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FRQ: Gravitational Field Strength — The Theron Survey Probe

Assessments aligned to 2026 AP Physics 1 standards

Translation Between Representations (TBR)  |  CHALLENGE  |  12 points  |  30 min

▤ Scenario

A survey probe is mapping the gravitational field of Theron, a newly catalogued airless dwarf planet lying far from its star and from every other massive body. Theron is modelled as a non-rotating, uniform sphere of mass M_T and radius R_T.

Working in Theron’s rest frame, which may be treated as inertial here, the probe uses its thrusters to hold station at a series of fixed distances r from Theron’s centre, as shown in Figure 1. At each station it releases an untethered 2.0 kg calibration mass from rest and tracks that mass optically, obtaining the magnitude of its acceleration from the recorded motion. Repeated runs at the same station agree to within about 2%. At the lowest station, at a height above Theron’s surface small enough to be neglected, the tracked acceleration is 1.60 m/s²; take this as the surface value g_T.

Figure 1 — The probe holds station a distance r from Theron’s centre (not to scale).

Every station is at r ≥ R_T. All motion is one-dimensional along the radial line joining the calibration mass to Theron’s centre; take the direction pointing away from Theron’s centre as positive. The probe’s thruster exhaust does not reach the calibration mass, and the gravitational forces exerted on that mass by the probe, by its instruments and by every body other than Theron are negligible.

The axes shown in Figure 2 are used in part C.

Figure 2 — Axes for part C: a_g against the distance r from Theron’s centre.

✎ Free Response Questions

Part A

Consider the 2.0 kg calibration mass immediately after it is released from rest at a station a distance r from Theron’s centre.

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

Figure 3 — The dot representing the calibration mass.

Part B

Starting with Newton’s second law, derive an expression for the magnitude a_g of the acceleration of a calibration mass of mass m immediately after it is released from rest at a distance r from Theron’s centre. Express your final answer in terms of m, M_T, r, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

The 2.0 kg calibration mass is replaced by a 50.0 kg calibration mass, released from rest at the same station. Using your expression, predict a_g for the 50.0 kg mass. Justify your response.

Part C

On the axes shown in Figure 2, sketch a graph of a_g as a function of the distance r from Theron’s centre, from r = R_T to r = 4R_T. Then use your expression from part B, together with the surface value g_T, to mark and label the values of a_g at r = 2R_T and at r = 4R_T.

Part D

At one station, the probe’s radar altimeter measures the height of that station above Theron’s surface as h = 1.50R_T, with an uncertainty negligible beside the 2% spread of the tracking runs, and the tracking data give a_g = 0.254 m/s² there.

Using your expression from part B together with the surface value g_T, predict a_g at that station. Then predict a_g at that same station if the field instead fell off as the inverse of the distance from Theron’s centre, with the same surface value g_T.

Indicate whether the reported measurement supports the dependence given by your expression from part B, the inverse dependence, or neither.

______ The dependence from part B

______ The inverse dependence

______ Neither

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 — Model Answer

Only one force is exerted on the calibration mass immediately after release: the gravitational force exerted on it by Theron. Theron is airless, so there is no drag. Once released, the untethered mass touches nothing, and the probe’s exhaust does not reach it, so no contact force is exerted on it. The gravitational forces exerted on it by the probe, its instruments and every other body are negligible.

Figure 4 — The one force on the mass immediately after release; the arrow shows direction only.

By Newton’s law of universal gravitation the force is attractive and is exerted along the line joining the two centres of mass, so it points toward Theron’s centre. Because it is the only force exerted on the mass, it is also the net force. It is still labelled as the gravitational force, not as F_net alone: a free-body diagram labels each force as the force it is, not as the sum of the forces. The mass is momentarily at rest, but a zero velocity does not mean a zero force, and nothing pushes the mass once it has left the probe. At a larger r the force would be smaller, but the diagram would still carry exactly one arrow.

Scoring (2 points):

▸ A1 — 1 point: For drawing exactly one force arrow, starting on the dot and directed toward Theron’s centre.

Scoring Note: Arrows of any nonzero length can earn A1. A1 does not depend on any other part.

▸ A2 — 1 point: For labelling an arrow directed toward Theron’s centre as the gravitational force.

Scoring Note: Examples of appropriate labels include F_g, F_G, F_grav, W, mg, “gravitational force”, “weight”, or a label naming Theron, such as F_Theron on mass; m·g_T and m·a_g, which like mg are the mass times the gravitational field strength, are also appropriate. The labels g, g_T, a_g and G are not. A label that names only the net force, or the mass times an unspecified acceleration (F_net, ma), is not appropriate, but one that also names the gravitational force, such as F_g = F_net, is. The response is not required to name the object exerting the force. A2 does not depend on any other part.

⚠︎ Common error (partial credit): Draws a correctly labelled gravitational arrow toward Theron’s centre and a second arrow, labelled “motion”, “inertia” or “force of release”, or drawn as a push from the probe — earns 1 of 2 points, A2. Once released, the untethered mass is touched by nothing, and neither motion nor inertia is a force.

⚠︎ Common error (partial credit): Draws exactly one arrow toward Theron’s centre but labels it g, g_T or a_g, or only F_net or ma — earns 1 of 2 points, A1. A field strength and an acceleration are not forces, and F_net and ma name what the forces on the mass add up to, not a force; the gravitational force is the mass times the field strength, which F_g or mg names.

✗ Common error (no credit): Draws a single arrow directed away from Theron, labelled as the force the calibration mass exerts on Theron. That force is the other member of the third-law pair: it is exerted on Theron and belongs on Theron’s diagram. A1 needs an arrow toward Theron’s centre, and there is none for A2 to score.

Part B — Model Answer

Let m be the mass of the calibration mass. The gravitational force exerted by Theron is the only force on it (part A), so that force is the net force. Newton’s second law, with Newton’s law of universal gravitation for the force and r measured from Theron’s centre, gives

F_net = m·a_g
G·M_T·m/r² = m·a_g
a_g = G·M_T/r²

The calibration mass m cancels, so a_g depends only on M_T and r, never on the object placed there. The quantity G·M_T/r² is Theron’s gravitational field strength at that distance, a property of Theron and of the location.

For the 50.0 kg mass released at the same station, the expression predicts the same acceleration, G·M_T/r², as for the 2.0 kg mass. Theron exerts 25 times as much gravitational force on the heavier mass, but that mass also has 25 times the inertia, and the acceleration is the ratio of the two. The cancellation is not an algebraic coincidence: the m in the law of universal gravitation is the object’s gravitational mass, which sets how strongly it takes part in the interaction, and the m in Newton’s second law is its inertial mass, which sets how strongly its motion resists change. The two have been verified experimentally to be equivalent, which is what lets them cancel.

Scoring (4 points):

▸ B1 — 1 point: For a multistep derivation that includes Newton’s second law.

Scoring Note: Newton’s second law may be applied as an equation, such as G·M_T·m/r² = m·a_g or a_g = F_g/m, or in words. B1 is met by a derivation that includes it even if the steps that follow are incorrect. A final expression for a_g preceded only by words, or only by unapplied equations from the reference information, is not a multistep derivation. B1 carries no consistency credit. B1 to B4 are scored independently of the response in part A.

▸ B2 — 1 point: For indicating that G·M_T·m/r² is the gravitational force exerted on the calibration mass, or that G·M_T/r² is the gravitational field strength at distance r.

Scoring Note: B2 carries no consistency credit.

▸ B3 — 1 point: For a correct expression for a_g, consistent with the indicated expression for the gravitational force or field strength: where B2 was not earned, an expression that follows correctly by Newton’s second law from the response’s own gravitational-force or field-strength expression in M_T and r, such as a_g = G·M_T/r from G·M_T·m/r or from a field of G·M_T/r, also earns this point.

Scoring Note: A correct, isolated, final expression for a_g earns points B2 and B3; a correct final expression preceded only by words, or only by unapplied equations from the reference information, is treated as isolated.

▸ B4 — 1 point: For predicting that the 50.0 kg mass has the same acceleration as the 2.0 kg mass, justified by the cancellation of m from the expression or by the gravitational force and the inertia both being proportional to m; or for the prediction that follows correctly from the response’s own expression for a_g, justified by how that expression depends on m.

Scoring Note: A bare statement that all objects fall at the same rate, or that a_g is unchanged, with no reference to the expression or to the force and the inertia scaling together, is not sufficient for B4.

⚠︎ Common error (partial credit): Derives a_g = G·M_T/r² correctly but justifies the prediction only by asserting that “all objects fall at the same rate” — earns 3 of 4 points, B1, B2 and B3.

⚠︎ Common error (partial credit): Writes G·M_T·m/r² = m·a_g, then reports a_g = G·M_T·m/r² without cancelling m, and predicts a 25 times larger acceleration for the 50.0 kg mass because that expression is proportional to m — earns 3 of 4 points, B1, B2 and B4. The prediction follows from the response’s own expression, but the expression is wrong: dividing both sides by m removes it from the right-hand side too.

✗ Common error (no credit): Writes only the near-surface relation F_g = m·g_T and asserts that a_g = 1.60 m/s² at every station and for the 50.0 kg mass “because g is the same for everything”. Newton’s second law is never used, no gravitational force or field in terms of M_T and r is written, and the prediction rests on a bare assertion, so no point is earned. The field weakens with distance from Theron; it is nearly uniform only over heights small beside R_T.

Part C — Model Answer

The expression from part B contains the product G·M_T, which is not given. Eliminate it with the surface measurement: at r = R_T the same expression gives g_T = G·M_T/R_T², so G·M_T = g_T·R_T² and

a_g = g_T·(R_T/r)²

Doubling r divides a_g by four, and quadrupling it divides a_g by sixteen:

at r = 2R_T:  a_g = (1.60 m/s²)/4 = 0.400 m/s²
at r = 4R_T:  a_g = (1.60 m/s²)/16 = 0.100 m/s²

Because a_g is proportional to 1/r², the curve leaves 1.60 m/s² at r = R_T steeply and then flattens: it decreases and is concave up across the whole range. Each doubling of the distance from Theron’s centre divides the field by four. Continued beyond the window, the curve would approach the axis without ever reaching it, since a gravitational field has no outer edge, though the sketch is not asked to show that.

Figure 5 — a_g against r, with the surface value and the two predicted values marked.

Scoring (3 points):

▸ C1 — 1 point: For a curve of a_g against r that decreases over the whole range and is concave up, steepest near r = R_T and flattening as r increases, and is not a straight line.

Scoring Note: C1 is scored against the physics and does not depend on the response in part B.

▸ C2 — 1 point: For a curve that begins at a_g = 1.60 m/s² at r = R_T.

Scoring Note: A curve that starts within 0.05 m/s² (half a grid division) of 1.60 m/s² meets C2. C2 is scored against the physics and does not depend on the response in part B.

▸ C3 — 1 point: For marking a_g = 0.400 m/s² at r = 2R_T and a_g = 0.100 m/s² at r = 4R_T, or the values at those distances that follow correctly from the response’s own part B expression once its value at r = R_T is set equal to g_T.

Scoring Note: For C3, each value may be shown by a mark within 0.05 m/s² (half a grid division) of it, or by a mark at that distance labelled with the value; values of 0.40 m/s² and 0.10 m/s² are accepted.

⚠︎ Common error (partial credit): With a correct inverse-square expression in part B, draws a decreasing, concave-up curve from 1.60 m/s² at r = R_T but halves a_g each time r doubles, marking 0.800 m/s² at r = 2R_T and 0.400 m/s² at r = 4R_T — the inverse-versus-inverse-square confusion — earns 2 of 3 points, C1 and C2.

⚠︎ Common error (partial credit): Draws a straight line from 1.60 m/s² at r = R_T down to zero at r = 4R_T, as if Theron’s field ran out at some distance, marking 1.07 m/s² at r = 2R_T and zero at r = 4R_T — earns 1 of 3 points, C2. A gravitational field has no outer edge, and an inverse-square fall-off is not linear.

Part D — Model Answer

The altimeter reports a height above the surface, but the r in Newton’s law of universal gravitation is the distance between the two centres of mass, so the height must be converted first:

r = R_T + h = R_T + 1.50R_T = 2.50R_T

With G·M_T eliminated by the surface value, as in part C, the dependence from part B and an inverse fall-off with the same surface value predict

inverse square:  a_g = g_T·(R_T/r)² = (1.60 m/s²)(1/2.50)² = 0.256 m/s²
inverse:  a_g = g_T·(R_T/r) = (1.60 m/s²)/2.50 = 0.640 m/s²

The dependence from part B

The reported 0.254 m/s² differs from the inverse-square prediction by about 0.8%, inside the 2% spread of the repeated runs, while the inverse prediction is two and a half times the reported value. The measurement therefore supports the dependence from part B, the inverse square, and rules out the inverse one.

Figure 6 — The two candidate dependences; the reported value lies on the inverse-square curve.

Both curves start from g_T at r = R_T. Only a reading at a second, independently measured distance can separate two dependences that share a surface value.

Scoring (3 points):

▸ D1 — 1 point: For converting the altimeter height into a distance from Theron’s centre, r = R_T + h = 2.50R_T.

Scoring Note: D1 is met by r = 2.50R_T, or by 1 + 1.50 = 2.50, used within either prediction. D1 does not depend on any other part.

▸ D2 — 1 point: For predicting a_g from the part B expression with its value at r = R_T set equal to g_T — 0.256 m/s² at r = 2.50R_T — or the value that follows correctly at the response’s own r from its own part B expression, so anchored.

Scoring Note: A prediction of 0.26 m/s² is accepted. D2 does not depend on the response in part C.

▸ D3 — 1 point: For predicting the inverse value g_T·(R_T/r) — 0.640 m/s² at r = 2.50R_T, or the value at the response’s own r — and indicating the dependence the measurement supports, justified by comparing each prediction with the reported 0.254 m/s². The indication is “The dependence from part B” for the values above (0.254 m/s² is within 1% of 0.256 m/s², far from 0.640 m/s²), and otherwise the one the response’s own two predictions support.

Scoring Note: A prediction of 0.64 m/s² is accepted, and the 2% spread need not be cited. A prediction is supported when the measurement lies within a few percent of it, and not otherwise; “Neither” follows when neither prediction is supported, as for 0.711 m/s² and 1.07 m/s² against 0.254 m/s².

⚠︎ Common error (partial credit): Uses the altimeter height as the distance from Theron’s centre, r = 1.50R_T, then predicts correctly at that distance, 0.711 m/s² from the part B expression and 1.07 m/s² for the inverse, and indicates “Neither” because both lie far from 0.254 m/s² — earns 2 of 3 points, D2 and D3. The r in the law is measured between the centres of mass, so r = R_T + h.

⚠︎ Common error (partial credit): Finds r = 2.50R_T and the part B prediction of 0.256 m/s² but never evaluates the inverse dependence, justifying “The dependence from part B” only by the values being close — earns 2 of 3 points, D1 and D2. Agreement with one model does not rule out the other.

✗ Common error (no credit): With a correct inverse-square expression in part B, ignores it: takes the altimeter height as the radius and divides the surface value by the ratio of the distances, not its square, to get 1.07 m/s², then indicates “Neither” and reports the tracking data as faulty. The radius is wrong, 1.07 m/s² does not follow from the part B expression, and only one prediction is compared with the data, so no point is earned; the data, not a model, are rejected.

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