Angular momentum, covered in the last several part of rotational dynamics, is also critically important in planetary motion. For example, the law of conservation of angular momentum applies to orbital motion  whether of satellites about the Earth or planets around the Sun.
For conservation of angular momentum L in an elliptical
orbit (with r_{a} the
radius vector at aphelion, and with r_{p} the radius vector at
perihelion)
L = mv_{a}
r_{a} = mv_{p}
r_{p}
Or: v_{a} r_{a} = v_{p} r_{p}
In more concise polar form:
L = mr ^{2 }dq /dt = r p _{q}_{ }= r p sin q
Or:
L = rmv_{q}
The angular momentum of a planet moving around the Sun is constant. This is none other than a restatement of Kepler's 2nd law.
We can also write: L = m v_{a}_{ }r_{a} sin q = mv_{p} r_{p}
For conservation of energy (Kinetic or mechanical K, and gravitational potential V):
E _{tot} = K + V = mv^{2 }/ 2  GMm/r =  GMm/2r
The centripetal acceleration is also a key quantity in orbital motion:
We can refer to the diagram below for the origin of the centripetal acceleration and the related force:
The way that the centripetal acceleration (a_{c} = v^{2}/ r) arises is via the change in direction of the velocity vector, v. Thus, the acceleration is: D v/r or: (v’ – v)/ r, but the magnitude of each vector is:
v = rq/ t = r w.
By similar triangles one would obtain:
D
v/v = s/ r and D v = v(s/r) but s = (rq)/ t
So: D v = v(q/ t) = vw
And since: w = v/r then:
a _{c} = D v/ r = vw/ r = = v^{2}/
r
A key relation is that the force of centripetal acceleration is directly provided by the Newtonian force of gravitational attraction, with M the solar mass and m the mass of a given planet:
And:
GMm/ r ^{2} = m v^{2}/ r
But Newton realized:
GM/ R^{2} = v^{2}/R and let: v = 2π/P,
P being the period, whence:
GM/R^{2} = (2π/P)^{2} 1/R
Or, in terms of P^{2}: P^{2} = (4π^{2}/ GM) R^{3}
Which is just the Newtonian statement of Kepler’s 3rd or Harmonic law.
The preceding can also be applied to the Earth Moon system, or indeed any planetsatellite system. For this we simply replace the mass M of the Sun with the mass of the Earth, M_{E} then if we set the weight (w =mg) of an object on Earth's surface equal to the force of gravitational attraction, F, we obtain:
mg = GM_{E} m/ r^{2}
Or: g = GM_{E}/r^{2}
In other words, g is independent of the mass m on the Earth's surface. Now, what about objects actually orbiting the Earth, say like artificial satellites? In this case we understand that what keeps the objects orbiting is the centripetal (or centerdirected) force, which is defined as:
F_{c
}= mv^{2}/r
Or:
mRw^{2} = gr^{2} m/R^{2}, so that for the angular velocity w:
w^{2} = gr^{2}/R^{3}, and:
R^{3}
= gr^{2}/w^{2}
Interesting application:
This would be to find R (= r + h), and thence h (altitude) of the satellite if the period is known to be one day or 86,400 secs. Then, T = 86,400s and, solving for R (using same magnitude for r, g as before):
R = [g r^{2}/w^{2}]^{1/3}
R = [(10 m/s^{2})(6.4 x 10^{6} m)^{2} (86400s)^{2})/ 4p^{2}]^{1/3}
R = 4.24 x 10^{7} m = 42 400 km
But we know r = 6400 km so h = R  r
And h = 42 400 km  6400 km = 36 000 km
Or h » 22 500 miles above the Earth.
We
call such an orbit geosynchronous or "geostationary" because
the orbiting body retains an essentially fixed position above a point on the
Earth and its motion (velocity) in orbit matches the rate of Earth's rotation.
To find velocity v we have v = wr = (2p/T) r
= { 2p x 42.4 x 10^{6} m} / 86400 s.
v = 3100 m/s
Note that what we have discussed applies to circular orbits for which the radius is constant. But what about elliptical? Go back to the conservation of angular momentum at the top of this post and how it confirms Kepler's 2nd law, and see if you can arrive at the connections. For help you can go to this link:
Gyroscopes, Torque & Angular Momentum:
The gyroscope is interesting in that it incorporates angular momentum, torque. We focus on the diagram below:
Here, a gyroscope is situated on a double gimbal mount which permits any orientation of the axis. If vertical forces are applied to the ends as shown (F up and F down in (b)) the axis will precess in a horizontal plane, always moving in a direction perpendicular to the forces.
Now to fix ideas, let each of the two forces applied have a magnitude 20 N. (As shown in (b)) These forces produce a torque on the system denoted by the symbol t with direction perpendicular to the plane of the forces. There is in fact a torque vector perpendicular to both the axle and vertical direction, so effectively a torque pair. If the magnitude of each torque is Fd where d is the distance shown then the total torque is t = 2F d.
Because of this torque the angular momentum L must change. To see how we change perspective and look at the flywheel from above. Diagrams showing: (a) the top view of the gyroscope and (b) the angular momentum vector diagram is shown below:
Here. (a) depicts one instant with the angular momentum L direction shown. At the same time position a shows a force direction into the page and b a force direction out of the page. The torque vector is perpendicular to L. In the small time interval Dt the angular momentum L changes by an amount D L . This is given approximately by:
D L = t Dt
The instantaneous precession in the angular velocity is then given by:
W = dq/ dt
And the average change by: Dq/ Dt
If Dt is sufficiently small we have:1) Consider the table shown below, applied to circular satellite orbits around the Earth:
Distance from 
2 r 
3 r 
3r/ 2 
4 r 
5 r/2 
5 r 
Acceleration of gravity 
g/4 





Complete the Table above given that r = 6.4 x 10 ^{6} m. Hence or otherwise, deduce the acceleration of gravity g at a distance 10r.
2) Newton’s version of Kepler’s 3^{rd} law is usually written as:(m1 + m2)P^{2} = 4p^{2}/G (r1 + r2)^{2}
where G is the Newtonian gravitational constant:
(G = 6.7 x 10^{11} Nm^{2}/kg^{2})
and (r1 + r2) is the distance between the centers of the two bodies.
Use this to find the period P of the Moon if its mass m1 = m2/80 where Earth’s mass m2 = 6.7 x 10^{ 24 }kg and (r1 + r2) = 384, 000 km. Compare this to the value obtained by using Kepler’s simpler version of the 3^{rd} law with (r1 + r2) = a1= 0.0025 AU.
3) Verify the conservation of angular momentum applies for a spacecraft in orbit around the Earth if its velocity at perigee is 10.7 km/ sec, its distance from Earth at perigee is 6.6 x 10^{ 3} km, its velocity at apogee is 0.75 km/ sec and its distance at apogee is: 9.3 x 10^{ 4} km. Find the period of the spacecraft.
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