Physics 321
Hour 22
Central Force Problems
Straight-line Motion in Spherical or Cylindrical
Coordinates Coordinates
y
r
r
x
a
t
We want to treat r like a Cartesian coordinate, but if we do, if
appears that there’s a force on the mass even when there is
no force!
Understanding the “Fictitious” Force
Think of r like x :
𝐹
0
a
The force acts in the +r direction.
r
Understanding the “Fictitious” Force
𝑥=𝑎
𝑦 = 𝑣𝑡
→ 𝑟=
𝑎2 + 𝑣 2 𝑡 2
2𝑣 2 𝑡 𝑣 2 𝑡
𝑟=
=
2𝑟
𝑟
𝑣 2 𝑣 2𝑡 𝑣 2𝑡 𝑣 2
𝑣 2𝑡2
𝑟=
− 2
=
1− 2
𝑟
𝑟 𝑟
𝑟
𝑟
𝑣 2 𝑟2 − 𝑣 2𝑡 2
𝑣 2 𝑎2
𝑣 2 𝑎2
=
=
= 3
2
2
𝑟
𝑟
𝑟 𝑟
𝑟
Understanding the “Fictitious” Force
𝑚𝑣 2 𝑎2
𝐹 = 𝑚𝑟 =
𝑟3
𝑙 = 𝑚𝑣𝑎
𝑙2
𝐹=
𝑚𝑟 3
We can define a corresponding potential
energy:
𝑈=−
𝐹𝑑𝑟 = −
𝑙2
𝑙2
𝑑𝑟 = +
3
𝑚𝑟
2𝑚𝑟 2
Before we even consider real forces, we have
to “add” a repulsive centrifugal force or
centrifugal potential to the problem:
𝑙2
𝐹=
𝑚𝑟 3
𝑙2
𝑈=+
2𝑚𝑟 2
Centrifugal Potential
𝑈𝑐𝑒𝑛𝑡
ℓ2
=+
2μ𝑟 2
𝑈𝑒𝑓𝑓 = 𝑈𝑐𝑒𝑛𝑡 + 𝑈𝑔
𝑈𝑔 = −𝐺
𝑀𝑚
𝑟
Central Force
• Force is in the radial direction in spherical
coordinates
• The curl is always zero – even if the force isn’t
inverse square
• Source at the origin
𝛼
𝛼𝑟
𝐹 = 𝑛 𝑟 = 𝑛+1
𝑟
𝑟
• Source at 𝑟1
𝛼(𝑟2 − 𝑟1 )
𝐹21 =
|𝑟2 − 𝑟1 |𝑛+1
Central Force
• Source at 𝑟2
𝐹12
• Potential energy
𝛼(𝑟1 − 𝑟2 )
=
|𝑟2 − 𝑟1 |𝑛+1
𝛼 𝑛−1
𝑈 𝑟 =
,
𝑛−1
𝑟2 − 𝑟1
𝑛≠1
Inverse-square Force
• Source at 𝑟1
𝐹21
• Potential energy
𝛼(𝑟2 − 𝑟1 )
=
|𝑟2 − 𝑟1 |3
𝛼
𝑈 𝑟 =
𝑟2 − 𝑟1
Example
moon_orbit.nb