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

FRQ: Gravitational Potential Energy in Free Fall — The Sentinel Payload Drop

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A disabled, untethered Sentinel payload of mass m = 4.00 × 10² kg is released from rest at r = 4R_E, where r is the payload’s distance from Earth’s centre and R_E = 6.37 × 10⁶ m is Earth’s radius, as shown in Figure 1. Its thrusters are inoperative, and it falls directly toward Earth along a radius. Once released, it touches nothing; air resistance is negligible at every point in the fall, and so are the gravitational forces exerted on the payload by the Moon, the Sun and every other body except Earth. The axes shown in Figure 2 are used in part C.

Figure 1 — The payload is released from rest at r = 4R_E (distances to scale, payload enlarged).

Figure 2 — Axes for part C: energy against the distance r of the payload from Earth’s centre.

Take Earth’s mass to be M_E = 5.97 × 10²⁴ kg and use GM_E = 3.98 × 10¹⁴ N·m²/kg. Earth’s own motion may be neglected: use a reference frame in which Earth’s centre is at rest and which does not rotate with Earth, and treat it as inertial. All motion is one-dimensional along a radius from Earth’s centre; take radially outward as the positive direction, and take the gravitational potential energy of the Earth–payload system to be zero at infinite separation.

✎ Free Response Questions

Part A

For this payload, the quantity GM_E m/R_E² has the value 3.92 × 10³ N. The side of each grid square in Figure 3 represents a force of magnitude 245 N.

On the dots in Figure 3 that represent the payload at r = 4R_E and at r = 2R_E, draw and label the forces (not components) that are exerted on the payload.

•   Each arrow must start on, and point away from, each dot.

•   Draw the length of each arrow to represent the magnitude of each force, consistent with the scale used in Figure 3.

Figure 3 — The payload at r = 4R_E (upper dot) and r = 2R_E (lower dot); one square is 245 N.

Part B

Starting with conservation of energy, derive an expression for the speed v_2 of the payload when it reaches r = 2R_E. Express your final answer in terms of m, M_E, R_E, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

For this payload, the quantity GM_E m/R_E has the value 2.50 × 10¹⁰ J.

On the axes shown in Figure 2, do the following for r = R_E to r = 6R_E.

•   Sketch and label a line or curve that represents the gravitational potential energy U of the Earth–payload system as a function of r.

•   Sketch and label a line or curve that represents the total mechanical energy E of the Earth–payload system as a function of r during the drop.

•   Sketch and label a line or curve that represents the kinetic energy K of the payload as a function of r during the drop.

If a line or curve is not defined over part of this range, draw it only where it is defined.

Part D

Released from rest at r = 4R_E, the payload gains kinetic energy ΔK_4 as it falls from r = 2R_E to r = R_E. Suppose instead that the same payload is released from rest at r = 6R_E, and that it gains kinetic energy ΔK_6 as it falls from r = 2R_E to r = R_E. Use the lines and curves you drew on Figure 2 in part C to compare ΔK_6 and ΔK_4.

Indicate whether ΔK_6 is greater than, less than, or equal to ΔK_4 by writing one of the following.

•   ΔK_6 > ΔK_4

•   ΔK_6 < ΔK_4

•   ΔK_6 = ΔK_4

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 payload at either position: the gravitational force exerted on it by Earth, directed toward Earth’s centre (to the left in Figure 3).

Figure 4 — One force at each position; the r = 2R_E arrow is four times the r = 4R_E arrow.

The thrusters are off, nothing touches the payload, and air resistance and the pulls of the Moon, the Sun and every other body are negligible, so each diagram carries exactly one arrow.

By Newton’s law of universal gravitation, F_g = GM_E m/r², so the force at a distance r is the supplied 3.92 × 10³ N multiplied by (R_E/r)². Halving the distance from Earth’s centre quadruples the gravitational force, so the r = 2R_E arrow is four times as long as the r = 4R_E arrow. At r = 4R_E the force is (3.92 × 10³ N)/16 = 245 N, one grid square; at r = 2R_E it is (3.92 × 10³ N)/4 = 980 N, four grid squares. Computed directly from the Scenario’s values, GM_E m/(2R_E)² = 981 N, the same to within rounding. The payload’s mass is the same at both positions, but the force also depends on the separation.

Scoring (3 points):

▸ A1 — 1 point: For drawing, on each dot, exactly one arrow, which starts on the dot and is directed toward Earth’s centre.

Scoring Note: Arrows of any nonzero length can earn A1. An arrow representing velocity, acceleration or motion counts as an additional arrow.

▸ A2 — 1 point: For labelling, on each dot, an arrow directed toward Earth’s centre as the gravitational force.

Scoring Note: Appropriate labels include F_g, F_G, F_grav, W, mg, “gravitational force”, “weight”, GM_E m/r², or a label naming Earth, such as F_Earth on payload; mg is appropriate with g read as the gravitational field strength at the payload’s position. The labels g, a_g and G are not; nor is a magnitude alone, such as 245 N, or a label naming only the net force, such as F_net or ma. The response is not required to name the object exerting the force.

▸ A3 — 1 point: For drawing, on each dot, an arrow directed toward Earth’s centre that is one grid square long at r = 4R_E and four grid squares long at r = 2R_E.

Scoring Note: Each length within half a grid square, with the r = 2R_E arrow between three and five times as long as the r = 4R_E arrow, meets A3. Where a dot also carries a velocity, acceleration or motion arrow directed toward Earth’s centre, A3 and its ratio are judged on the other arrow. A3 does not depend on the labels. A1 to A3 do not depend on any other part.

⚠︎ Common error (partial credit): Draws one correctly labelled arrow toward Earth’s centre on each dot but uses the supplied 3.92 × 10³ N at both dots, because the payload’s mass has not changed, or scales it by R_E/r instead of (R_E/r)², drawing four squares at r = 4R_E and eight at r = 2R_E — earns 2 of 3 points, A1 and A2. The force depends on the separation as well as on the masses, and it varies as 1/r².

⚠︎ Common error (partial credit): Draws one arrow on each dot, toward Earth’s centre and one and four squares long, but labels them g or a_g, or only F_net or ma — earns 2 of 3 points, A1 and A3. A field strength and an acceleration are not forces, and F_net names what the forces add up to; the gravitational force is the mass times the field strength.

⚠︎ Common error (partial credit): Draws the correct gravitational arrows but adds a second arrow on a dot — a velocity or acceleration arrow, a “force of motion”, a normal force, or an outward “centrifugal force” — earns 2 of 3 points, A2 and A3. Velocity and acceleration are not forces, nothing touches the payload, and in an inertial frame nothing pushes it outward: gravity is the only force.

✗ Common error (no credit): Draws each arrow pointing away from Earth, labelled as the force the payload exerts on Earth. That force is the other member of the third-law pair; it is exerted on Earth and belongs on Earth’s diagram. A1 and A3 need an arrow directed toward Earth’s centre, and there is none for A2 to score, so no point is earned.

Part B — Model Answer

The payload is released from rest. No force from outside the Earth–payload system does work on it: nothing else acts on the payload, and Earth’s motion is neglected. The gravitational interaction within the system is conservative, so the system’s mechanical energy is constant between r = 4R_E and r = 2R_E:

K_1 + U_1 = K_2 + U_2
0 + (−GM_E m/(4R_E)) = ½mv_2² + (−GM_E m/(2R_E))
½mv_2² = GM_E m/(2R_E) − GM_E m/(4R_E) = GM_E m/(4R_E)
v_2 = √(GM_E/(2R_E))

The payload’s mass cancels, so any object released from rest at r = 4R_E reaches r = 2R_E with the same speed. For the Sentinel payload, v_2 = √((3.98 × 10¹⁴ N·m²/kg)/(2 × 6.37 × 10⁶ m)) = 5.59 × 10³ m/s.

Scoring (3 points):

▸ B1 — 1 point: For a multistep derivation that includes conservation of energy.

Scoring Note: Conservation of energy may be written as K_1 + U_1 = K_2 + U_2, as ΔK + ΔU = 0, or as ΔK = −ΔU. A final expression for v_2 preceded only by words, or only by unapplied equations from the reference information, is not a multistep derivation. B1 to B3 do not depend on the response in part A.

▸ B2 — 1 point: For expressions for the gravitational potential energy of the Earth–payload system at r = 4R_E and at r = 2R_E, −GM_E m/(4R_E) and −GM_E m/(2R_E), or for a correct expression for its change between those positions, ΔU = −GM_E m/(4R_E).

▸ B3 — 1 point: For a correct expression for v_2.

Scoring Note: A correct, isolated, final expression for v_2 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. Any algebraic equivalent is accepted, such as √(gR_E/2) with g = GM_E/R_E², the field strength at Earth’s surface, and so is the velocity −√(GM_E/(2R_E)), whose sign shows the inward direction.

⚠︎ Common error (partial credit): Writes the kinetic energy as mv_2² instead of ½mv_2², reaching v_2 = √(GM_E/(4R_E)), 3.95 × 10³ m/s — earns 2 of 3 points, B1 and B2.

⚠︎ Common error (partial credit): Takes U = 0 at the release point but U = −GM_E m/(2R_E) at r = 2R_E, reaching v_2 = √(GM_E/R_E) — earns 1 of 3 points, B1. The zero of U may be placed anywhere, but it must be the same zero at both positions; the Scenario puts it at infinite separation.

⚠︎ Common error (partial credit): Uses the near-surface relation ΔU = mgΔy with the surface field over a fall of 2R_E, reaching v_2 = √(4gR_E) — earns 1 of 3 points, B1. That relation holds only where the field is nearly constant; between r = 4R_E and r = 2R_E it grows fourfold.

✗ Common error (no credit): Uses the constant-acceleration relation v_2² = 2g(2R_E) with the surface value of g. No conservation of energy is written, U = −GM_E m/r is never used, and the acceleration is not constant over the fall, so no point is earned.

Part C — Model Answer

The gravitational potential energy of the Earth–payload system is U = −GM_E m/r. With GM_E m/R_E = 2.50 × 10¹⁰ J,

U = −(2.50 × 10¹⁰ J)(R_E/r)
r = R_E: U = −2.50 × 10¹⁰ J
r = 2R_E: U = −1.25 × 10¹⁰ J
r = 4R_E: U = −0.625 × 10¹⁰ J

U is negative at every finite r because its zero is at infinite separation and gravity is attractive: moving the payload outward toward that zero takes energy. Doubling r halves the magnitude of U, so the curve rises steeply near Earth and flattens as r increases, approaching zero without reaching it (−0.417 × 10¹⁰ J at r = 6R_E). U depends only on the separation, so it is defined at every r in the range.

The system’s mechanical energy is constant (part B) and is fixed at release, where K = 0:

E = K + U = 0 + (−0.625 × 10¹⁰ J) = −0.625 × 10¹⁰ J

so E is a horizontal line at −0.625 × 10¹⁰ J. The kinetic energy at each r is K = E − U, the vertical gap from the U curve up to the E line: zero at r = 4R_E, 0.625 × 10¹⁰ J at r = 2R_E, which is ½mv_2² from part B, and 1.875 × 10¹⁰ J at r = R_E.

Figure 5 — U, E and K against r; K is the gap from the U curve up to the E line.

The kinetic energy at each r is the gap between the E line and the U curve, so the payload can only be where the E line lies on or above the U curve. For r > 4R_E, U is greater than E, and K = E − U would be negative, which is impossible: the payload, released from rest and falling inward, never reaches those separations, so K is not defined there. E is the system’s constant energy, so a line stopped at r = 4R_E and one continued across the range are both accepted. The E line meets the U curve at r = 4R_E, where the payload is released from rest.

Scoring (4 points):

▸ C1 — 1 point: For sketching a line or curve, labelled as the gravitational potential energy, that is negative from r = R_E to r = 6R_E, increases as r increases, and becomes less steep as r increases without reaching zero.

▸ C2 — 1 point: For a U curve at −2.50 × 10¹⁰ J at r = R_E, −1.25 × 10¹⁰ J at r = 2R_E and −0.625 × 10¹⁰ J at r = 4R_E.

Scoring Note: Each value within 0.125 × 10¹⁰ J (half a grid division) meets C2. C1 and C2 do not depend on any other part.

▸ C3 — 1 point: For sketching a horizontal line, labelled as the total mechanical energy, at −0.625 × 10¹⁰ J or at the value of the response’s own U curve at r = 4R_E, continuous from r = R_E to at least r = 4R_E.

Scoring Note: A line within 0.125 × 10¹⁰ J (half a grid division) of either value meets C3, and a line that continues past r = 4R_E also meets C3.

▸ C4 — 1 point: For sketching a line or curve, labelled as the kinetic energy, that is zero at r = 4R_E, increases as r decreases, and is not drawn for r > 4R_E, with values of 0.625 × 10¹⁰ J at r = 2R_E and 1.875 × 10¹⁰ J at r = R_E, or with values equal at each r to the vertical distance from the response’s own U curve up to its own E line.

Scoring Note: Each value within 0.125 × 10¹⁰ J (half a grid division) meets C4. C1 to C4 do not depend on the response in part B.

⚠︎ Common error (partial credit): Draws U through the three correct values but lets it reach or cross zero before r = 6R_E, or stops it at r = 4R_E as if U, like K, were undefined where the payload never goes, with E and K correct — earns 3 of 4 points, C2, C3 and C4. U depends only on the separation, and with its zero at infinite separation it is negative at every finite r.

⚠︎ Common error (partial credit): Joins −2.50 × 10¹⁰ J at r = R_E and −0.625 × 10¹⁰ J at r = 4R_E with a straight line, as if U changed by the same amount for each metre of the fall, with E and K consistent with it — earns 2 of 4 points, C3 and C4. That line gives −1.875 × 10¹⁰ J at r = 2R_E, not −1.25 × 10¹⁰ J, and reaches zero at r = 5R_E; the field weakens with distance from Earth.

⚠︎ Common error (partial credit): Draws U positive and decreasing, as though U = +GM_E m/r, with E horizontal at that curve’s value at r = 4R_E and K correct — earns 2 of 4 points, C3 and C4. With its zero at infinite separation, the potential energy of an attracting pair is negative at every finite r.

⚠︎ Common error (partial credit): Draws U and K correctly but makes E rise as r decreases, “because the payload speeds up, so it gains energy” — earns 3 of 4 points, C1, C2 and C4. The payload gains kinetic energy, but the system loses exactly as much potential energy; nothing outside the system does work, so E cannot change.

⚠︎ Common error (partial credit): Draws U correctly but draws E as a horizontal line at zero, “because the payload starts from rest”, and K as the gap from the U curve up to it — earns 2 of 4 points, C1 and C2. At release K is zero but U is not, so E equals U at r = 4R_E; with E at zero, K would not be zero at the release point.

⚠︎ Common error (partial credit): Draws U and E correctly and draws K as E − U over the whole range, so that K falls below zero for r > 4R_E — earns 3 of 4 points, C1, C2 and C3. Kinetic energy cannot be negative; the payload is never at those separations.

Part D — Model Answer

ΔK_6 = ΔK_4

In either drop the system’s mechanical energy is constant, so the kinetic energy the payload gains between two positions equals the decrease in U between them. U depends only on the separation r: the U curve drawn on Figure 2 is the same whatever the release point, and from r = 2R_E to r = R_E it falls by 1.25 × 10¹⁰ J in both drops. Released from r = 6R_E, the E line sits higher, at U(6R_E) = −0.417 × 10¹⁰ J, and the K curve lies 0.208 × 10¹⁰ J higher at every r from R_E to 4R_E, but the gap between the E line and the U curve grows by the same 1.25 × 10¹⁰ J over the interval.

Figure 6 — From either release, U falls by the same 1.25 × 10¹⁰ J between r = 2R_E and r = R_E.

The kinetic energy gained over a stretch of the fall is set by how far U falls over that stretch, so it does not depend on where the fall began. The payload released from r = 6R_E does cross the interval faster and gains less speed there, 3.75 × 10³ m/s against 4.09 × 10³ m/s, but it gains the same kinetic energy.

Scoring (2 points):

▸ D1 — 1 point: For indicating ΔK_6 = ΔK_4.

Scoring Note: D1 does not depend on the response in part C.

▸ D2 — 1 point: For a justification that the kinetic energy gained over the interval is the same for both release points, either because it equals the decrease in U over the interval (the system’s mechanical energy being constant in each drop) and that decrease depends only on r, or because it equals the work done by the gravitational force over the interval, which is the same for both because that force depends only on r.

Scoring Note: The justification may be made with the graph — the E line is higher for the release from r = 6R_E, but the U curve is the same, so the gap between them grows by the same amount over the interval — in words, or in symbols. Stating that U at r = 2R_E and at r = R_E does not depend on the release point meets the second element. D2 does not depend on the response in part C.

⚠︎ Common error (partial credit): Indicates ΔK_6 = ΔK_4 but justifies it only by saying that energy is conserved, with no link between the kinetic energy gained and the fall in U over the interval — earns 1 of 2 points, D1.

✗ Common error (no credit): Indicates ΔK_6 > ΔK_4 because the payload released from r = 6R_E reaches r = 2R_E already moving faster, or because that drop has more energy to convert. The kinetic energy it already has is not converted again; only the fall in U over the interval adds to it, so no point is earned.

✗ Common error (no credit): Indicates ΔK_6 < ΔK_4 because the faster payload spends less time crossing the interval, so gravity acts on it for less time. Less time does mean a smaller change in speed, but the kinetic energy gained is set by the force and the distance over which it acts, both unchanged, so no point is earned.

Keep going:

Equations & Traps deck · Mind Maps bundle