·M·m)·d(1/r) =

= d(−G·M·m/r)

Since

*is a full differential of some function, integral of it along a path from*

**F·**d**r***A*to

*B*equals to a difference of the values of this function at the points of limits of integration

∫

_{[AB]}

*∫*

**F·**d**r**=_{[AB]}

*d(−G·M·m/r) =*

= −G·M·m/r(B)+G·M·m/r(A) =

= G·M·m/r(A) − G·M·m/r(B)

= −G·M·m/r(B)+G·M·m/r(A) =

= G·M·m/r(A) − G·M·m/r(B)

which proves that work performed by the force of gravity on an object moving along some trajectory is independent of the trajectory, that is it proves that gravitational force is conservative.