GRE Physics Companion: Orbital Energy, Escape Speed, and Bound versus Unbound Motion

GRE Physics Companion: Orbital Energy, Escape Speed, and Bound versus Unbound Motion

The specific orbital energy is

(1)

The sign of

(2)

classifies gravitational motion.

1 High-value GRE facts

1. 𝜀 = v²∕2 − μ∕r.
2. Circular orbit: 𝜀 = −μ∕(2r).
3. Bound energy: 𝜀 < 0.
4. Escape threshold: 𝜀 = 0.
5. Unbound energy: 𝜀 > 0.
6. v_esc = .
7. v_esc = v_circ at the same radius.
8. Unbound motion has 𝜀 = v_∞²∕2.
9. For unbound motion, v² = v _∞² + v _esc².
10. Exact two-body relative motion uses μ = G(m₁ + m₂).

Part I: Original GRE-style problems

Problem 1: circular specific energy

For a circular orbit of radius r, the specific orbital energy is

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

Problem 2: bound classification

A trajectory with 𝜀 < 0 is

1. necessarily circular
2. gravitationally bound
3. necessarily radial
4. at escape threshold
5. unbound

Problem 3: escape threshold

At the minimum escape condition, 𝜀 equals

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

Problem 4: escape speed

The escape speed at radius r is

1. 
2. 
3. μ∕r
4. 2μ∕r
5. 

Problem 5: escape/circular ratio

At one radius, v_esc∕v_circ equals

1. 1∕
2. 1
3. 
4. 2
5. 2π

Problem 6: positive specific energy

For an unbound orbit with asymptotic speed v_∞, 𝜀 is

1. −v_∞²∕2
2. zero
3. v_∞²∕2
4. v_∞²
5. 2v_∞²

Problem 7: local unbound speed

For an unbound trajectory, v² equals

1. v_∞² − v _esc²
2. v_∞² + v _esc²
3. v_∞v_esc
4. v_esc²∕2
5. v_∞²

Problem 8: escape-speed scaling

If radius increases by a factor of four, escape speed changes by a factor

1. 4
2. 2
3. 1
4. 1∕2
5. 1∕4

Problem 9: radial turning point

For purely radial bound motion, at the outer turning point the

1. radial speed is zero and total speed is zero
2. total speed equals escape speed
3. energy is zero
4. potential energy is zero
5. radius is infinite

Problem 10: circular-to-escape increment

The ideal instantaneous speed increment from circular speed to local escape speed is

1. (2 − 1)v_circ
2. ( − 1)v_circ
3. v_circ∕2
4. v_circ
5. 2v_circ

Problem 11: exact two-body parameter

For exact relative motion of masses m₁ and m₂, the gravitational parameter is

1. Gm₁
2. Gm₂
3. G(m₁ + m₂)
4. Gm₁m₂
5. G(m₁ − m₂)

Problem 12: conic-energy classification

In the ideal inverse-square two-body problem, positive specific energy corresponds to

1. a circular orbit
2. an elliptic orbit
3. a parabolic orbit
4. a hyperbolic orbit
5. no motion

Part II: Complete worked solutions

Solution 1

For circular motion,

(3)

Therefore

(4)

Answer: (C).

Solution 2

Negative total energy means the body cannot reach infinity with real speed. Answer: (B).

Solution 3

Minimum escape arrives at infinity with zero final speed and therefore zero total specific energy. Answer: (C).

Solution 4

Set 𝜀 = 0:

(5)

Answer: (B).

Solution 5

Since

(6)

the ratio is . Answer: (C).

Solution 6

At infinity the potential term vanishes, so

(7)

Answer: (C).

Solution 7

Energy conservation gives

(8)

Answer: (B).

Solution 8

Since v_esc ∝ r^−1∕2, multiplying radius by four divides speed by two. Answer: (D).

Solution 9

For purely radial motion, all velocity is radial. At the outer turning point the radial speed, and therefore total speed, is zero. Answer: (A).

Solution 10

The required change is

(9)

Answer: (B).

Solution 11

The relative acceleration depends on the sum of the masses, giving

(10)

Answer: (C).

Solution 12

Positive specific energy corresponds to an unbound hyperbola. Answer: (D).

2 GRE checklist

1. Compute 𝜀 = v²∕2 − μ∕r before classifying the motion.
2. Compare with zero, not with the potential energy alone.
3. Use local radius in the escape-speed formula.
4. Keep circular speed and escape speed distinct.
5. Use v_∞ only for positive-energy unbound motion.
6. Do not use a radial turning-point formula when angular momentum is nonzero.
7. Use G(m₁ + m₂) for exact relative two-body motion.

References

References