Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.

FRQ: Gravitational Field Strength — The Theron Survey Probe

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Translation Between Representations (TBR)  |  CHALLENGE  |  12 points

▤ 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. At each station it releases a 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.

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 survey probe holds station at a fixed distance r from Theron's centre, where it releases the 2.0 kg calibration mass from rest.

✎ Free Response Questions

(a) Draw and label a free-body diagram for the 2.0 kg calibration mass at the instant it is released at a station a distance r from Theron's centre. Represent the mass as a dot, and draw every force exerted on it as an arrow originating at the dot, labelled with the type of force and the object exerting it.

(b) Starting from Newton's second law and Newton's law of universal gravitation, derive an expression for the magnitude a_g of the calibration mass's acceleration in terms of G, M_T and r. Then state what your expression predicts for a_g if the 2.0 kg mass is replaced by a 50.0 kg mass released at the same station, and justify that prediction.

(c) Sketch a graph of a_g against r, with r on the horizontal axis running from r = R_T to r = 4R_T. Label the value of a_g at r = R_T. Then use your expression from part (b), together with the surface value g_T, to mark the values of a_g at r = 2R_T and at r = 4R_T.

(d) At one station the probe's radar altimeter measures the height of that station above Theron's surface as h = 1.50R_T, 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 what a_g would be at that same station if the field instead fell off as the inverse of the distance from Theron's centre, and justify which of the two dependences the reported measurement supports.

❖ Answer Key & Scoring Guide

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

Part (a) — Model Answer

Theron has no atmosphere, so no drag force is exerted on the calibration mass, and once released the mass touches nothing, so no contact force is exerted on it either. The only force exerted on the mass is the gravitational force exerted on it by Theron.

Free-body diagram of the calibration mass at the instant of release.

By Newton's law of universal gravitation that force acts along the line joining the two centres of mass — that is, toward Theron's centre. Because only one force is exerted on the mass, that force is also the net force: F_net = F_g. Redrawn at a larger r the arrow would be shorter, but the diagram would still carry exactly one arrow.

Scoring (2 points):

▸ 1 point: Exactly one force arrow is drawn, originating at the dot and directed toward Theron's centre.

▸ 1 point: That arrow is labelled with both the type of force and the object exerting it — F_g (Theron on the mass), "gravitational force of Theron on the mass", or an equivalent that names Theron as the exerting body.

⚠︎ Common error (partial credit): A second arrow labelled "motion", "inertia" or "the force of release" is drawn alongside a fully labelled gravitational arrow — the diagram no longer shows exactly one force, so the first criterion is lost while the second is earned: earns 1 of 2 points.

⚠︎ Common error (partial credit): Exactly one arrow is drawn toward Theron's centre but is labelled only "weight", "mg" or "F_g", naming no exerting object — earns the first criterion only: 1 of 2 points.

✗ Common error (no credit): A single arrow is drawn directed away from Theron and labelled as the force the calibration mass exerts on Theron — the other member of the third-law pair, which is exerted on Theron and belongs on Theron's diagram. Neither criterion is earned.

Part (b) — Model Answer

Let m be the mass of the calibration object. The gravitational force is the only force exerted on it, so it is also the net force, and Newton's second law applies to it directly.

F_g = G·M_T·m / r²

F_net = m·a_g

m·a_g = G·M_T·m / r²

a_g = G·M_T / r²

The calibration mass m cancels, so a_g depends only on G, M_T and r — never on the object placed at that point. Replacing the 2.0 kg mass with a 50.0 kg mass at the same station therefore changes nothing: 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 mass in the law of universal gravitation measures how strongly the object takes part in the gravitational interaction; the mass in Newton's second law measures how strongly its motion resists change. Those two masses are experimentally equivalent, which is what lets them cancel. The quantity a_g = G·M_T/r² is Theron's gravitational field strength at that location — a property of Theron and of the point in space, not of whatever is placed there.

Scoring (4 points):

▸ 1 point: Begins from fundamental principles — quotes both F_g = G·m_1·m_2/r² and F_net = m·a from the equation sheet before any manipulation. This point is earned even if the algebra that follows is incorrect.

▸ 1 point: Identifies the gravitational force as the net force on the calibration mass and equates it to m·a_g.

▸ 1 point: Reaches the correct expression a_g = G·M_T/r², with the calibration mass cancelled so that only G, M_T and r remain.

▸ 1 point: States that a_g is unchanged for the 50.0 kg mass and justifies it by noting that the gravitational force and the inertia both scale with the object's mass, so their ratio does not.

⚠︎ Common error (partial credit): Reaches a_g = G·M_T/r² correctly but justifies the 50.0 kg prediction only by asserting that "all objects fall at the same rate", with no reference to the cancellation or to force and inertia scaling together — earns the first three criteria: 3 of 4 points.

⚠︎ Common error (partial credit): Writes m·a_g = G·M_T·m/r² and then reports a_g = G·M_T·m/r², failing to cancel the calibration mass, and so predicts an acceleration 25 times larger for the 50.0 kg mass — earns the first two criteria: 2 of 4 points.

✗ Common error (no credit): Uses the near-surface weight relationship alone: writes F_g = m·g_T and states without further equations that a_g = g_T = 1.60 m/s² at every station, including for the 50.0 kg mass, "because g is the same for everything". Universal gravitation is never quoted, nothing is equated to m·a_g, the expression carries no r, and the unchanged prediction rests on no mechanism, so no criterion is earned.

Part (c) — Model Answer

The expression from part (b) contains the product G·M_T, which is not given. Eliminate it using the surface measurement: at r = R_T that 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

Evaluating at the three marked distances:

at r = R_T:  a_g = 1.60 m/s²

at r = 2R_T:  a_g = 1.60 / 4 = 0.400 m/s²

at r = 4R_T:  a_g = 1.60 / 16 = 0.100 m/s²

Because a_g is proportional to 1/r², the curve leaves (R_T, 1.60 m/s²) steeply and then flattens: it decreases and is concave up across the whole range. Continued beyond the plotted window it would approach the horizontal axis without ever reaching it — a gravitational field has no outer edge — though the sketch is not asked to show that.

a_g against r for Theron, from r = R_T to r = 4R_T. The marked points are the surface value g_T and the two values calculated from a_g = g_T(R_T/r)².

Scoring (3 points):

▸ 1 point: The sketch is a single curve that decreases and is concave up over the whole range — steep near r = R_T, flattening as r increases — and is not a straight line.

▸ 1 point: The curve passes through the surface value at r = R_T and that value is labelled a_g = 1.60 m/s².

▸ 1 point: Both marked values are correct — 0.400 m/s² at r = 2R_T and 0.100 m/s² at r = 4R_T. Accept values obtained by correct substitution into the student's own expression from part (b).

⚠︎ Common error (partial credit): a_g is halved when r doubles, so 0.800 m/s² is marked at 2R_T and 0.400 m/s² at 4R_T — the inverse-versus-inverse-square confusion. The curve is still decreasing and concave up and the surface value is marked correctly, so only the third criterion is lost: 2 of 3 points. If the student carried an inverse expression forward from part (b), these are the values that expression gives and the third criterion is earned under its consistency clause.

⚠︎ Common error (partial credit): The curve is brought down to meet the horizontal axis at r = 4R_T, on the reasoning that Theron's field "runs out" at some distance, so zero is marked there while the surface value and the 2R_T value are both correct. A gravitational field has no outer edge, so the third criterion is lost: 2 of 3 points.

Part (d) — Model Answer

The altimeter reports a height above the surface. The gravitational force is exerted along the line joining the two centres of mass, so the r in the expression from part (b) is measured from Theron's centre, and the height must be converted before it is used:

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

Using the surface value to eliminate G·M_T, as in part (c):

a_g = g_T·(R_T / r)² = 1.60 × (1 / 2.50)² = 1.60 / 6.25 = 0.256 m/s²

If the field instead fell off as the inverse of the distance from Theron's centre, the same surface value would give

a_g = g_T·(R_T / r) = 1.60 / 2.50 = 0.640 m/s²

The two candidate dependences over the surveyed range. The reported value at r = 2.50R_T lies on the inverse-square curve and nowhere near the inverse one.

The reported 0.254 m/s² is within about 0.8% of the inverse-square prediction of 0.256 m/s², inside the 2% spread of the repeated runs, while the inverse prediction of 0.640 m/s² is two and a half times the reported value. The measurement therefore supports the inverse-square dependence and rules the inverse one out. Both candidates are anchored to the same surface value, so they agree at r = R_T by construction; it is the reading at a second, independently measured distance that separates them.

Scoring (3 points):

▸ 1 point: Converts the reported height into a distance from Theron's centre: r = R_T + h = 2.50R_T.

▸ 1 point: Predicts a_g under both dependences at that station — 0.256 m/s² for the inverse square and 0.640 m/s² for the inverse — by eliminating G·M_T with the surface value. Accept values correctly computed from the student's own expression from part (b) and the student's own r.

▸ 1 point: Justifies the choice by comparing both predictions with the reported 0.254 m/s² and stating which dependence the data support. Award this criterion whenever the student compares both of their own predicted values with the reported one and draws the conclusion those numbers warrant.

⚠︎ Common error (partial credit): The altimeter height is taken as the distance from Theron's centre, so r = 1.50R_T. Both dependences are then evaluated correctly on that radius — 0.711 m/s² and 1.07 m/s² — and compared with the reported value. The second and third criteria are earned on a consistency basis and only the conversion criterion is lost: 2 of 3 points.

⚠︎ Common error (partial credit): r = 2.50R_T and the inverse-square prediction of 0.256 m/s² are both correct, but the inverse dependence is never evaluated and the justification stops at "the predicted and reported values are close" — earns the first criterion and no more, since the second asks for both predictions: 1 of 3 points.

✗ Common error (no credit): The surface value is scaled by the ratio of the distances rather than by its square, and the altitude is used as the radius: a_g = 1.60/1.50 = 1.07 m/s², from which the probe's tracking data are reported as faulty. The radius is wrong, neither dependence is evaluated from the part (b) expression, and the conclusion rejects the data instead of the model, so no criterion is earned.