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502

 

 

Week 12: Gravity

The gravitational field produced by this same shell equals the usual

 

~g(~r) =

G M

rˆ

(1054)

 

r2

outside of the shell. As a consequence the field outside of any spherically symmetric distribution of mass is just

~g(~r) =

G M

rˆ

(1055)

r2

These two results can be proven by direct integration or by using Gauss’s Law for the gravitational field (using methodology developed next semester for the electrostatic field). The latter is so easy that it is hardly worth the time to learn the former for this special case.

Note well the most important consequence for our purposes in the homework of this rule is that when we descend a tunnel into a uniformly dense planet, the gravity will diminish as we are only pulled down by the mass inside our radius. This means that the gravitational field we experience is:

~g(~r) =

G M (r)

rˆ

(1056)

r2

where M (r) = ρ4πr3/3 for a uniform density, something more complicated in cases where the density itself changes with r. You will use this expression in several homework problems.

12.5: Gravitational Potential Energy

W t = 0

W

12

 

 

B

A

W

r W t = 0

12

1

r2

 

Figure 148: A crude illustration of how one can show the gravitational force to be conservative (so that the work done by the force is independent of the path taken between two points), permitting the evaluation of a potential energy function.

If you examine figure 148 above, and note that the force is always “down” along ~r, it is easy to conclude that gravity must be a conservative force. Gravity produced by some (spherically symmetric or point-like) mass does work on another mass only when that mass is moved in or out along ~r connecting them; moving at right angles to this along a surface of constant radius r involves no gravitational work. Any path between two points near the source can be broken up into approximating segments parallel to ~r and perpendicular to ~r at each point, and one can make the approximation as good as you like by choosing small enough segments.

This permits us to easily compute the gravitational potential energy as the negative work done

Week 12: Gravity

moving a mass m from a reference position ~r0 to a final position ~r:

U (r) = Zr0 F~

· d~r

 

 

 

 

 

 

r

 

 

 

 

 

 

 

 

=

Zr0

 

 

r2

dr

 

 

 

 

r

 

GM m

 

=

(

GM m

GM m

 

 

 

 

 

 

 

)

 

r

 

 

r0

 

GM m GM m

 

=

 

 

 

+

 

 

 

 

 

r

 

 

r0

 

503

(1057)

(1058)

(1059)

(1060)

Note that the potential energy function depends only on the scalar magnitude of ~r0 and ~r, and that r0 is in the end the radius of an arbitrary point where we define the potential energy to be zero.

By convention, unless there is a good reason to gravitational potential energy function to be at r0 =

choose otherwise, we require the zero of the . Thus:

U (r) =

GM m

(1061)

r

Note that since energy in some sense is more fundamental than force (the latter is the negative derivative of the former) we could just as easily have learned Newton’s Law of Gravitation directly as this scalar potential energy function and then evaluated the force by taking its negative gradient (multidimensional derivative).

The most important thing to note about this function is that it is always negative. Recall that the force points in the direction that the potential energy decreases most strongly in. Since U (r) is negative and gets larger in magnitude for smaller r, gravitation (correctly) points down to smaller r where the potential energy is “smaller” (more negative).

The potential energy function will be very useful to us when we wish to consider things like escape velocity/energy, killer asteroids, energy diagrams, and orbits. Let’s start with energy diagrams and orbits.

12.6: Energy Diagrams and Orbits

Let’s write the total energy of a particle moving in a gravitational field in a clever way that isolates the radial kinetic energy:

Etot =

1

mv2

GM m

 

 

 

 

(1062)

 

 

 

 

 

 

 

 

 

 

 

2

 

r

 

 

 

 

=

1

mvr2

+

1

mvt2

GM m

 

(1063)

 

 

 

 

 

 

 

 

2

2

r

 

=

1

mvr2

+

1

(mvtr)2

GM m

(1064)

 

 

 

 

 

2

2mr2

r

 

1

mvr2

 

 

L2

GM m

 

 

=

 

 

+

 

 

 

 

(1065)

 

2

2mr2

r

 

=

 

1

mvr2

+ Ue (r)

 

 

 

 

(1066)

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In this equation, 12 mvr2 is the radial kinetic energy, and

 

L2

GM m

 

Ue (r) =

 

 

(1067)

2mr2

r

is the radial potential energy plus the rotational kinetic energy of the orbiting particle, formed out of the transverse velocity vt as Krot = 12 mvt2 = L2/2mr2. If we plot the e ective potential (and its pieces) we get a one-dimensional radial energy plot as illustrated in figure 149.