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[parent] GRE Physics Companion: Gravitational Fields, Superposition, and Continuous Mass Distributions

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GRE Physics Companion: Gravitational Fields, Superposition, and Continuous Mass Distributions

The Gravitational Field of a point mass is

|-------------|
g =  − GM  r- |
-----------r3--
(1)

when the source is at the origin.

For several masses,

|-----------|
|    ∑      |
g =     gj. |
------j------
(2)

PIC

Figure 1. A compact field strategy: identify the source geometry, choose the correct density, exploit symmetry, integrate only surviving components, and check limiting behavior.

1 High-value GRE facts

  1. Gravitational field has units N∕kg = m∕s2.
  2. Field is force per unit test mass.
  3. Point-mass field magnitude is GM∕r2.
  4. Fields add as vectors.
  5. dm = λdℓ for a line.
  6. dm = Σ dA for a surface.
  7. dm = ρdV for a volume.
  8. Symmetry should be used before integration.
  9. A uniform ring has zero field at its center.
  10. Far from a localized source, the leading field usually approaches point-mass form.

Part I: Original GRE-style problems

Problem 1: field units

The SI units of gravitational field are

  1. N
  2. N m
  3. N∕kg
  4. kg∕N
  5. kg m∕s

Problem 2: field and force

At a fixed point in a gravitational field, the test mass is doubled. The gravitational field at that point

  1. doubles
  2. halves
  3. remains unchanged
  4. becomes zero
  5. reverses direction

Problem 3: midpoint field

Two equal masses are placed symmetrically on opposite sides of the origin. The field at the origin is

  1. toward the left mass
  2. toward the right mass
  3. zero
  4. twice one field
  5. undefined

Problem 4: line density

For a thin rod with line density λ, the correct mass element is

  1. dm = λdV
  2. dm = λdA
  3. dm = λdℓ
  4. dm = dℓ∕λ
  5. dm = λr2 dr

Problem 5: ring center

The gravitational field at the exact center of a uniform ring is

  1. GM∕a2
  2. 2GM∕a2
  3. zero
  4. infinite
  5. GM∕a

Problem 6: ring far field

Far along the axis of a ring of total mass M, its field approaches

  1. GM∕z
  2. GM∕z2
  3. GM∕z3
  4. a constant
  5. zero identically

Problem 7: infinite-line scaling

The field magnitude of an infinite uniform line mass scales with perpendicular distance r as

  1. r
  2. constant
  3. 1∕r
  4. 1∕r2
  5. 1∕r3

Problem 8: infinite-sheet field

The magnitude of the field of an infinite uniform sheet with surface density Σ is

  1. GΣ∕r2
  2. 2πGΣ
  3. 4πGΣr
  4. GΣ∕r
  5. zero

Problem 9: disk far field

A finite uniform disk observed at distances much larger than its radius behaves to leading order like

  1. an infinite sheet
  2. an infinite line
  3. a point mass with the disk’s total mass
  4. zero mass
  5. a dipole only

Problem 10: ring near center

For small axial displacement z from the center of a uniform ring,

gz
(3)

is approximately proportional to

  1. 1∕z2
  2. 1∕z
  3. a constant
  4. −z
  5. z2

Problem 11: density dimensions

Volume mass density has units

  1. kg∕m
  2. kg∕m2
  3. kg∕m3
  4. N∕m3
  5. m3∕kg

Problem 12: symmetry

For a point on the symmetry axis of a uniform ring, transverse field components

  1. add constructively
  2. cancel in pairs
  3. are larger than the axial component
  4. are undefined
  5. produce a torque only

Part II: Complete worked solutions

Solution 1

By definition,

g = F ∕m,
(4)

so the units are

N∕kg.
(5)

Answer: (C).

Solution 2

The field is created by the source masses and is independent of the test mass. Answer: (C).

Solution 3

The two equal field vectors have equal magnitude and opposite direction. Answer: (C).

Solution 4

A one-dimensional mass model uses

dm  = λ dℓ.
(6)

Answer: (C).

Solution 5

Every ring element has a partner producing an equal and opposite field at the center. Answer: (C).

Solution 6

At z ≫ a,

gz = − GM  z∕(a2 + z2)3∕2 ≈ − GM ∕z2.
(7)

Answer: (B).

Solution 7

The infinite-line limit gives

g = 2G λ∕r.
(8)

Answer: (C).

Solution 8

The infinite-sheet limit of the disk result is

g = 2 πG Σ.
(9)

Answer: (B).

Solution 9

Any localized source looks like its total point mass at sufficiently large distance to leading order. Answer: (C).

Solution 10

For |z|≪ a,

gz ≈ − GM--z.
        a3
(10)

Answer: (D).

Solution 11

volume density is mass per volume:

           3
[ρ] = kg∕m  .
(11)

Answer: (C).

Solution 12

Rotational symmetry pairs transverse components with equal and opposite partners. Answer: (B).

2 GRE checklist

  1. Separate field from force.
  2. Use vectors for superposition.
  3. Pick λ, Σ, or ρ from source dimension.
  4. Use symmetry before integration.
  5. Check center, far-field, and infinite-source limits.
  6. Verify that a localized extended source approaches point-mass behavior far away.

References

References

[1]   PhysicsLibrary, M07-02, Gravitational Fields, Superposition, and Continuous Mass Distributions.

[2]   PhysicsLibrary, M07-01, Newton’s Law of Universal Gravitation.

[3]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.


"GRE Physics Companion: Gravitational Fields, Superposition, and Continuous Mass Distributions" is owned by bloftin.
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Other names:  M07-02G
Also defines:  gravitational field, field superposition, line mass density, surface mass density, volume mass density
Keywords:  GRE physics, gravitational field, superposition, continuous mass distribution, ring, rod, disk, density

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Cross-references: dimension, superposition, volume density, torque, displacement, volume, vectors, magnitude, force, units, mass, The Gravitational Field
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This is version 1 of GRE Physics Companion: Gravitational Fields, Superposition, and Continuous Mass Distributions, born on 2026-10-31.
Object id is 1552, canonical name is GREPhysicsCompanionGravitationalFieldsSuperpositionAndContinuousMassDistributions.
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Classification:
Physics Classification: 45.20.Dd (Newtonian mechanics)
 45.50.-j (Dynamics and kinematics of a particle and a system of particles)

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