Angular momentum, along with Kepler's laws, 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 ra the radius vector at aphelion, and with rp the radius vector at perihelion)
L = mva ra = mvp rp
Or: va ra = vp rp
In more concise polar form:
L = mr 2 dq /dt = r p q = r p sin q
Or:
L = rmvq
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 va ra sin q = mvp rp
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:
ac = 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 (ac = 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, ME 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 = GME m/ r2
Or: g = GME/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:
Fc = 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.
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.
Supplemental Problems:
1) Find the velocity of said geosynchronous satellite which matches the Earth's rotation rate.
2) 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.
See Also:
Practical Astronomy Focus: A Deep Dive Into Kepler's 2nd Law Of Planetary Motion
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