GRE Physics Companion: Effective Potential and Radial Motion in Central Gravity

GRE Physics Companion: Effective Potential and Radial Motion in Central Gravity

For specific angular momentum

(1)

the radial energy equation is

(2)

1 High-value GRE facts

1. Φ_eff = h²∕(2r²) − μ∕r.
2. Allowed radial motion requires 𝜀 ≥ Φ_eff.
3. radial turning points satisfy 𝜀 = Φ_eff.
4. For h≠0, Φ_eff → +∞ as r → 0.
5. Circular orbit: r_c = h²∕μ.
6. Circular specific energy: 𝜀_c = −μ∕(2r_c).
7. The Newtonian circular orbit is a stable effective-potential minimum.
8. At periapsis and apoapsis, ṙ = 0 but tangential speed is generally nonzero.
9. For h = 0, the angular-momentum barrier disappears.
10. The radial equation is r = h²∕r³ − μ∕r².

Part I: Original GRE-style problems

Problem 1: effective potential

The specific effective potential for Newtonian gravity is

1. h²∕(2r) − μ∕r²
2. h²∕(2r²) − μ∕r
3. h²∕r² + μ∕r
4. −μ∕r
5. h∕r − μ∕r²

Problem 2: allowed region

Radial motion is allowed where

1. 𝜀 < Φ_eff
2. 𝜀 = Φ_eff only
3. 𝜀 ≥ Φ_eff
4. Φ_eff = 0
5. h = 0

Problem 3: turning point

At a radial turning point,

1. ṙ = 0
2. 𝜃 = 0
3. total speed must be zero
4. angular momentum must vanish
5. energy must vanish

Problem 4: circular radius

For fixed h, the circular-orbit radius is

1. μ∕h²
2. h∕μ
3. h²∕μ
4. μ²∕h
5. 2h²∕μ

Problem 5: circular energy

At the circular-orbit radius,

(3)

is

1. +μ∕(2r_c)
2. zero
3. −μ∕(2r_c)
4. −μ∕r_c
5. +μ∕r_c

Problem 6: stability

The Newtonian circular orbit is radially stable because at r_c

1. dΦ_eff∕dr≠0
2. d²Φ _eff∕dr² < 0
3. d²Φ _eff∕dr² > 0
4. Φ_eff = 0
5. h = 0

Problem 7: barrier

For nonzero h, as r → 0 the specific effective potential

1. approaches −∞
2. approaches zero
3. approaches +∞
4. remains constant
5. oscillates

Problem 8: zero angular momentum

If h = 0, the effective potential becomes

1. +μ∕r
2. −μ∕r
3. h²∕(2r²)
4. zero
5. −μ∕r²

Problem 9: radial acceleration

The radial equation can be written

1. r = h²∕r³ − μ∕r²
2. r = −h²∕r³ − μ∕r²
3. r = h∕r² − μ∕r
4. r = μ∕r²
5. r = 0

Problem 10: two bound turning points

For fixed nonzero h, a negative energy above the effective-potential minimum generally gives

1. no turning radii
2. one finite turning radius
3. two finite turning radii
4. infinitely many turning radii
5. only r = 0

Problem 11: periapsis speed

At periapsis for nonzero angular momentum,

1. total speed must vanish
2. radial velocity vanishes but transverse speed remains
3. angular momentum vanishes
4. potential energy vanishes
5. radial acceleration must vanish

Problem 12: small radial oscillation frequency

For a Newtonian circular orbit,

(4)

for small radial oscillations is

1. μ∕r_c³
2. μ∕r_c²
3. 2μ∕r_c³
4. h∕r_c²
5. zero

Part II: Complete worked solutions

Solution 1

Angular-momentum conservation converts the transverse kinetic energy into

(5)

so

(6)

Answer: (B).

Solution 2

Since

(7)

the right-hand side must be nonnegative. Answer: (C).

Solution 3

A radial turning point is defined by

(8)

For nonzero angular momentum, transverse motion can remain. Answer: (A).

Solution 4

Set

(9)

This gives

(10)

and therefore

(11)

Answer: (C).

Solution 5

At the circular minimum,

(12)

Answer: (C).

Solution 6

The second derivative is

(13)

The extremum is therefore a minimum. Answer: (C).

Solution 7

The positive term

(14)

dominates over

(15)

at sufficiently small radius. Answer: (C).

Solution 8

When

(16)

the angular term disappears and only

(17)

remains. Answer: (B).

Solution 9

Differentiating the effective potential gives

(18)

Answer: (A).

Solution 10

A horizontal negative-energy line above the minimum intersects the effective-potential curve twice. Answer: (C).

Solution 11

At periapsis,

(19)

but

(20)

is generally nonzero. Answer: (B).

Solution 12

The curvature of the effective potential at the circular minimum is

(21)

Thus

(22)

Answer: (A).

2 GRE checklist

1. Convert transverse kinetic energy using conserved angular momentum.
2. Use the energy level to identify allowed radial regions.
3. Set ṙ = 0 at turning points, not necessarily v = 0.
4. Find circular orbits from dΦ_eff∕dr = 0.
5. Test stability with the second derivative.
6. Remember that the small-radius barrier disappears for h = 0.

References

References