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FRQ: Gravitational Forces Between Objects — The Deep-Space Alignment Bench

Assessments aligned to 2026 AP Physics 1 standards

Mathematical Routines (MR)  |  CHALLENGE  |  10 points  |  25 min

▤ Scenario

A free-flying research module coasts through deep space with its engines shut down, far from any star or planet. Inside the module, three uniform spheres are clamped along a straight rail whose scale is marked in metres, so that their centres of mass lie on a single line. Sphere X, of mass 800 kg, is clamped at x = 0. Sphere Y, of mass 5.00 kg, is clamped at x = 2.00 m. Sphere Z, of mass 1600 kg, is clamped at x = 6.00 m, as shown in Figure 1.

Figure 1 — The three spheres clamped on the rail; sphere sizes are not drawn to scale.

The clamps hold each sphere at rest. The gravitational forces exerted on the spheres by the rail, by the clamps and by the module structure are negligible.

All three centres of mass lie on the x-axis, and every force considered in this question is exerted along that axis. Treat the module as an inertial reference frame, with the +x direction, from X toward Z, as positive. Use G = 6.67 × 10⁻¹¹ N·m²/kg².

✎ Free Response Questions

Part A

i.    On the following dot that represents Sphere Y, draw and label the forces (not components) that are exerted on Sphere Y. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.

Figure 2 — The dot representing Sphere Y.

ii.   Calculate the magnitude of the net gravitational force exerted on Sphere Y, and state its direction.

iii.  Spheres X and Z stay clamped where they are. Suppose Sphere Y were unclamped and placed between them. Let L be the distance between the centres of Spheres X and Z, and let m_X, m_Y and m_Z be the masses of Spheres X, Y and Z. Derive an expression for the distance d between the centres of Spheres X and Y at which the net gravitational force exerted on Sphere Y would be zero. Express your final answer in terms of m_X, m_Y, m_Z, L, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part B

All three spheres are again clamped in the positions given in the Scenario. A thick lead slab, whose mass is comparable to that of Sphere X, is inserted into the gap between Sphere X and Sphere Y, touching neither sphere, and is held at rest there. The slab’s centre of mass lies on the x-axis.

Indicate whether the magnitude of the net gravitational force exerted on Sphere Y increases, decreases, or remains constant when the slab is inserted.

______ Increases

______ Decreases

______ Remains constant

Justify your response.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠︎ common error, partial credit  ✗ common error, no credit

Points are independent and there are no deductions. ▸ criteria follow College Board’s published scoring standard; ⚠︎ and ✗ lines are TheScienceCube teaching commentary, not College Board rubric.

Part A (i) — Model Answer

Three forces are exerted on Sphere Y. Spheres X and Z each exert a gravitational force on it, and gravitation is attractive, so each pulls Y toward itself along the line joining the centres: X toward −x and Z toward +x. The clamp holding Y on the rail exerts a contact force on it along the rail. No planet is nearby, so there is no weight toward a floor, and the gravitational forces exerted by the rail, the clamps and the module structure are negligible, so no arrow is drawn for them.

Y is at rest in an inertial frame, so the net force on it is zero, and the clamp’s force must balance the two gravitational forces together. Its direction depends on which pull is larger. With r_X = 2.00 m and r_Z = 4.00 m the distances from Y’s centre to X’s and to Z’s, the ratio of the two pulls can be taken before any value is worked out:

F_Z / F_X = (m_Z / m_X)(r_X / r_Z)² = (2)(2.00 m / 4.00 m)² = 1/2

Z is twice as massive as X but also twice as far from Y, and the separation enters squared. Doubling the separation divides the force by four, which outweighs doubling the mass. X’s pull is therefore twice Z’s, the two gravitational forces leave a net pull toward X, and the clamp pulls Y toward +x with a force equal in size to Z’s pull.

Figure 3 — The forces on Sphere Y, to one scale: Z’s pull and the clamp’s together balance X’s.

Scoring (2 points):

▸ A1 — 1 point: For drawing two gravitational forces as labelled arrows starting on the dot, one directed toward −x (toward Sphere X) and one directed toward +x (toward Sphere Z).

▸ A2 — 1 point: For drawing a contact force directed toward +x as a distinct labelled arrow starting on the dot and not drawn on top of any other arrow, with no force arrow starting on the dot other than this one and the two gravitational forces of A1.

Scoring Note: Any label that identifies each force is accepted. For the gravitational forces: F_g, F_G, gravitational force, or a label naming the exerting sphere such as F_X or F_X on Y; the same label may be used on both. For the contact force: F_N, F_C, F_clamp, F_rail, f, f_s or contact force — any label that identifies a contact force. The response is not required to name the object exerting each force, so a misnamed exerting sphere does not by itself lose A1, and the contact force may be attributed to the clamp or to the rail. Arrow lengths are not scored. An arrow for the combined gravitational force, for a weight toward the floor of the module, or for a force from the rail perpendicular to the x-axis counts as another force arrow. A1 and A2 do not depend on any other part.

⚠︎ Common error (partial credit): Draws only the two gravitational forces, with no contact force — earns 1 of 2 points, A1. Y is at rest and the two gravitational pulls do not balance each other, so the clamp must be pulling on Y. (A response that does not square the separations reaches the same diagram: in that model 800/2 = 1600/4, and the pulls would balance.)

⚠︎ Common error (partial credit): Draws the two gravitational forces correctly but directs the contact force toward −x, reasoning that Z’s pull is the larger because Z has twice X’s mass — earns 1 of 2 points, A1. Z is also twice as far away, so its pull is half X’s.

✗ Common error (no credit): Draws both gravitational forces toward +x, toward Z, the more massive sphere, with the contact force opposing them toward −x. Gravitation is attractive, so X pulls Y toward X: with no gravitational force toward −x, A1 is not earned, and with the contact force toward −x, A2 is not earned.

Part A (ii) — Model Answer

Each outer sphere attracts Y separately, with a force set by that pair’s masses and centre-to-centre separation: 2.00 m for X and Y, and 6.00 m − 2.00 m = 4.00 m for Z and Y. From Newton’s law of universal gravitation, F_g = G·m_1·m_2/r²,

F_X = (6.67 × 10⁻¹¹ N·m²/kg²)(800 kg)(5.00 kg)/(2.00 m)² = 6.670 × 10⁻⁸ N

F_Z = (6.67 × 10⁻¹¹ N·m²/kg²)(1600 kg)(5.00 kg)/(4.00 m)² = 3.335 × 10⁻⁸ N

F_X points toward −x, toward X, and F_Z toward +x, toward Z. With +x positive, the two gravitational forces are added with their signs, and the result is rounded only at the end:

ΣF_g = (−6.670 × 10⁻⁸ N) + (+3.335 × 10⁻⁸ N) = −3.335 × 10⁻⁸ N

The net gravitational force exerted on Sphere Y has magnitude 3.34 × 10⁻⁸ N and is directed toward −x, toward Sphere X. Each pair of objects interacts separately, with its own centre-to-centre separation, and the forces then add as vectors. Because X’s pull is exactly twice Z’s, the net gravitational force equals Z’s pull in size, which is why the clamp’s force in part A (i) matches Z’s.

Figure 4 — The two gravitational forces on Sphere Y added tip to tail along the x-axis.

Scoring (2 points):

▸ A3 — 1 point: For both pairwise interactions, each from Newton’s law of universal gravitation with that pair’s own centre-to-centre separation (2.00 m and 4.00 m): as forces, 6.67 × 10⁻⁸ N and 3.34 × 10⁻⁸ N; as gravitational fields at Y’s position, 1.33 × 10⁻⁸ N/kg and 6.67 × 10⁻⁹ N/kg; or as the two terms of a single expression.

▸ A4 — 1 point: For a net gravitational force of 3.34 × 10⁻⁸ N directed toward −x (toward Sphere X), obtained by combining the two forces as opposing vectors, or by combining the two fields as opposing vectors and multiplying by m_Y; or the magnitude and direction that follow correctly, by either route, from the two pairwise quantities the response calculated (where that magnitude is zero, no direction is required).

Scoring Note: A3 and A4 accept values that round to the two-figure values of the key — 6.7 and 3.3 × 10⁻⁸ N for the forces, or 1.3 × 10⁻⁸ and 6.7 × 10⁻⁹ N/kg for the fields; for the net gravitational force, 3.33 × 10⁻⁸ N (Z’s force rounded before subtracting) and 3.34 × 10⁻⁸ N are both accepted. The direction may be given as −x, as toward X, or as the sign of a signed result. A3 and A4 do not depend on the response in part A (i).

⚠︎ Common error (partial credit): Uses 6.00 m for Z’s force, measuring from Sphere X rather than from Sphere Y — earns 1 of 2 points, A4. Z’s force then comes out 2.25 times too small, 1.48 × 10⁻⁸ N, and the net gravitational force 5.19 × 10⁻⁸ N toward −x, which combines the response’s own values correctly.

⚠︎ Common error (partial credit): Adds the two magnitudes to obtain 1.00 × 10⁻⁷ N — earns 1 of 2 points, A3. Both magnitudes are right, but the forces point in opposite directions and must be combined with their signs.

⚠︎ Common error (partial credit): Does not square the separations, using F = G·m_1·m_2/r, and obtains 1.33 × 10⁻⁷ N for each force and a net gravitational force of zero — earns 1 of 2 points, A4. The force falls off with the square of the separation; only in a first-power model would these two pulls balance.

✗ Common error (no credit): Adds the two outer masses and applies the law once, with Y’s separation from X, as F = G(m_X + m_Z)m_Y/(2.00 m)² = 2.00 × 10⁻⁷ N. No pairwise interaction is calculated, so neither A3 nor A4 is earned: each pair of objects interacts separately, with its own separation.

Part A (iii) — Model Answer

Let the centre of Sphere Y be a distance d from the centre of X, so that it is a distance L − d from the centre of Z. Between the spheres the two gravitational forces on Y point in opposite directions, so the net gravitational force is zero when their magnitudes are equal:

G·m_X·m_Y/d² = G·m_Z·m_Y/(L − d)²

G and m_Y appear on both sides and cancel, leaving a relation between the geometry and the outer masses alone:

(L − d)²/d² = m_Z/m_X

Between the spheres both d and L − d are positive, so the positive square root applies:

(L − d)/d = √(m_Z/m_X)

d = L/(1 + √(m_Z/m_X))

The middle sphere’s mass cancels, so the zero-force point is fixed by X and Z alone. It is the point where X and Z together produce no gravitational field, no force per unit mass, so an object of any mass placed there feels no net gravitational force from them.

Figure 5 — The zero-force point lies 2.49 m from X, closer to the less massive sphere.

For the spheres in the Scenario, L = 6.00 m and m_Z/m_X = 1600/800 = 2.00, so d = (6.00 m)/(1 + √2) = 2.49 m. The point lies at x = 2.49 m, closer to the less massive sphere and 0.49 m beyond Y’s clamp, so Y sits on X’s side of it, where X’s pull wins, as part A (ii) found.

Scoring (3 points):

▸ A5 — 1 point: For a multistep derivation that includes Newton’s law of universal gravitation for the force exerted on Sphere Y by each outer sphere, or for the gravitational field each produces at Y’s position, with the net set to zero or the two magnitudes set equal. Each force or field must be inverse-square in its separation; using a wrong separation, such as L in place of L − d, does not by itself lose this point. This point is earned for the starting equation alone, even if the steps that follow are incorrect; a final expression for d on its own is not a multistep derivation.

▸ A6 — 1 point: For a correct relation between d, L and the outer masses that contains neither G nor m_Y — m_X/d² = m_Z/(L − d)², (L − d)/d = √(m_Z/m_X), or any equivalent relation of that kind — or a correct relation in which the square root has been taken and m_Y remains only as a factor that cancels identically. A relation that follows from an incorrect starting equation, or that contradicts the response’s own equations, does not earn this point.

▸ A7 — 1 point: For a correct expression for d that places the point between the spheres, d = L/(1 + √(m_Z/m_X)), or any algebraically equivalent expression, such as d = L√m_X/(√m_X + √m_Z); an expression in which m_Y appears but cancels identically is equivalent. Where A5 was earned but A6 was not, an expression that follows correctly, with no further error, from the response’s A5 equation also earns this point.

Scoring Note: A correct, isolated, final expression for d earns points A6 and A7; a correct final expression preceded only by an unapplied equation from the reference information, or only by words, is treated as isolated. The value 2.49 m is not required. A5, A6 and A7 do not depend on the responses in parts A (i) and A (ii).

⚠︎ Common error (partial credit): Measures Z’s separation from X instead of from Y, writing G·m_X·m_Y/d² = G·m_Z·m_Y/L², and reaches d = L√(m_X/m_Z), 4.24 m here — earns 2 of 3 points, A5 and A7. The relation is not correct, so A6 is not earned, but the expression follows correctly from the response’s own equation.

⚠︎ Common error (partial credit): Sets the two magnitudes equal correctly and reaches (L − d)²/d² = m_Z/m_X, then takes the negative square root and reports d = L/(1 − √(m_Z/m_X)), which is −14.5 m — earns 2 of 3 points, A5 and A6. Outside the pair the two forces point the same way — at x = −14.5 m both toward +x — so they cannot cancel there.

⚠︎ Common error (partial credit): Sets the two magnitudes equal correctly, then cancels the outer masses along with m_Y and reports the midpoint, d = L/2 — earns 1 of 3 points, A5. Equal separations balance the two forces only when the outer masses are equal.

✗ Common error (no credit): Balances the masses instead of the forces, m_X·d = m_Z·(L − d), reaching d = m_Z·L/(m_X + m_Z), 4.00 m here, the centre of mass of X and Z. No gravitational force is written, so A5 is not earned, the relation is not one of A6’s, and A7’s carry-through needs an equation that earns A5.

✗ Common error (no credit): Does not square the separations, G·m_X·m_Y/d = G·m_Z·m_Y/(L − d), reaching d = m_X·L/(m_X + m_Z), 2.00 m, Y’s own clamped position. No inverse-square force or field is written, so A5 is not earned, the relation is not one of A6’s, and A7’s carry-through needs an equation that earns A5.

Part B — Model Answer

The magnitude of the net gravitational force exerted on Sphere Y increases.

The slab does not change the force exerted on Y by X. The gravitational force between two objects depends only on their masses and the separation of their centres, and matter placed between them does not block, absorb or weaken it. X still pulls Y with 6.67 × 10⁻⁸ N toward −x, and Z still pulls it with 3.34 × 10⁻⁸ N toward +x.

What the slab adds is a gravitational force of its own. It has mass, so it attracts Y toward itself, and because it lies between X and Y, that pull is directed toward −x, the same direction as X’s pull and as the net gravitational force already exerted on Y. Added to the sum of Figure 4, it carries the total further toward −x.

Figure 6 — With the slab in place its pull adds toward −x, and the net gravitational force grows.

The net gravitational force on Y therefore grows by the size of the slab’s pull and still points toward X, and the clamp must now pull Y toward +x harder to hold it at rest. Matter between two objects cannot block gravitation, but it exerts a gravitational force of its own. The two tempting answers each miss one half of this: “Decreases” treats the slab as a gravity shield, which does not exist, and “Remains constant” gets the shielding right but ignores the pull of the slab’s own mass.

Scoring (3 points):

▸ B1 — 1 point: For indicating “Increases”.

▸ B2 — 1 point: For indicating that the gravitational force exerted on Sphere Y by Sphere X is unchanged, because gravitation is not blocked or weakened by matter between two objects, or because that force depends only on the two masses and their centre-to-centre separation, which the slab does not change.

▸ B3 — 1 point: For indicating that the slab exerts its own gravitational force on Sphere Y, directed toward the slab — that is, toward −x, the direction of X’s pull and of the net gravitational force already exerted on Y.

Scoring Note: For B3, “the slab attracts Y”, “the slab pulls Y toward itself” or any of the directions listed gives the direction; a statement only that the slab “has its own gravity” or “also exerts a gravitational force on Y” does not. The size of the slab’s force is not required. B1, B2 and B3 are scored independently of the responses in part A and of one another.

⚠︎ Common error (partial credit): Indicates “Increases”, stating that X’s force on Y is unchanged because gravitation cannot be blocked, and that the slab also exerts a gravitational force on Y, but not which way it acts — earns 2 of 3 points, B1 and B2. The slab’s pull raises the net gravitational force only because it points the same way as the net gravitational force already does.

⚠︎ Common error (partial credit): Indicates “Remains constant”, reasoning that lead cannot block gravity, so nothing about the forces on Y changes — earns 1 of 3 points, B2. The first half is right, but the slab is itself an object with mass, and it pulls on Y.

✗ Common error (no credit): Indicates “Decreases”, reasoning that the lead slab blocks part of X’s gravitational pull on Y. Gravitational shielding does not exist, so B2 is not earned; the slab’s own pull on Y is never identified, so B3 is not earned; and B1 is not earned.