Application to Planetary Motion:
We begin with the polar
coordinates of a point mass in an orbit of radius r under gravitational
attraction to the Sun:
a r = dr
2/ dt 2 - r
(dq /dt
) 2
a q = r (d2 q /dt
2) + 2 dr/dt ((dq /dt )
In polar
coordinates we are also interested in the angular momentum, e.g.
L = mr 2 dq
/dt = r p q = r p sin q
Note that this
angular momentum for a planet orbiting the Sun is constant! The force
components in polar coordinates:
F r =
m[dr 2/ dt 2 - r
(dq /dt
) 2 ]
F q = m [r
(d2 q /dt
2) + 2 dr/dt ((dq /dt ) ]
If we take: d/dt (r p q ) we
get:
r d p q /dt
= r F q = dL/ dt
The rate of change
of angular momentum which is also called the torque of the force F about the
point 0.
t = dL/
dt
The diagram below shows how torque is derived:
The change in angular momentum is then:
dL/ dt = r F q =
m [r
(d2 q /dt
2) + 2 dr/dt ((dq
/dt ) ]
= d/ dt m r 2 (dq
/dt)
Note the angular
momentum per unit mass m is just:
h
= L/ m
=
r 2 dq /dt
= r p q / m
This
quantity is exactly the h used in the Kepler 2nd law, and a
constant of the motion. In effect, with a bit of further working:
r 2 dq
/dt = 2 dA/ dt = h or dA/dt = h/ 2
The preceding can
be integrated to obtain:
A =
½ h t + c
Then the area
swept out by the radius vector in time:
t - t 0
according to Kepler’s 2nd law would be:
A =
½ h (t - t 0)
Given a period of
revolution P, i.e. for elliptic motion about a fixed origin, when t = t + P,
and P > 0, the area swept out is:
A + p ab
= ½ h (t +
P -
t 0)
b = a
(1 -
e 2 ) 1/2
Subtracting the first equation above from the
second, i.e.
½ h (t +
P -
t 0)
- A + p
ab
And solving for P,
one arrives at:
P = 2 p ab
h -1
An alternative
approach uses:
dq = h dt/ r
2
Then integrate to
obtain:
q = q 0 + ò t 0 h
dt/ r 2 (t)
Problems:
1) Prove the relations below:
i) dA/dt = h/ 2
ii) P = 2 p ab
h -1
And show: P
= 2 p Ö (a 3 ) / Öm
Where: m = G
(m1 + m2)
2) Because of the
Earth’s rotation a plumb bob does not hang precisely in the direction of
Earth’s gravitational attraction, i.e. in the direction of its weigh mg. Given the radius of the Earth = 6.4 x 10 6 m and your latitude is 45 degrees, find the
angle by which the bob deviates from the direction of g. (Hint: Sketch a diagram with forces, angle shown to assist in solution.)
3) An asteroid moves in an elliptic orbit around the Sun. The lengths of the major and minor axes are 2a and 2b, respectively. If the asteroid’s velocity at the point of closest approach (where it crosses the major axis) is v o then how much time is needed for the object to make one complete orbit? (Hint: Take the area of an ellipse to be: p ab )
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