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FRQ: Escape Speed and Gravitational Potential Energy — The Callisto Return Capsule

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A return capsule sits at rest on the surface of Callisto, one of Jupiter’s moons. Model Callisto as an airless, non-rotating sphere of mass M = 1.08 × 10²³ kg and radius R = 2.41 × 10⁶ m, as shown in Figure 1. The capsule’s engine fires briefly and shuts off while the capsule is still at the surface, at r = R, where r is the capsule’s distance from Callisto’s centre. After the burn, the capsule’s mass is m = 850 kg. At the instant of engine cut-off, the capsule is moving radially outward with speed v_esc, exactly the minimum speed needed for it never to return to Callisto.

Figure 1 — The capsule at Callisto’s surface at engine cut-off (capsule enlarged).

Treat the capsule–Callisto system as isolated: ignore the gravitational forces exerted on the capsule and on Callisto by Jupiter, the Sun and every other body outside the system. After cut-off, the gravitational force exerted by Callisto is then the only force exerted on the capsule, which touches nothing. The motion of Callisto caused by the capsule is negligible; use a reference frame in which Callisto’s centre is at rest, and treat it as inertial. Take the gravitational potential energy of the capsule–Callisto system to be zero when the capsule is infinitely far from Callisto, and use G = 6.67 × 10⁻¹¹ N·m²/kg². All motion is one-dimensional along a radius from Callisto’s centre; take radially outward as the positive direction.

✎ Free Response Questions

Part A

Figure 2 shows two incomplete energy bar charts for the capsule–Callisto system: one for the instant of engine cut-off, and one for when the capsule is so far from Callisto that its distance from Callisto’s centre can be treated as infinite. Each chart has a column for the kinetic energy K of the capsule, the gravitational potential energy U_G of the capsule–Callisto system, and the total mechanical energy E of the capsule–Callisto system.

Draw shaded rectangles to complete both energy bar charts in Figure 2.

•   Positive energy values are above the zero-energy line (“0”), and negative energy values are below the zero-energy line.

•   Shaded regions should start at the dashed line representing zero energy.

•   Represent any energy that is equal to zero with a distinct line on the zero-energy line.

•   The relative height of each shaded region should reflect the magnitude of the respective energy consistent with the scale shown.

Figure 2 — Energy bar charts for part A: at engine cut-off, and very far from Callisto.

Part B

Derive an expression for the speed v_esc of the capsule at the instant of engine cut-off. Express your final answer in terms of m, M, R, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Part C

Figure 3 shows axes for the capsule’s speed v as a function of its distance r from Callisto’s centre. On the axes shown in Figure 3, sketch a line or curve that represents v as a function of r after engine cut-off, from r = R to r = 10R.

Figure 3 — Axes for part C: the capsule’s speed v against its distance r from Callisto’s centre.

Part D

Suppose instead that the capsule’s engine shuts off at the surface when the capsule is moving radially outward with speed 2v_esc, and that everything else is as described in the Scenario. Let v_far be the capsule’s speed when it is so far from Callisto that its distance from Callisto’s centre can be treated as infinite.

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

•   v_far > v_esc

•   v_far < v_esc

•   v_far = v_esc

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

At engine cut-off the capsule is at r = R, so the gravitational potential energy of the capsule–Callisto system is U_G = −GMm/R. It is negative because its zero is at infinite separation and gravity is attractive: pulling the capsule away from Callisto raises U_G toward that zero.

The capsule has exactly the minimum speed needed never to return. If the system’s mechanical energy E were below zero, the capsule would stop at the finite distance where U_G = E and fall back; with E zero or positive, K = E − U_G stays positive at every finite r, so it never stops. The minimum speed makes E exactly zero, so at cut-off the kinetic energy K = +GMm/R just cancels U_G = −GMm/R. For this capsule, GMm/R = 2.54 × 10⁹ J.

Figure 4 — At cut-off K and U_G cancel, so E = 0; very far away all three energies are zero.

After cut-off the only force exerted on the capsule is Callisto’s gravitational force, which is internal to the capsule–Callisto system, so E stays zero. Very far from Callisto, U_G has risen to zero, so K = E − U_G is zero as well: the capsule’s speed falls to zero at infinite separation. All three energies are drawn as lines on the zero-energy line.

Scoring (3 points):

▸ A1 — 1 point: For drawing, in the chart for engine cut-off, a shaded K bar above the zero-energy line and a shaded U_G bar below it.

Scoring Note: The heights of the bars are not required to earn A1.

▸ A2 — 1 point: For representing K as a distinct line on the zero-energy line in the chart for very far from Callisto.

▸ A3 — 1 point: For representing E as a distinct line on the zero-energy line in both charts, with the K bar at engine cut-off as far above the zero-energy line as the U_G bar is below it.

Scoring Note: For A3, the K and U_G bars count as equal in height when their heights differ by no more than half a grid division. A1 to A3 do not depend on any other part.

⚠︎ Common error (partial credit): Draws K and U_G correctly in both charts but draws E at engine cut-off as a positive bar equal in height to K, “because the engine gave the capsule that much energy”, and E far away as a line on zero — earns 2 of 3 points, A1 and A2. E is the sum K + U_G, which is zero at cut-off; and with gravity the only force exerted on the capsule, E cannot fall between the two instants.

⚠︎ Common error (partial credit): Draws U_G at engine cut-off as a positive bar the same height as K, with E zero in both charts and all three energies zero far away — earns 1 of 3 points, A2. With its zero at infinite separation, the potential energy of an attracting pair is negative at every finite separation, and E must equal K + U_G.

⚠︎ Common error (partial credit): Draws the K bar at cut-off taller than the U_G bar is deep, so that E is positive, “because escaping takes extra energy”, and keeps that energy far away as a positive K — earns 1 of 3 points, A1. That capsule escapes, but faster than it needs to; the minimum speed leaves nothing over.

⚠︎ Common error (partial credit): Places the zero of U_G at Callisto’s surface: U_G = 0 and E = K at cut-off, then a positive U_G bar holding all of E far away, with K = 0 — earns 1 of 3 points, A2. The Scenario puts the zero at infinite separation, so U_G at the surface is negative and E is zero.

Part B — Model Answer

Take the capsule–Callisto system. After cut-off the only force exerted on the capsule is Callisto’s gravitational force, which is internal to the system and conservative; every other body is ignored, so no force from outside the system does work on it, and the system’s mechanical energy is constant. Callisto’s motion is negligible, so the system’s kinetic energy is the capsule’s. Compare the instant of cut-off with the capsule infinitely far away, where U_G = 0 and, because v_esc is the minimum speed, K = 0:

K_1 + U_1 = K_2 + U_2
½mv_esc² + (−GMm/R) = 0 + 0
½mv_esc² = GMm/R
v_esc = √(2GM/R)

The capsule’s mass cancels, so any object launched from Callisto’s surface needs the same escape speed. For Callisto, v_esc = √(2(6.67 × 10⁻¹¹ N·m²/kg²)(1.08 × 10²³ kg)/(2.41 × 10⁶ m)) = 2.45 × 10³ m/s.

Scoring (4 points):

▸ B1 — 1 point: For a multistep derivation that includes either conservation of energy or the escape condition, that the mechanical energy of the capsule–Callisto system at engine cut-off is zero.

Scoring Note: Conservation of energy may be written as K_1 + U_1 = K_2 + U_2, as ΔK + ΔU_G = 0, or as the capsule’s kinetic energy at cut-off equal to the increase in U_G from r = R to infinite separation; the work–energy theorem, with the work done by gravity equal to −ΔU_G, is equivalent. A final expression for v_esc preceded only by words, only by equations into which nothing specific to this situation has been substituted (such as K_1 + U_1 = K_2 + U_2 or K + U_G = 0), or only by correct expressions for K or U_G that are never combined into an equation, is isolated and is not a multistep derivation.

▸ B2 — 1 point: For an expression for the gravitational potential energy of the capsule–Callisto system at engine cut-off, −GMm/R, or for its increase from engine cut-off to infinite separation, GMm/R, obtained from U_G = −Gm_1m_2/r with r equal to R, the capsule’s distance from Callisto’s centre.

▸ B3 — 1 point: For using that, at the minimum speed, no kinetic energy is left at infinite separation, where U_G = 0: setting the system’s mechanical energy at engine cut-off, K + U_G, equal to zero, or the capsule’s kinetic energy at cut-off equal to the increase in U_G out to infinite separation, with the response’s own expressions for K and U_G.

▸ B4 — 1 point: For a correct expression for v_esc.

Scoring Note: A correct, isolated, final expression for v_esc earns points B2, B3 and B4. Any algebraic equivalent is accepted, such as √(2g_C R) with g_C = GM/R², the gravitational field strength at Callisto’s surface. B1, B2 and B4 carry no consistency credit, and B1 to B4 do not depend on the response in part A.

⚠︎ Common error (partial credit): Writes the kinetic energy as mv_esc² instead of ½mv_esc², reaching v_esc = √(GM/R) — earns 3 of 4 points, B1, B2 and B3. That is the speed of a capsule skimming the surface in a circular orbit, 1.73 × 10³ m/s here; escaping takes √2 times as much.

⚠︎ Common error (partial credit): Uses −GMm/R², with r² in place of r, for the potential energy, reaching √(2GM/R²) — earns 2 of 4 points, B1 and B3. GMm/R² is the magnitude of the gravitational force, not an energy, and the result does not have the units of speed.

⚠︎ Common error (partial credit): Treats escape as a climb of height R in a uniform field of 10 N/kg, writing ½mv_esc² = mgR and reaching √(2gR) = 6.94 × 10³ m/s — earns 1 of 4 points, B1. Escape means reaching infinite separation, not a height R; the field is not uniform over the climb (at r = 2R it is a quarter of its surface value); and Callisto’s surface field is GM/R² = 1.24 N/kg, not Earth’s 10 N/kg. With Callisto’s own field mgR does equal GMm/R, which is why √(2g_C R) is accepted as a final expression.

✗ Common error (no credit): Sets Callisto’s gravitational force on the capsule equal to mv²/R, as for a circular orbit, reaching v = √(GM/R). No energy equation is written, no potential energy is found from U_G = −Gm_1m_2/r, and the capsule is not moving in a circle, so no point is earned.

Part C — Model Answer

The system’s mechanical energy is zero at every instant after cut-off (parts A and B), so at any distance r,

½mv² + (−GMm/r) = 0
v = √(2GM/r) = v_esc √(R/r)
r = R: v = v_esc     r = 2R: v = 0.71v_esc     r = 4R: v = 0.50v_esc
r = 6R: v = 0.41v_esc     r = 10R: v = 0.32v_esc

The speed is proportional to 1/√r, so quadrupling the capsule’s distance from Callisto’s centre halves its speed.

Figure 5 — v = v_esc √(R/r): the speed halves by r = 4R and never reaches zero.

The curve falls steeply at first and then less and less steeply, because v is proportional to 1/√r: the speed halves between R and 4R, and halves again only between 4R and 16R. It never reaches zero at a finite distance: K = GMm/r is positive wherever r is finite, and the speed approaches zero only as r becomes infinitely large. Callisto’s gravitational force also approaches zero as r grows, as 1/r², but it is never zero at any finite distance: it acts on the capsule for the whole journey, and it is what keeps slowing the capsule down.

Scoring (3 points):

▸ C1 — 1 point: For a line or curve that begins at v = v_esc at r = R.

Scoring Note: A starting value within 0.05v_esc (half a grid division) of v_esc meets C1.

▸ C2 — 1 point: For a curve that decreases throughout from r = R to r = 10R, less steeply as r increases, and is still above v = 0 at r = 10R.

Scoring Note: A curve that never rises and is lower at r = 10R than at r = 6R counts as decreasing throughout, even if it looks level over part of that stretch; the correct curve falls by less than one grid division between those two distances, from 0.41v_esc to 0.32v_esc.

▸ C3 — 1 point: For a line or curve at v = 0.5v_esc at r = 4R.

Scoring Note: A value within 0.05v_esc (half a grid division) of 0.5v_esc at r = 4R meets C3. C1 to C3 do not depend on the response in any other part.

⚠︎ Common error (partial credit): Draws v proportional to 1/r, falling to 0.25v_esc at r = 4R and 0.10v_esc at r = 10R — earns 2 of 3 points, C1 and C2. It is the kinetic energy, like the magnitude of U_G, that is proportional to 1/r; the speed goes as its square root, so as 1/√r. A part B potential energy of −GMm/r² leads to this same curve.

⚠︎ Common error (partial credit): Draws the curve through (4R, 0.5v_esc) but has it reach v = 0 at about r = 8R and run along the axis beyond it, as though the capsule stopped there — earns 2 of 3 points, C1 and C3. On a minimum-speed launch the system’s mechanical energy is zero, so K = GMm/r is positive at every finite r; the capsule stops only at infinite separation, and a speed of zero cannot last over a stretch of r.

⚠︎ Common error (partial credit): Draws a straight line from (R, v_esc) to (10R, 0) — earns 1 of 3 points, C1. A straight line means the speed drops by the same amount over each metre and reaches zero at a finite distance; energy gives v = v_esc √(R/r), which halves by r = 4R and then falls more and more slowly. Even a constant slowing force would make v², not v, fall by the same amount per metre.

Part D — Model Answer

v_far > v_esc

Only the launch speed has changed, so the capsule still starts at r = R, and U_G at cut-off is still −GMm/R. From part B, ½mv_esc² = GMm/R, so the kinetic energy at cut-off is now ½m(2v_esc)² = 4(½mv_esc²) = 4GMm/R, and the system’s mechanical energy is E = 4GMm/R − GMm/R = +3GMm/R, constant after cut-off. Very far from Callisto U_G = 0, so all of E is kinetic:

½mv_far² = 3GMm/R = 3(½mv_esc²)
v_far = √3 v_esc ≈ 1.73v_esc

Figure 6 — Launched at 2v_esc: E = +3GMm/R, all of it kinetic far away, so v_far = √3 v_esc.

Escaping costs the capsule a fixed amount of kinetic energy, GMm/R, not a fixed amount of speed, so it keeps √3 v_esc rather than v_esc. The same loss of kinetic energy takes less speed from a faster capsule, because kinetic energy grows as the square of the speed.

Scoring (2 points):

▸ D1 — 1 point: For indicating v_far > v_esc.

▸ D2 — 1 point: For one of the following justifications that the capsule is moving faster than v_esc when far away: that its kinetic energy there, ½m(2v_esc)² − GMm/R = 3(½mv_esc²), is greater than ½mv_esc² (in speeds, v_far² = (2v_esc)² − v_esc² = 3v_esc²); that escaping removes the same kinetic energy, GMm/R = ½mv_esc², as it does from the capsule launched at v_esc, and that, because kinetic energy is proportional to the square of the speed, this takes less than v_esc of speed from a capsule moving faster than v_esc; or that the gravitational force at each distance is the same for both capsules, but the faster capsule spends less time at each distance, so it receives a smaller impulse than the capsule launched at v_esc, which loses all of its speed.

Scoring Note: The justification may be made with an energy equation, with energy bar charts (K at cut-off four times as tall as the U_G bar is deep, so E is three times that depth, all of it kinetic far away), with impulse, or in words. D1 and D2 do not depend on the responses in parts A to C.

⚠︎ Common error (partial credit): Indicates v_far > v_esc because the capsule “has more energy than it needs to escape”, so it is still moving when far away, with no comparison of its remaining kinetic energy with ½mv_esc² and no reason why it loses less speed than the capsule launched at v_esc — earns 1 of 2 points, D1. That shows only that v_far is greater than zero.

⚠︎ Common error (partial credit): Indicates v_far > v_esc because a capsule launched faster than the escape speed is not slowed at all, so v_far = 2v_esc — earns 1 of 2 points, D1. Callisto’s pull slows every capsule on the way out; this one loses GMm/R of kinetic energy, and its speed falls from 2v_esc to √3 v_esc.

✗ Common error (no credit): Indicates v_far = v_esc because escaping uses up v_esc of the capsule’s speed, leaving 2v_esc − v_esc. Escaping uses up a fixed amount of kinetic energy, not of speed, so no point is earned.

✗ Common error (no credit): Indicates v_far < v_esc because the graph in part C shows the speed falling to zero far from Callisto. That graph describes the capsule launched at exactly v_esc, for which the mechanical energy of the capsule–Callisto system is zero; for this launch it is positive, so no point is earned.

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