Kepler’s Problem - II

Deduction of the Kepler’s 2nd law

For central force motion, angular momentum 𝐽 is a conserved quantity.
But angular momentum 𝐽 = 𝑚𝑟 2 𝜃ሶ ------ (1)
So 𝑚𝑟 2 𝜃ሶ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 -------- (2)
1 1 2
Taking 𝑚 = , we get 𝑟 𝜃ሶ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ------ (3)
2 2
1 2
Let us show that 𝑟 𝜃ሶ represents areal velocity.
2
If the radius vector 𝒓 rotates through a small angle 𝑑𝜃
-
in time 𝑑𝑡, then the area swept out in time 𝑑𝑡 is given by
1 1 2
𝑑𝐴 = 𝑟(𝑟 𝑑𝜃) = 𝑟 𝑑𝜃 ------ (4)
2 2
𝑑𝐴 1 2 𝑑𝜃 1 2
Thus areal velocity becomes = 𝑟 = 𝑟 𝜃ሶ ------- (5)
𝑑𝑡 2 𝑑𝑡 2
𝑑𝐴
Hence using equation (3), we can write Areal velocity = = Constant.
𝑑𝑡
i.e. radius vector sweeps out equal areas in equal times, which is Kepler’s 2nd law.
Deduction of the Kepler’s 3rd law
➢For an elliptic orbit, the semimajor axis is one-half of the sum of two apsidal
distances 𝑟1 and 𝑟2 .
➢By definition, the radial velocity is zero at these points.
1 𝐽2
➢So, the expression for energy conservation equation 𝐸 = 𝑚𝑟ሶ 2 + + 𝑉(𝑟) at
2 2𝑚𝑟 2
these points become
𝐽2
𝐸= + 𝑉(𝑟)
2𝑚𝑟 2

𝑘 𝐽2 𝑘
➢ For Kepler’s problem, 𝑉 𝑟 = − . So 𝐸 = −
𝑟 2𝑚𝑟 2 𝑟

𝐽2 𝑘 𝐽2
➢Thus 𝐸 − + =0 or 𝑟2𝐸 − + 𝑘𝑟 = 0
2𝑚𝑟 2 𝑟 2𝑚

𝑘 𝐽2
➢Such that 𝑟2 + 𝑟 − = 0 ------------ (1)
𝐸 2𝑚𝐸
➢The two roots of the above equation (1) are two apsidal distances 𝑟1 and 𝑟2 .
𝑘 𝑟1 +𝑟2
➢Hence the semimajor axis is given by 𝑎 = =− ------- (2)
2𝐸 2
(Since the sum of two roots of a quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is −(𝑏/ 𝑎).)
𝑘
➢From equation (2), we have 𝐸 = − ------- (3)
2𝑎
2𝐸𝐽2
➢But we know that eccentricity is 𝑒 = 1+ ------- (4)
𝑚𝑘 2

𝐽2
➢ For present case, 𝑒 = 1− ---------(5)
𝑚𝑘𝑎
➢From the conservation of angular momentum, the areal velocity is constant and is
given by
𝑑𝐴 1 1 𝐽 𝐽
= 𝑟 2 𝜃ሶ = 𝑟 2 = ---------- (6)
𝑑𝑡 2 2 𝑚𝑟 2 2𝑚

➢The area of the orbit 𝐴, is to be found by integrating equation (6) over a complete
𝜏 𝑑𝐴 𝜏 𝐽 𝐽
period 𝜏, A = ‫׬‬0 𝑑𝑡 𝑑𝑡= ‫׬‬0 2𝑚 𝑑𝑡 = 𝜏 --------- (7)
2𝑚
➢But if 𝑎 and 𝑏 are semimajor and semiminor axes respectively, then area of the
ellipse is A = π𝑎𝑏 -------- (8)
➢From equations (7) and (8), we have
𝐽 2𝑚
𝜏 = π𝑎𝑏 or 𝜏= π𝑎𝑏 --------- (9)
2𝑚 𝐽

➢Again 𝑎 and 𝑏 are related through the eccentricity 𝑒 as 𝑏 = 𝑎 1 − 𝑒 2 -------- (9)

𝐽2
➢But from equation (4), we have 𝑒 = 1−
𝑚𝑘𝑎

𝐽2 𝐽2 𝐽2
➢ So 𝑏 =𝑎 1− 1 − =𝑎 = 𝑎1/2 -------- (10)
𝑚𝑘𝑎 𝑚𝑘𝑎 𝑚𝑘

2𝑚 2𝑚 𝐽2
➢Now equation (9) becomes 𝜏 = π𝑎𝑏 = π𝑎 𝑎1/2
𝐽 𝐽 𝑚𝑘

𝑚
or 𝜏 = 2π 𝑎3/2 ------- (11) i.e. 𝜏 2 ∝ 𝑎3 -------- (12)
𝑘
➢Thus the square of time period of a planet is proportional to the cube of its
semimajor axis. This is Kepler’s 3rd law.
➢But motion of a planet around sun is a two-body problem and 𝑚 must be replaced
𝑚𝑀
by reduced mass 𝜇 = , where 𝑚 is mass of planet and 𝑀 is mass of Sun.
𝑚+𝑀
𝑚𝑀
➢Further, for the attractive gravitational force 𝑓 = −𝐺 (i.e. 𝑘 = 𝐺𝑚𝑀)
𝑟2

➢Type equation here.Under these conditions, time period of the planet becomes
1 2π 𝑎3/2
𝜏 = 2π 𝑎3/2 ≈ ----------- (13)
𝐺(𝑚+𝑀) 𝐺𝑀

➢Here we neglect the mass of the planet compared to the sun.

➢Thus 𝜏 ∝ 𝑎3/2 with the same constant of proportionality for all planets.
 The Laplace-Runge-Lenz vector
❖The Kepler problem is also distinguished by the existence of an additional
conserved vector besides the angular momentum.
❖For a general central force, Newton’s second law of motion can be written
ሶ 𝑟Ԧ
vectorially as 𝑝Ԧ = 𝑓 𝑟 --------- (1)
𝑟
𝑟Ԧ
❖ Now 𝑝Ԧሶ × 𝐽Ԧ = 𝑓 𝑟 × 𝑟Ԧ × 𝑝Ԧ
𝑟
𝑟Ԧ
=𝑓 𝑟 × 𝑟Ԧ × 𝑚𝑟Ԧሶ
𝑟
𝑚𝑓 𝑟
= 𝑟Ԧ × 𝑟Ԧ × 𝑟Ԧሶ
𝑟
𝑚𝑓 𝑟
= Ԧ 𝑟Ԧሶ − 𝑟 2 𝑟Ԧሶ
𝑟Ԧ 𝑟. ------------ (2)
𝑟
 ሶ 1 𝑑 1 𝑑
But 𝑟.
Ԧ 𝑟Ԧ = 𝑟.
Ԧ 𝑟Ԧ = 𝑟 2 = 𝑟𝑟ሶ -------------- (3)
2 𝑑𝑡 2 𝑑𝑡

Since 𝐽Ԧ is a constant, equation (2) can be written as

𝑑 𝑚𝑓 𝑟
Ԧ
𝑝Ԧ × 𝐽 = 𝑟Ԧ 𝑟𝑟ሶ − 𝑟 2 𝑟Ԧሶ
𝑑𝑡 𝑟

𝑑 𝑟Ԧ𝑟ሶ 𝑟Ԧሶ
or 𝑝Ԧ × 𝐽Ԧ = 𝑚𝑓 𝑟 2
𝑟 2 −
𝑑𝑡 𝑟 𝑟

𝑑 Ԧሶ 𝑟Ԧ𝑟ሶ
𝑟 𝑟−
or 𝑝Ԧ × 𝐽Ԧ = −𝑚𝑓 𝑟 𝑟 2
𝑑𝑡 𝑟2

𝑑 𝑑 𝑟Ԧ
or 𝑝Ԧ × 𝐽Ԧ = −𝑚𝑓 𝑟 𝑟 2 ------------ (4).
𝑑𝑡 𝑑𝑡 𝑟
𝑘
But for Kepler’s problem, 𝑓(𝑟) = − . So above equation becomes
𝑟2
𝑑 𝑑 𝑟Ԧ
𝑝Ԧ × 𝐽Ԧ = 𝑚𝑘
𝑑𝑡 𝑑𝑡 𝑟
 𝑑 𝑑 𝑚𝑘𝑟Ԧ
or 𝑝Ԧ × 𝐽Ԧ =
𝑑𝑡 𝑑𝑡 𝑟

𝑑 𝑑 𝑚𝑘 𝑟Ԧ
or 𝑝Ԧ × 𝐽Ԧ − =0
𝑑𝑡 𝑑𝑡 𝑟

𝑑 𝑚𝑘𝑟Ԧ
or [ 𝑝Ԧ × 𝐽Ԧ − ]=0 --------- (5)
𝑑𝑡 𝑟

➢The above equation (5) tells that for the Kepler problem, there exist a constant
vector 𝐴Ԧ defined by
𝑚𝑘𝑟Ԧ
Ԧ Ԧ
𝐴 = 𝑝Ԧ × 𝐽 − ----------- (6)
𝑟

This vector 𝐴Ԧ is known as Laplace-Runge-Lenz vector.

➢Since 𝐽Ԧ is perpendicular to 𝑝Ԧ × 𝐽Ԧ and 𝑟Ԧ is perpendicular to 𝐽Ԧ = 𝑟Ԧ × 𝑝 , then

Ԧ 𝐽Ԧ =0
𝐴. ---------- (7)
It follows from the orthogonality of 𝐴Ԧ to 𝐽Ԧ that 𝐴Ԧ must be some fixed vector in the plane
of the orbit. If 𝜃 be the angle between 𝑟Ԧ and the fixed direction of 𝐴Ԧ , then
𝑚𝑘𝑟Ԧ
Ԧ 𝑟Ԧ = 𝐴𝑟𝑐𝑜𝑠𝜃 = 𝑝Ԧ × 𝐽Ԧ . 𝑟Ԧ −
𝐴. . 𝑟Ԧ
𝑟

or Ԧ 𝑝Ԧ × 𝐽Ԧ − 𝑚𝑘𝑟 -------- (8)

𝐴𝑟𝑐𝑜𝑠𝜃 = 𝑟.

Ԧ 𝑝Ԧ × 𝐽Ԧ = 𝐽.Ԧ 𝑟Ԧ × 𝑝Ԧ = 𝐽.Ԧ 𝐽Ԧ = 𝐽2
But from vector algebra, we have 𝑟.
Now equation (8) gives 𝐴𝑟𝑐𝑜𝑠𝜃 = 𝐽2 − 𝑚𝑘𝑟
𝐽2
⇒ 𝐴𝑐𝑜𝑠𝜃 = − 𝑚𝑘
𝑟
𝐽2
⇒ = 𝐴𝑐𝑜𝑠𝜃 + 𝑚𝑘
𝑟
1 𝐴𝑐𝑜𝑠𝜃+𝑚𝑘
⇒ =
𝑟 𝐽2
1 𝑚𝑘 𝐴
⇒ = 1+ 𝑐𝑜𝑠𝜃 --------------- (9)
𝑟 𝐽2 𝑚𝑘
➢ Thus the Laplace-Runge-Lenz vector provides another way of deriving the orbit equation for the
Kepler problem.
➢ In comparison with the orbit equation for Kepler’s problem
1 𝑚𝑘
= 1 + e cos 𝜃 − 𝜃 ′ with equation (9) and using equation (6), we can say
𝑟 𝐽2

the Laplace-Runge-Lenz vector 𝐴Ԧ is in the direction vector to the perihelion point on

the orbit and has a magnitude 𝐴 = 𝑚𝑘e ------- (10)
2𝐸𝐽2 2𝐸𝐽2
➢ Since 𝑒 = 1+ , 𝐴 = 𝑚𝑘 1 + and hence 𝐴2 = 𝑚2 𝑘 2 + 2𝑚𝐸𝐽2 --------- (11)
𝑚𝑘 2 𝑚𝑘 2

➢ Thus for the Kepler’s problem, we have two vector constants of the motion 𝐽Ԧ and 𝐴Ԧ and a scalar E.
➢ Since a vector must have all three independent components, this corresponds to seven conserved
quantities in all.
➢ In general orbits for central force motion are not closed. But if the orbits are degenerate, there exist
an additional conserved quantity that is an algebraic function of 𝑟Ԧ and 𝑝,Ԧ such as Laplace-Runge-
Lenz vector.
➢ Thus, the existence of an additional constant or integral of the motion, beyond 𝐽Ԧ and E, that is a
simple algebraic function of the motion like Laplace-Runge-Lenz vector, is sufficient to indicate that
in Kepler’s problem, the motion is degenerate and the bounded orbits are closed.