Equation of motion under central
force and first integrals
Equation of motion under central force and first integrals
Let us consider a particle of mass ‘m’ moving under central force.
We know that (Conservative) central force is a function of 𝑟 only
𝜕𝑉(𝑟)
So 𝐹Ԧ = 𝑟𝑓
Ƹ 𝑟 = −𝑟Ƹ
𝜕𝑟
where 𝑉 𝑟 is the potential energy.
Using plane polar coordinates (𝑟, 𝜃), the Lagrangian of the system can be written as
1
𝐿 = 𝑇 − 𝑉 = 𝑚 𝑟ሶ 2 + 𝑟 2 𝜃ሶ 2 − 𝑉(𝑟) ------------- (1)
2
Here the Lagrangian does not contain 𝜃 and therefore 𝜃 is a cyclic coordinate.
𝜕𝐿
So =0.
𝜕𝜃
𝜕𝐿
Canonical momentum corresponds to coordinate 𝜃 is 𝑝𝜃 = = 𝑚𝑟 2 𝜃ሶ
𝜕𝜃ሶ
Now Lagrange’s equation in 𝜃 becomes
𝑑 𝜕𝐿 𝜕𝐿
− =0
𝑑𝑡 𝜕𝜃ሶ 𝜕𝜃
𝑑
or 𝑚𝑟 2 𝜃ሶ = 0
𝑑𝑡
or 𝑚𝑟 2 𝜃ሶ = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝐽 ---------- (2)
Where J is the constant magnitude of Angular Momentum and it is a conserved quantity
Equation (2) represents one first integral of motion.
Again the Lagrange’s equation in ‘𝑟’ is
𝑑 𝜕𝐿 𝜕𝐿
− =0
𝑑𝑡 𝜕𝑟ሶ 𝜕𝑟
𝜕𝐿 𝜕𝑉(𝑟) 𝜕𝐿
But ሶ
= 𝑚𝑟𝜃 −2 and = 𝑚𝑟ሶ
𝜕𝑟 𝜕𝑟 𝜕𝑟ሶ
Now Lagrange’s equation become
𝑑 𝜕𝑉(𝑟)
ሶ
𝑚𝑟ሶ − 𝑚𝑟𝜃 +2 =0
𝑑𝑡 𝜕𝑟
𝜕𝑉(𝑟)
or ሶ 2
𝑚𝑟ሷ − 𝑚𝑟𝜃 = −
𝜕𝑟
𝐽 2 𝜕𝑉(𝑟)
or 𝑚𝑟ሷ − 𝑚𝑟 =− (Since J= 𝑚𝑟 2 𝜃)ሶ
𝑚𝑟 2 𝜕𝑟
𝐽2
or 𝑚𝑟ሷ − = 𝑓(𝑟) -------------- (A)
𝑚𝑟 3
𝐽2
or 𝑚𝑟ሷ = − 𝑓(𝑟)
𝑚𝑟 3
𝜕 𝐽2 𝜕𝑉(𝑟)
or 𝑚𝑟ሷ = − −
𝜕𝑟 2𝑚𝑟 2 𝜕𝑟
𝜕 𝐽2
or 𝑚𝑟ሷ = − + 𝑉(𝑟) ------------- (3)
𝜕𝑟 2𝑚𝑟 2
Multiplying 𝑟ሶ in both sides of above equation, we get
𝜕 𝐽2
or 𝑚𝑟ሷ 𝑟ሶ = − + 𝑉(𝑟) 𝑟ሶ
𝜕𝑟 2𝑚𝑟 2
𝑑 1 𝑑 𝐽2
or 𝑚𝑟ሶ 2 = − + 𝑉(𝑟)
𝑑𝑡 2 𝑑𝑡 2𝑚𝑟 2
𝑑 1 2 𝐽2
or [ 𝑚𝑟ሶ + + 𝑉(𝑟)] = 0
𝑑𝑡 2 2𝑚𝑟 2
1 𝐽 2
or 𝑚𝑟ሶ 2 + + 𝑉 𝑟 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ------------- (4)
2 2𝑚𝑟 2
But we know that total energy E is
𝐸 =𝑇+𝑉
1
= 𝑚 𝑟ሶ 2 + 𝑟 2 𝜃ሶ 2 + 𝑉(𝑟)
2
1 1
= 𝑚𝑟ሶ 2 + 𝑚𝑟 2 𝜃ሶ 2 + 𝑉(𝑟)
2 2
1 1 𝐽 2
= 𝑚𝑟ሶ 2 + 𝑚𝑟 2 + 𝑉(𝑟)
2 2 𝑚𝑟 2
1 2 𝐽2
= 𝑚𝑟ሶ + +𝑉 𝑟 ------------ (5)
2 2𝑚𝑟 2
1 2 𝐽2
Thus 𝐸 = 𝑚𝑟ሶ + + 𝑉 𝑟 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ----------- (6)
2 2𝑚𝑟 2
i.e. The total energy of the system is a conserved quantity.
Equation (6) represents another first integral of motion.
Differential equation of orbit
The equation involving the dependence of 𝑟 upon 𝜃 by eliminating the parameter 𝑡 is called
equation of orbit.
𝐽2
We have 𝑓 𝑟 = 𝑚𝑟ሷ − 3 -------- (1)
𝑚𝑟
1
Let us put 𝑟 =
𝑢
𝑑 1 𝑑 1 𝑑𝑢 1 𝑑𝑢
So 𝑟ሶ = = = − 2
𝑑𝑡 𝑢 𝑑𝑢 𝑢 𝑑𝑡 𝑢 𝑑𝑡
1 𝑑𝑢 𝑑𝜃 1 𝑑𝑢
or 𝑟ሶ = − =− 𝜃ሶ
𝑢2 𝑑𝜃 𝑑𝑡 2
𝑢 𝑑𝜃
1 𝐽 𝑑𝑢
or 𝑟ሶ = − (Since J= 𝑚𝑟 2 𝜃)ሶ
𝑢2 𝑚𝑟 2 𝑑𝜃
𝐽 𝑑𝑢
or 𝑟ሶ = − ---------------- (2) (since 𝑢2 𝑟 2 = 1)
𝑚 𝑑𝜃
𝐽 𝑑 𝑑𝑢 𝐽 𝑑 𝑑𝑢 𝑑𝜃
Now 𝑟ሷ = − = −
𝑚 𝑑𝑡 𝑑𝜃 𝑚 𝑑𝜃 𝑑𝜃 𝑑𝑡
𝐽 𝑑 𝑑𝑢
or 𝑟ሷ = − 𝜃ሶ
𝑚 𝑑𝜃 𝑑𝜃
𝐽 𝐽 𝑑 𝑑𝑢
or 𝑟ሷ = − (Since J= 𝑚𝑟 2 𝜃)ሶ
𝑚 𝑚𝑟 2 𝑑𝜃 𝑑𝜃
𝐽2 𝑑 2 𝑢
or 𝑟ሷ = − 2 2 2
𝑚 𝑟 𝑑𝜃
𝐽 2 𝑢2 𝑑 2 𝑢
or 𝑟ሷ = − 2 2 ----------- (3)
𝑚 𝑑𝜃
Now equation (1) can be written as
1 𝐽 2 𝑢2 𝑑 2 𝑢 𝐽 2 𝑢3
𝑓 =𝑚 − 2 2 −
𝑢 𝑚 𝑑𝜃 𝑚
1 𝐽 2 𝑢2 𝑑 2 𝑢 𝐽 2 𝑢3
or 𝑓 = − −
𝑢 𝑚 𝑑𝜃 2 𝑚
1 𝐽 2 𝑢2 𝑑 2 𝑢
or 𝑓 = − +𝑢
𝑢 𝑚 𝑑𝜃 2
𝑑2 𝑢 𝑚 1
or +𝑢 = − 2 2 𝑓 ------------ (4)
𝑑𝜃 2 𝐽 𝑢 𝑢
➢ This equation (4) is the differential equation for the orbit if the force law 𝑓 𝑟 is known.
➢ Conversely if the equation of the orbit is known, i.e. 𝑟 is given as a function of 𝜃, then one can
work back and obtain the force law 𝑓 𝑟 .
➢ Equation (4) tells that the orbit is symmetrical about turning points. Because equation (4) remains
invariant under 𝜃 = −𝜃 transformation, where 𝜃 is the turning angle.
➢ The orbit is therefore invariant under reflection about the apsidal vectors. This means that
complete orbit can be traced, if the portion of the orbit between two turning point is known.
