GRE Physics Companion: Orbit Equation, Conic Sections, and Kepler's First Law

GRE Physics Companion: Orbit Equation, Conic Sections, and Kepler’s First Law

The Newtonian gravitational orbit equation is

(1)

with

(2)

1 High-value GRE facts

1. u = 1∕r.
2. Newtonian Binet equation: u′′ + u = μ∕h².
3. Orbit equation: r = p∕(1 + e cos ν).
4. p = h²∕μ.
5. e = 0 circle.
6. 0 < e < 1 ellipse.
7. e = 1 parabola.
8. e > 1 hyperbola.
9. Ellipse: r_p = a(1 − e) and r_a = a(1 + e).
10. e² = 1 + 2𝜀h²∕μ².

Part I: Original GRE-style problems

Problem 1: reciprocal radius

Binet’s substitution defines

1. u = r
2. u = r²
3. u = 1∕r
4. u = ṙ
5. u = 𝜃∕r

Problem 2: Newtonian Binet equation

For inverse-square gravity, the orbit equation for u(𝜃) is

1. u′′ + u = μ∕h²
2. u′′− u = μ∕h²
3. u′′ + u = μu²∕h²
4. u′′ = μ∕r²
5. u′′ + h²u = μ

Problem 3: semilatus rectum

For a gravitational conic,

(3)

equals

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

Problem 4: ellipse condition

A bound noncircular ellipse has eccentricity

1. e < 0
2. e = 0
3. 0 < e < 1
4. e = 1
5. e > 1

Problem 5: parabolic condition

The escape-threshold conic has

1. e = 0
2. 0 < e < 1
3. e = 1
4. e > 1
5. imaginary e

Problem 6: periapsis radius

For

(4)

the periapsis radius is

1. p∕(1 − e)
2. p∕(1 + e)
3. p∕e
4. p(1 + e)
5. p

Problem 7: ellipse eccentricity from apsides

For an ellipse,

(5)

equals

1. (r_a + r_p)∕(r_a − r_p)
2. (r_a − r_p)∕(r_a + r_p)
3. r_a∕r_p
4. r_p∕r_a
5. 

Problem 8: semilatus rectum of ellipse

For an ellipse,

(6)

equals

1. a(1 − e)
2. a(1 + e)
3. a(1 − e²)
4. a∕e
5. ae

Problem 9: energy and eccentricity

The correct energy-eccentricity relation is

1. e² = 1 + 2𝜀h²∕μ²
2. e = 1 + 2𝜀h∕μ
3. e² = 1 − 2𝜀μ²∕h²
4. e² = 2𝜀∕μ
5. e = 𝜀h²∕μ²

Problem 10: focus location

In a Keplerian ellipse, the attracting body lies

1. at the ellipse center
2. at one focus
3. at a vertex
4. outside the orbital plane
5. at the other body’s instantaneous position

Problem 11: hyperbolic asymptote

For a hyperbolic orbit, the asymptotic true anomaly satisfies

1. cos ν_∞ = e
2. cos ν_∞ = 1∕e
3. cos ν_∞ = −1∕e
4. sin ν_∞ = −1∕e
5. ν_∞ = π∕2 for every hyperbola

Problem 12: first law

Kepler’s first law for ideal Newtonian bound two-body motion states that the orbit is

1. always circular
2. an ellipse with the attracting body at one focus
3. a parabola
4. a hyperbola
5. an arbitrary closed curve

Part II: Complete worked solutions

Solution 1

Binet’s substitution is

(7)

Answer: (C).

Solution 2

Inverse-square gravity makes the right-hand side constant:

(8)

Answer: (A).

Solution 3

Comparing the conic solution with Binet’s solution gives

(9)

Answer: (C).

Solution 4

A noncircular bound ellipse has

(10)

Answer: (C).

Solution 5

The parabolic escape threshold is

(11)

Answer: (C).

Solution 6

Periapsis occurs at

(12)

so

(13)

Answer: (B).

Solution 7

Using

(14)

gives

(15)

Answer: (B).

Solution 8

For an ellipse,

(16)

Answer: (C).

Solution 9

Substitution into the energy equation yields

(17)

Answer: (A).

Solution 10

The polar origin of the Keplerian conic is the occupied focus, not the ellipse center. Answer: (B).

Solution 11

The hyperbolic radius diverges when

(18)

so

(19)

Answer: (C).

Solution 12

Negative-energy inverse-square gravitational motion is elliptic, with the attracting body at a focus. Answer: (B).

2 GRE checklist

1. Use u = 1∕r before differentiating.
2. Remember p = h²∕μ.
3. Classify the orbit from e or from the sign of 𝜀.
4. Use periapsis at ν = 0.
5. For ellipses, connect r_p, r_a, a, e, and p.
6. Keep the occupied focus distinct from the ellipse center.

References

References