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FRQ: Newton's Law of Universal Gravitation — The Deep-Space Alignment Bench

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Mathematical Routines (MR)  |  CHALLENGE  |  10 points

▤ 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.

The three spheres clamped on the rail inside the drifting module. 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. Take the module as the reference frame, with the +x direction, from X toward Z, as positive. Use G = 6.67 × 10⁻¹¹ N·m²/kg².

✎ Free Response Questions

(a) Represent Sphere Y as a single labelled dot. Draw and label every gravitational force exerted on Sphere Y, using arrow lengths that correctly represent the relative magnitudes of those forces.

(b) Calculate the magnitude of the net gravitational force exerted on Sphere Y at x = 2.00 m, and state its direction.

(c) Spheres X and Z stay clamped where they are. Suppose Sphere Y were unclamped and moved to a different position between them. Derive an expression for the distance d from Sphere X at which the net gravitational force exerted on Sphere Y would be zero, in terms of the separation L between X and Z and the masses m_X and m_Z. Calculate that distance, and state whether it would be different if Sphere Y were replaced by a sphere of a different mass. Justify your statement.

(d) With all three spheres back in the positions given in the Scenario, a thick lead slab is inserted into the gap between Sphere X and Sphere Y, touching neither sphere and held at rest there. The slab's centre of mass lies on the x-axis.

Student M claims: “The force that X exerts on Y is now smaller, because the lead slab blocks some of the gravity travelling from X to Y.”

Student N replies: “Lead cannot block gravity, so nothing about the forces on Y changes at all.”

Evaluate both claims. Identify what is correct and what is incorrect in each, and state what happens to the gravitational force exerted on Y by X and to the net gravitational force exerted on Y, including the direction of that net force.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Gravitation is attractive, so Sphere Y is pulled toward every other mass. Only X and Z exert appreciable gravitational forces on it, so exactly two arrows are drawn on the dot representing Y: one in the −x direction, toward X, labelled F_(on Y by X); and one in the +x direction, toward Z, labelled F_(on Y by Z).

The relative lengths follow from the ratio of the two forces, which can be taken before any values are worked out:

F_(on Y by Z) / F_(on Y by X) = (m_Z / m_X) × (r_X / r_Z)²
= (2) × (2.00 / 4.00)² = 1/2

Sphere Z is twice as massive as X, but it is also twice as far away, and the separation enters as its square — so the factor of 2 in mass is beaten by the factor of 4 in r², and X's pull is the stronger of the two.

The two gravitational forces exerted on Sphere Y, drawn from a single dot. Arrow lengths are to scale.

Scoring (2 points):

▸ 1 point: Exactly two force arrows are drawn on a single dot labelled Y — one directed toward X and one directed toward Z — each labelled by the object that exerts it.

▸ 1 point: The arrow toward X is drawn clearly longer than the arrow toward Z, in a ratio of approximately 2 : 1.

⚠︎ Common error (partial credit): Drawing the arrow toward Z longer because Z is the more massive sphere — earns 1 of 2 points; the directions are right, but the inverse-square dependence on separation has not been weighed against the mass.

⚠︎ Common error (partial credit): Adding a third arrow for the clamp that holds Y, or for the module structure — earns 1 of 2 points; the first criterion asks for exactly two arrows and only gravitational forces are wanted, but a correct 2 : 1 ratio still earns the second point.

✗ Common error (no credit): Drawing both arrows in the same direction, toward the more massive sphere Z — earns 0 of 2 points; gravitation is attractive, so the force exerted by X on Y is directed toward X, and with no arrow toward X the relative lengths cannot be right either.

Part (b) — Model Answer

Treat the two interactions separately, then add the forces as vectors along the x-axis. Sphere Y is 2.00 m from X and 6.00 − 2.00 = 4.00 m from Z, measured centre to centre:

F_(on Y by X) = G·m_X·m_Y / r², with r = 2.00 m
= (6.67 × 10⁻¹¹)(800)(5.00) / (2.00)² = 6.670 × 10⁻⁸ N, in the −x direction
F_(on Y by Z) = G·m_Z·m_Y / r², with r = 4.00 m
= (6.67 × 10⁻¹¹)(1600)(5.00) / (4.00)² = 3.335 × 10⁻⁸ N, in the +x direction

The two forces are exerted in opposite directions, so they are added with their signs, and the answer is rounded only at the end:

ΣF = (−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 in the −x direction — that is, toward Sphere X. Because X's pull is exactly twice Z's, the net force on Y is equal in magnitude to the pull of Z alone.

The two pairwise forces exerted on Sphere Y and their vector sum, drawn to a common scale.

Scoring (2 points):

▸ 1 point: Sets up the two pairwise forces separately from F_g = G·m_1·m_2 / r², one for the interaction with X and one for the interaction with Z. This point is awarded for the correct starting relationship alone, before any values are substituted.

▸ 1 point: Uses the correct centre-to-centre separations of 2.00 m and 4.00 m, obtains 6.67 × 10⁻⁸ N and 3.34 × 10⁻⁸ N, and combines them as opposing vectors to give 3.34 × 10⁻⁸ N — accept 3.33 × 10⁻⁸ N from rounding the second force first — directed in the −x direction, toward X.

⚠︎ Common error (partial credit): Using r = 6.00 m for the force exerted by Z, measuring from Sphere X rather than from Sphere Y — earns 1 of 2 points; that force then comes out 2.25 times too small at 1.48 × 10⁻⁸ N, and the net force becomes 5.19 × 10⁻⁸ N.

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

✗ Common error (no credit): Adding the two outer masses and applying the law once, as F = G(m_X + m_Z)·m_Y / r² — earns 0 of 2 points; each pair of objects interacts separately, with its own centre-to-centre separation, and the two resulting forces are then added as vectors.

Part (c) — Model Answer

Let Sphere Y sit a distance d from X, so that it is a distance (L − d) from Z, where L is the separation between X and Z. Write its mass as m. The two forces on it are exerted in opposite directions, so the net force is zero when their magnitudes are equal:

G·m_X·m / d² = G·m_Z·m / (L − d)²

G and m appear on both sides and cancel, which leaves a relation between the geometry and the two outer masses alone. Taking the root between X and Z:

(L − d)² / d² = m_Z / m_X     ⇒     (L − d) / d = √(m_Z / m_X)
d = L / (1 + √(m_Z / m_X))

With L = 6.00 m and m_Z / m_X = 1600 / 800 = 2.00, this gives d = 6.00 / (1 + √2) = 6.00 / 2.414 = 2.49 m. The net gravitational force is zero at x = 2.49 m, which is 0.49 m further from X than Sphere Y's clamped position — and that agrees with part (b), since Y sits on the X side of that point.

The point of zero net gravitational force lies at d = 2.49 m, not at the midpoint of the rail.

The distance would not be different for a sphere of different mass. The mass m of the middle sphere is a factor in both pairwise forces and cancels when they are set equal, so it does not appear in the derived expression at all — d is fixed by L and by the ratio m_Z / m_X. The point of zero force is a property of the gravitational field the two outer spheres create at that location, which is the force per unit mass of whatever object is placed there, rather than a property of the object itself.

Scoring (3 points):

▸ 1 point: Sets the two opposing pairwise gravitational forces equal in magnitude, or equivalently sets the net force to zero, with each force written from F_g = G·m_1·m_2 / r². This point is awarded for the correct starting principle alone, and is earned even if the algebra that follows is incorrect.

▸ 1 point: Carries the symbolic pathway through correctly — separations written as d and (L − d), G and the middle mass m cancelled, and the square root taken — reaching d = L / (1 + √(m_Z / m_X)) or an equivalent form.

▸ 1 point: Calculates d = 2.49 m — accept 2.5 m, or any value consistent with the student's own derived expression — and states that this position is the same whatever the mass of the middle sphere, justified by the cancellation of m.

⚠︎ Common error (partial credit): Setting the two magnitudes equal correctly and then placing the point at the midpoint of the rail, x = 3.00 m — earns 1 of 3 points; equal separations balance the two forces only when the outer masses are equal.

⚠︎ Common error (partial credit): Taking the other root, (L − d) = −√(m_Z / m_X)·d, and reporting d = 6.00 / (1 − √2) = −14.5 m — earns 1 of 3 points; outside the two spheres both forces are exerted in the same direction, so they cannot cancel anywhere there.

⚠︎ Common error (partial credit): Reaching 2.49 m but stating that the position would move if a heavier sphere were put in Y's place — earns 2 of 3 points; the cancellation of m is the substance of this part.

✗ Common error (no credit): Balancing the masses instead of the forces, writing m_X·d = m_Z·(L − d) and obtaining d = 4.00 m — earns 0 of 3 points; that condition locates the centre of mass of X and Z, and it drops the inverse-square dependence on separation entirely.

Part (d) — Model Answer

Both students are partly right, and the disagreement turns on the difference between a single pairwise force and the net force.

Student M is wrong. There is no gravitational shielding: placing matter between two objects does not reduce the gravitational force they exert on each other, because that force depends only on the two masses and on the separation of their centres of mass, and the slab has changed none of them.

F_(on Y by X) = G·m_X·m_Y / r² = 6.67 × 10⁻⁸ N, unchanged

Student N is right about shielding and wrong about the conclusion drawn from it. “Lead cannot block gravity” is correct, but “nothing about the forces on Y changes at all” does not follow, because the slab is itself an object with mass. It attracts Y toward itself, and it lies between X and Y, so that attraction is exerted in the −x direction.

A third gravitational force now acts on Sphere Y; the arrow for the slab's force shows direction only.

That is a third pairwise force, one that did not act before, and the net force on Y is the vector sum of every force exerted on it:

ΣF_on Y = F_(on Y by X) + F_(on Y by Z) + F_(on Y by slab)

So the force exerted on Y by X is exactly unchanged at 6.67 × 10⁻⁸ N in the −x direction, while the net gravitational force on Y increases in magnitude and remains directed in the −x direction, because the slab's pull adds to X's rather than opposing it. Student M reaches “something changes” by a mechanism that does not exist; Student N applies the right principle and then forgets that the shield has mass.

Scoring (3 points):

▸ 1 point: States that Student M is incorrect and that the gravitational force exerted on Y by X is unchanged, because gravitational forces are not blocked or absorbed by intervening matter — the force depends only on the two masses and their centre-to-centre separation.

▸ 1 point: States that Student N's conclusion is incorrect because the lead slab itself has mass and therefore exerts its own gravitational force on Y — a third pairwise force that did not act before.

▸ 1 point: Separates the two quantities in the conclusion — the pairwise force F_(on Y by X) is unchanged while the net gravitational force on Y increases in magnitude — and gives the direction of that net force as −x, toward X, because the slab lies on the X side of Y.

⚠︎ Common error (partial credit): Agreeing with Student N in full — there is no shielding, therefore nothing changes — earns 1 of 3 points; this is the targeted error of treating a pairwise force and the net force as the same quantity, and it misses that the shield is itself a source of gravitation.

⚠︎ Common error (partial credit): Getting the first two criteria right but leaving the direction of the new net force unstated, or giving it as +x on the reasoning that the slab pushes Y away — earns 2 of 3 points; a net force is a vector, and the slab lies on the X side of Y, so Y is attracted toward −x.

✗ Common error (no credit): Agreeing with Student M, so that the force exerted on Y by X is reported as smaller — earns 0 of 3 points; gravitational shielding is not a physical phenomenon, and this response alters the one quantity that does not change while never identifying the slab's own mass as a source of gravitation