GRE Physics Companion: Angular Momentum, Areal Velocity, and Kepler’s Second
Law
For planar central force motion,
and
1 High-value GRE facts
- A central force gives zero torque about the force center.
- Zero torque gives constant angular momentum.
- L = mr2𝜃.
- h = L∕m = r2𝜃.
- dA =
r2d𝜃.
- Ȧ = h∕2 = L∕(2m).
- Equal times imply equal swept areas.
- Kepler’s second law applies to every central force.
- At apsides, rpvp = rava.
- Radial motion with L = 0 has zero areal velocity.
Part I: Original GRE-style problems
Problem 1: central force torque
For a central force, the torque about the force center is
- constant and nonzero
- zero
- proportional to r
- proportional to 1∕r2
- undefined
Problem 2: angular momentum
For planar motion, the angular-momentum magnitude is
- mr𝜃
- mr2𝜃
- mṙ2
- mrṙ
- m∕r2
Problem 3: areal velocity
If specific angular momentum is h, the areal velocity is
- h
- 2h
- h∕2
- h2∕2
- 1∕(2h)
Problem 4: angular rate
For fixed h, the angular rate is
- h∕r
- h∕r2
- hr
- hr2
- 1∕(hr2)
Problem 5: equal areas
Kepler’s second law states that equal swept areas correspond to equal
- distances
- speeds
- angular displacements
- time intervals
- radii
Problem 6: periapsis versus apoapsis
If ra = 2rp, then the apsidal speeds satisfy
- vp = va∕2
- vp = va
- vp = 2va
- vp = 4va
- vp =
va
Problem 7: generality of the area law
Constant areal velocity follows from
- inverse-square gravity only
- circular motion only
- any central force
- constant speed
- zero energy
Problem 8: doubled radius
At one point on an orbit the radius is twice that at another point. With the same h, the
angular-rate ratio is
- 1∕2
- 1∕4
- 1
- 2
- 4
Problem 9: radial motion
For purely radial central force motion,
- areal velocity is zero
- areal velocity is infinite
- angular momentum is nonzero
- angular rate is constant and nonzero
- Kepler’s second law fails
Problem 10: ellipse period
For a known ellipse of area πab and specific angular momentum h, the period is
- πab∕h
- 2πab∕h
- h∕(2πab)
- 2h∕(πab)
- πh∕(ab)
Problem 11: apsidal velocity direction
At periapsis of a nonradial orbit, the instantaneous velocity is
- purely radial
- purely transverse
- zero
- parallel to angular momentum
- undefined
Problem 12: two-body relative motion
In the exact two-body problem, the relative areal velocity is
- Lrel∕μr
- Lrel∕(2μr)
- 2Lrel∕μr
- G(m1 + m2)
- zero
Part II: Complete worked solutions
Solution 1
A central force is parallel or antiparallel to r, so
Answer: (B).
Solution 2
The transverse contribution gives
Answer: (B).
Solution 3
Since
and
we obtain
Answer: (C).
Solution 4
Angular-momentum conservation gives
Answer: (B).
Solution 5
Constant areal velocity means area is proportional to elapsed time. Answer: (D).
Solution 6
At the apsides,
If ra = 2rp, then vp = 2va. Answer: (C).
Solution 7
The proof requires only zero torque, which is true for every central force. Answer:
(C).
Solution 8
Because
doubling radius divides angular rate by four. Answer: (B).
Solution 9
Pure radial motion has
Therefore
Answer: (A).
Solution 10
One orbit sweeps the full area:
Thus
Answer: (B).
Solution 11
At periapsis,
so the velocity is transverse. Answer: (B).
Solution 12
The relative angular momentum is
Thus
Answer: (B).
2 GRE checklist
- Start with zero central torque.
- Convert angular momentum to h = r2𝜃.
- Keep the factor 1∕2 in the area formula.
- Use apsidal speed relations only where ṙ = 0.
- Remember that the area law is more general than inverse-square gravity.
- For exact two-body motion, use the relative coordinate and reduced mass.
References
References
[1] PhysicsLibrary, M07-07, Angular Momentum, Areal Velocity, and Kepler’s Second
Law.
[2] PhysicsLibrary, M07-06, Effective Potential and Radial Motion in Central Gravity.
[3] J. R. Taylor, Classical Mechanics, University Science Books, 2005.