(a) By examining the effective potential energy (8.32) find...

(a) By examining the effective potential energy (8.32) find the radius at which a planet (or comet) with angular momentum $\ell$ can orbit the sun in a circular orbit with fixed radius. [Look at $\left.d U_{\mathrm{eff}} / d r .\right](\mathbf{b})$ Show that this circular orbit is stable, in the sense that a small radial nudge will cause only small radial oscillations. [Look at $\left.d^{2} U_{\mathrm{eff}} / d r^{2} .\right]$ Show that the period of these oscillations is equal to the planet's orbital period.

Key Concepts

Effective Potential Energy

In central force problems, the effective potential energy is a tool used to combine the actual potential energy (such as the gravitational potential) with the 'centrifugal' potential due to angular momentum. This construction simplifies the study of radial motion by encapsulating both the attractive force and the inertial effects from angular momentum into a single function of radius.

Circular Orbit Condition

A circular orbit is achieved when the net radial force on the orbiting body is zero. In the effective potential framework, this condition is found by setting the first derivative of the effective potential with respect to radius to zero (dU_eff/dr = 0). This derivative condition determines the specific radius at which the centrifugal repulsion exactly balances the gravitational attraction.

Stability Analysis

To assess the stability of a circular orbit, one examines the second derivative of the effective potential with respect to the radius (d²U_eff/dr²). A positive value for this second derivative indicates that small perturbations in the radial direction lead to a restoring force that confines the particle to oscillate around the equilibrium radius, thereby ensuring the orbit's stability against small disturbances.

Radial Oscillations and Orbital Period

When a planet in a stable circular orbit is slightly perturbed in the radial direction, it undergoes small oscillations about the equilibrium orbit. The dynamics of these oscillations typically resemble a simple harmonic motion, and in certain central force fields (particularly the inverse-square law of gravitation), the frequency of these radial oscillations coincides with the orbital frequency. This result implies that the period of small radial oscillations is equal to the orbital period, linking the stability analysis directly to the motion characteristics of the orbit.

Transcript

0:00 Hi there.

00:01 So for this problem, we have the orbit of the earth that is about the sun.

00:06 And we consider that as almost circular.

00:10 Now, the closest and farthest distance are the radius that we are going to call one, and that is 1 .47 times 10 to the 8 kilometers, and r2 is equal to, and the radius of the farthest distance is 1 .52 times 10 to the 8 kilometers.

00:44 Now what we need to determine is the maximum variations for part a for the potential energy.

00:53 So we need to find that variation in the potential energy.

00:57 Now, we know that the potential energy in this case is going to be equal to the initial potential, well, the final one, we're going to call this, and that is about the gravitational constant times the mass of the sum times the mass of the earth over the, the distance and we are going to call this distance the this is for the closest distance so that is r1 and for the potential energy the initial one we have that that is the gravitational constant times the mass of the sum times the mass of the earth over the radius two now in here when we do the change in the potential energy.

02:02 We know that that is the final potential energy minus the initial potential energy.

02:08 So we substitute this presessions in here and we can take out which the terms that are common, which are the gravitational constants times the mass of the sum, times the mass of the earth.

02:26 And this times, times one over the radius 1 minus 1 over the radius 2.

02:34 And here we just need to simply substitute all of those values.

02:39 Now the gravitational, the gravitational constant has a value of 6 .67 times 10 to the minus 11, newtons times meters squared over kilograms squared.

02:59 And this times the mass of the sun that we know is 1 .99 times 10 to the 30 kilograms and the mass of the earth that is 5 .98 times 10 to the 24 kilograms and in here we have one over the closest distance which is 1 .47 times 10 to the 11 because we past these two meters and minus 1 over 1 .52 times 10 to the 11 meters.

03:39 So solving for this, we found that the change in the potential energy from these two distances is equal to 1 .78 times 10 to the 32 joules.

03:56 So this is a solution for part a of this problem...

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