The Basics of Rotational Dynamics (Part 4): Angular Momentum Applied To Orbital Motion, Torque & Gyroscopes

 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  = mv² / 2 -   GMm/r =  -   GMm/2r

The centripetal acceleration is also a key quantity in orbital motion:  

a_c = v²/r   = (r w)² /r =  r w ²

 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 = v²/ 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 = = v²/ 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 v²/ r

But Newton realized:

GM/ R² = v²/R  and let: v = 2π/P,

 P being the period, whence:

GM/R² = (2π/P)² 1/R

Or, in terms of P²:   P² = (4π²/ GM) R³

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 planet-satellite 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/ r²

Or: g = GM_E/r²

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 center-directed) force, which is defined as:

F_c = mv²/r 

Or:

mRw² = gr² m/R², so that for the angular velocity w:

w² = gr²/R³, and:

R³ = gr²/w²

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 r²/w²]^1/3

R = [(10 m/s²)(6.4 x 10⁶ m)² (86400s)²)/ 4p²]^1/3

R = 4.24 x 10⁷ 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 "geo-stationary" 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 10⁶ 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:

• Another Crankster Exposed With Kepler's 2nd Law

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:

Dq = D L/ L   =   t  Dt / L 

 With each small interval of time, the angular momentum vector continues to change in direction so that viewed from above (diagram (a)) the gyroscope as a whole rotates in a counterclockwise sense around its support point. This type of motion, where the axis of rotation changes direction with time is referred to as precession.  Dividing by  Dt   and taking the limit  as Dt  ® 0  the result is:

W   =   t  / L   

But we know already that: the torque magnitude is: 

t = 2F d.

And the magnitude of L is simply: L = Iw    where I is the moment of inertia of the flywheel about its axis and w is the magnitude of  the angular velocity of rotation about this axis. 

Therefore we can write:  W   = 2F d/ Iw

Selected Problems:

1) Consider the table shown below, applied to circular satellite orbits around the Earth:

Distance from Center of Earth	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)P² = 4p²/G (r1 + r2)² 

where G is the Newtonian gravitational constant: 

(G = 6.7 x 10^-11 N-m²/kg²) 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.

4) A gyroscope is made from a flywheel made to spin about a rotation axis (bar) as shown in the sample diagrams in the text. If each of the two forces applied = 20 N and the distance d is 0.15m, then find the magnitude of the torque t.

If the mass of the flywheel is 1.0 kg and its radius r = 0.1 m find the magnitude of the angular momentum if  w =  p rad/sec. Thence or otherwise find the rate of precession of the angular velocity W.

5) (Inquiry problem)

A proponent of Immanuel Velikovsky's "colliding worlds" nonsense claims that all that was needed for Earth's rotation to stop (when Venus came so close to Earth in passing from Jupiter to its present orbit) was a torque defined:  N = F_r X R_c.  as used in his diagram below:

Explain why this is impossible and can't work as claimed. 

(Science writer Isaac Asimov once wrote ('The Stars in Their Courses') if it had worked, all the Earth's rotational kinetic energy would have been converted into heat energy and melted the Earth's crust.)

6) Inquiry Problem

In the early morning hours of June 14, 2002, the Earth had a  remarkably close encounter with an asteroid (2002 MN) the size of a small city. It remained undetected until three days after it had passed the Earth. At its closest approach, the asteroid was 73,600 miles from the center of the Earth—about a third of the distance to the Moon. Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it had an average density of 3.3 g/ cm ³

------------------

Hint:  Use  E _tot = K + V  = mv² / 2 -   GMm/r =  -   GMm/2r

And take  the asteroid as coming from 'infinity' with r the distance to Earth at closest approach.