Lecture 19: Effective Potential, and Gravity • The expression for the energy of central-force motion was: 1 l 2 E = µr"2 +U (r) + 2 2µr2 • We can treat this as a one-dimensional problem if we define an effective potential: l 2 V (r) = U (r) + 2µr2 – “Effective” means that V acts like a potential energy, even though it isn’t one – there is no force corresponding to the gradient of the second term • With this definition, the energy becomes: 1 2 T is the “radial E = µr" +V (r) = T +V (r) r 2 r kinetic energy” Example: -k/r potential • Most of the remaining discussion of central-force motion will focus on motion under gravity • How much can we learn about the motion simply from the effective potential? • Start with a sketch: U E V r • We see that the motion in r for a given energy can be quite different for different values of l – if l = 0, the system can reach r = 0 – If there is a non-zero angular momentum, there will be an “excluded region” at small r. The separation between the two masses will never be smaller than this value • The minimum of the potential corresponds to the location of a circular orbit – i.e., this is an equilibrium point for radial motion – for 1/r potentials the equilibrium is stable (convenient for those of us who like to maintain a relatively constant distance from the sun!) – The position is given by: dV k l 2 = − = 0 dr r2 µr3 Increases rapidly as l 2 r = l increases µk Minimum Energy • We also see that for a given l there is a minimum energy the particle can have – This is the energy of a circular orbit k l 2 E = − + min µ 2 rmin 2 rmin 2 µ 2 µ 2 µ 2 = − k + l = − k + k = − k l 2 ’ ≈ 2 ’2 l 2 2l 2 2l 2 µ l ∆µ ÷ 2 ∆ ÷ « k ◊ «µk ◊ Gravitation • Now that we’ve learned some of the basic rules of central- force motion, we’ll see how they apply to the only force described by Newton: gravity – Now we know there are three other forces: 1. Electromagnetism (which you’ll study in another course, using math similar to what we use for gravity) 2. The weak force 3. The strong force • The last two can only really be understood in the context of quantum field theory • In one looks at the interaction between fundamental particles, gravity is – by far – the weakest of the forces • However, it is the only force that both acts over long distances, and is always attractive, making it the most significant influence on astronomical objects The Basic Rule • The force of gravity between two point masses is given by: r M • = − GMm F er er r2 F • m where G is a universal constant: 6.67 x 10-11 Nm2kg-2 • The force is linear: if there were several point masses in the problem, the total force on m would be the vector sum of all the individual forces • If there is a continuous distribution of matter, one needs to integrate to find the total force on m: ρ (r′) = − ′ F Gm— 2 dv V r The Gravitational Field • Our definition of the gravitational force requires it to act over large distances • Not as easy to understand as the force between two objects that “touch” each other • One conceptual way to think about it is that the presence of a mass alters the space around it, in a way that other masses can feel – In fact, such a picture is at the core of general relativity, the modern theory of gravity • We can represent this modification of space as adding a “gravitational field” to every point in space • The field due to a mass M is: −GM This is also the acceleration that g = e r2 r any mass in the field will have The Gravitational Potential • The lines of the gravitational field look like: – They don’t form closed loops • • More precisely, it’s easy to show that ∇ × g = 0 • This means that we can define a potential corresponding to the field – Note that this is not a potential energy, which corresponds to a force! • The required form is −GM Φ = + C r ∂Φ −GM −∇Φ = − e = e ∂r r r2 r • As with all potentials, the constant C is completely arbitrary • However, it’s usually convenient to treat the potential at infinite distances as zero, which corresponds to C = 0 • The relationship between gravitational potential and potential energy is the same as that between the gravitational field and gravitational force: – For a particle of mass m, the potential energy is U = mΦ Gravitational Self-Energy • It’s often interesting to know how much gravitational energy is “stored” in a given object (or collection of objects) – In other words, how much energy would we gain by pulling the object apart? If the value is negative (as it always will be for gravity!) that means we need to supply energy to pull the object apart. • To find the answer, we calculate the energy needed to assemble the object piece by piece • Simple example: three point masses – How much gravitational energy is stored in the following configuration? • m3 d13 d23 m1• • m2 d12 • We start with all three masses infinitely apart. Then we find the change in potential energy when m1 is moved to its final position – This is 0, since the other two masses are still infinitely far away • Now m2 is brought in to its final position. It now must move though the field generated by m1, so the change in potential energy is: Gm m ∆ = Φ = − 1 2 U2 m2 d12 • Similarly, when m3 is brought in, we have: Gm m Gm m ∆U = − 1 3 − 1 3 3 d d 13 23 Note that the answer • So the total self-energy is: doesn’t depend on the Gm m Gm m Gm m ∆U = − 1 2 − 1 3 − 1 3 order in which the d12 d13 d23 masses were assembled • In one of the homework problems, you’ll show that the self-energy of a uniform sphere of radius R and mass M is 3 GM 2 U = − 5 R • If the sphere’s radius were to become smaller, the gravitational self-energy would decrease – Thus, to conserve energy overall, the object would need to release energy (somehow…) • In the 1800’s, this was the accepted explanation for why the sun shines – i.e., it’s a ball of contracting gas, that converts gravitational energy to heat and light • Sounds great, but… – The current gravitational self-energy of the sun is − 2 3 (6.67 ×10 11 Nm2kg-2 )(2 ×1030 kg) U = − = −2.3×1041 J 5 (7 ×108 m) – Note that this represents the opposite of the total of all the energy released in the past • The sun’s luminosity was know to be about 4 x 1026W – But that implies that the sun could have been shining for at most 6 x 1014s, or less than 100 million years – And geologists had good evidence that the Earth was older than that! The geologists were right – it’s the strong nuclear force, not gravity, that powers the sun!.