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FRQ: Orbital Speed and Orbital Radius — The Ceres Survey Orbiter

Assessments aligned to 2026 AP Physics 1 standards

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

▤ Scenario

A survey orbiter of mass m travels at constant speed in a circular orbit of radius r around Ceres, the largest body in the asteroid belt, as shown in Figure 1. Model Ceres as an airless sphere of mass M and radius R, and measure every orbital radius from the centre of Ceres. The orbiter moves counterclockwise in Figure 1, and point P is marked on its orbit.

Figure 1 — The orbiter’s circular orbit around Ceres (orbiter enlarged), with point P marked.

The orbiter’s engines are off, and it touches nothing. Treat the orbiter–Ceres system as isolated: ignore the gravitational forces exerted on the orbiter and on Ceres by the Sun and by every other body outside the system. The orbiter’s mass is negligible compared with the mass of Ceres, so the motion of Ceres caused by the orbiter is negligible; use a reference frame in which the centre of Ceres is at rest, and treat it as inertial. All motion is in the plane of the orbit shown in Figure 1.

✎ Free Response Questions

Part A

Consider the orbiter at the instant it passes point P, shown in Figure 1.

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

Beside the dot, draw and label a separate arrow that shows the direction of the orbiter’s acceleration at that instant.

Figure 2 — The dot representing the orbiter at point P.

Part B

Starting with Newton’s second law, derive an expression for the orbital period T of the orbiter. Express your final answer in terms of m, M, r, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.

Part C

Orbiters can travel in circular orbits of different radii r around Ceres, each as described in the Scenario. Let v_R be the orbital speed of an orbiter in a circular orbit that just clears the surface of Ceres; take the radius of that orbit to be R. Figure 3 shows axes for the orbital speed v of an orbiter against the radius r of its circular orbit.

On the axes shown in Figure 3, sketch a line or curve that represents v as a function of r from r = R to r = 5R.

Figure 3 — Axes for part C: orbital speed v against the radius r of the circular orbit.

Part D

Suppose instead that the orbiters travel in circular orbits around a hypothetical body with the same radius R as Ceres but four times its mass, 4M, and that everything else is as described in the Scenario. Predict how your graph from part C would change, stating the factor by which the orbital speed at each radius would change, and predict the factor by which the orbital period at a given orbital radius r would change. 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 orbiter at P: the gravitational force exerted on it by Ceres, directed from the orbiter toward the centre of Ceres, to the left in Figure 2. The engines are off, the orbiter touches nothing, Ceres is airless and every other body is ignored. There is no forward force, since nothing pushes the orbiter along its path, and there is no separate “centripetal force” to draw. The gravitational force is itself the centripetal force, so it is drawn only once.

The acceleration points the same way as the net force, toward the centre of Ceres, perpendicular to the orbiter’s velocity. The orbiter’s speed is constant, but the direction of its velocity changes all the time, so the orbiter is accelerating.

Figure 4 — One force, toward the centre of Ceres; the acceleration points the same way.

Scoring (3 points):

▸ A1 — 1 point: For drawing exactly one arrow starting on the dot, which represents a force and is directed toward the centre of Ceres (to the left in Figure 2).

Scoring Note: Arrows of any nonzero length can earn A1. Every arrow that starts on the dot is counted. An arrow represents a force unless it is labelled only as a velocity, as “motion”, or as the orbiter’s acceleration, a or a_c; an unlabelled arrow, or one with any other label, including a label not appropriate for A2 (such as g, a_g, G or F_c), represents a force for A1.

▸ A2 — 1 point: For labelling an arrow directed toward the centre of Ceres as the gravitational force.

Scoring Note: Appropriate labels include F_g, F_G, F_grav, “gravitational force”, W, “weight”, mg (with g read as the gravitational field strength at the orbiter’s position), GMm/r², or a label naming Ceres, such as F_Ceres on orbiter. The labels F_c and “centripetal force” are not appropriate (they name the role of a net force in circular motion, not a force exerted by an object), nor are F_net, ma, g, a_g, G or a; a label that also names the gravitational force, such as F_g = F_c, is appropriate. The response is not required to name the object exerting the force.

▸ A3 — 1 point: For a separate arrow, labelled as the acceleration, directed toward the centre of Ceres.

Scoring Note: A3 may be earned by an acceleration arrow drawn beside the dot or starting on it; one that starts on the dot also counts as an additional arrow for A1. An arrow counted as a force for A1 cannot also earn A3. A1 to A3 do not depend on any other part.

⚠︎ Common error (partial credit): Draws the gravitational force correctly and the acceleration toward the centre, but puts a second arrow on the dot: a “centripetal force” toward the centre, a “thrust” or “force of motion” along the orbiter’s path, or the acceleration arrow itself — earns 2 of 3 points, A2 and A3. The gravitational force is the centripetal force, so a separate centripetal arrow counts it twice; nothing pushes the orbiter along its path; and the acceleration is not a force, so it is drawn beside the dot.

⚠︎ Common error (partial credit): Draws one arrow from the dot toward the centre of Ceres and the acceleration correctly, but labels the arrow F_c or “centripetal force” — earns 2 of 3 points, A1 and A3. The force is gravitational; “centripetal” names the direction of the net force, not a force of its own.

⚠︎ Common error (partial credit): Draws the gravitational force correctly but writes a = 0, or draws no acceleration arrow, because the orbiter’s speed is constant — earns 2 of 3 points, A1 and A2. Constant speed is not constant velocity: the direction of the velocity changes all the time, so the orbiter accelerates toward the centre.

⚠︎ Common error (partial credit): Draws the gravitational force toward the centre and an equal arrow away from it labelled “centrifugal force”, and writes a = 0 because the two forces balance — earns 1 of 3 points, A2. In an inertial frame nothing pushes the orbiter outward, and with balanced forces it would move in a straight line, not a circle.

✗ Common error (no credit): Draws one arrow along the orbiter’s direction of motion (up the page at P), labelled “thrust”, “force of motion” or “velocity”, and an acceleration arrow the same way, with no arrow toward the centre of Ceres. Velocity is not a force and no force acts along the path, so no point is earned.

Part B — Model Answer

By part A, the gravitational force exerted by Ceres is the only force on the orbiter, so it is the net force, and it produces the centripetal acceleration v²/r. Newton’s second law gives

F_net = ma_c

GMm/r² = mv²/r

v = √(GM/r)

In one period the orbiter travels once around its orbit, a distance 2πr, at constant speed, so v = 2πr/T. Setting the two expressions for v equal and solving for T:

2πr/T = √(GM/r)

T = 2πr√(r/GM) = 2π√(r³/GM)

The orbiter’s mass cancels, so the period depends only on the mass of Ceres and the radius of the orbit. Squaring gives T² = 4π²r³/GM, Kepler’s third law for circular orbits: T² is proportional to r³, with a constant of proportionality, 4π²/GM, set by the mass of the central body.

Scoring (3 points):

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

Scoring Note: Newton’s second law is applied by an equation that sets an expression for the gravitational force exerted on the orbiter equal to m times an expression for its centripetal acceleration, such as GMm/r² = mv²/r or GMm/r² = m(4π²r/T²), or an expression for the gravitational field strength equal to an expression for the centripetal acceleration; a statement of the law in words counts only if such an equation follows. Such an equation meets B1 on its own, even if the steps that follow are incorrect; F_net = ma with nothing substituted does not. A response that contains such an equation is a multistep derivation; a response that contains none is isolated, whether its final expression for T is preceded by words, by equations from the reference information, or by expressions for the force or the acceleration that are never combined.

▸ B2 — 1 point: For relating the orbital speed to the period, v = 2πr/T, or writing the centripetal acceleration in terms of the period, a_c = 4π²r/T², or any equivalent relation, such as T = 2πr/v, ω = 2π/T with v = rω or a_c = ω²r, or v = 2πrf with f = 1/T.

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

Scoring Note: A correct, isolated, final expression for T earns points B2 and B3. Any algebraic equivalent is accepted, such as T = 2πr√(r/GM), T = 2πr/√(GM/r) or T² = 4π²r³/GM. B1 to B3 do not depend on the response in part A.

⚠︎ Common error (partial credit): Starts correctly with GMm/r² = mv²/r and uses v = 2πr/T, but loses one or two factors of r in rearranging, reaching T = 2π√(r²/GM) or T = 2π√(r/GM) — earns 2 of 3 points, B1 and B2. A check of units catches the slip: GM has units of m³/s², so r³/GM is in s² and only √(r³/GM) is a time.

⚠︎ Common error (partial credit): Writes T² = 4π²r³/GM from memory as Kepler’s third law, with no derivation — earns 2 of 3 points, B2 and B3. The relationship is derived, not given in the reference information, so it is where the derivation must arrive rather than where it may begin.

⚠︎ Common error (partial credit): Sets ½mv² equal to GMm/r, as if for escape, reaching v = √(2GM/r) and then T = 2πr/v = 2π√(r³/2GM) — earns 1 of 3 points, B2. The energy that escape would require says nothing about the speed of a circular orbit; Newton’s second law with the centripetal acceleration does.

Part C — Model Answer

From part B, v = √(GM/r). At r = R this is v_R = √(GM/R), so v = v_R√(R/r), and v is proportional to 1/√r:

r = R:  v = v_R         r = 4R:  v = 0.50v_R

r = 5R:  v = v_R/√5 ≈ 0.45v_R

Figure 5 — v = v_R√(R/r): the speed falls from v_R at r = R to about 0.45v_R at r = 5R.

The orbital speed is proportional to 1/√r, so quadrupling the radius of an orbit halves the speed. The curve starts at its largest value at r = R and decreases as r increases, falling steeply at first and then more gradually: it is concave up and never reaches zero over the plotted range. By r = 5R the speed has fallen to about 45 percent of its value at r = R.

Scoring (3 points):

▸ C1 — 1 point: For a curve that decreases throughout from r = R to r = 5R.

Scoring Note: A curve that falls throughout from r = R to r = 4R and is lower at r = 5R than at r = 4R counts as decreasing throughout, even if it looks level between 4R and 5R; the correct curve falls by only about half a grid division there.

▸ C2 — 1 point: For a curve that falls less steeply as r increases, steepest near r = R, so that it is not a straight line.

▸ C3 — 1 point: For a line or curve that is at v_R at r = R and at a value from 0.40v_R to 0.50v_R at r = 5R.

Scoring Note: The correct value at r = 5R is v_R/√5 ≈ 0.45v_R. A value within 0.05v_R (half a grid division) of v_R at r = R counts as v_R, and a line or curve drawn beyond the range R to 5R is judged only over that range, for C1 to 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 from v_R at r = R to 0.20v_R at r = 5R — earns 2 of 3 points, C1 and C2. It is v², GM/r, that is proportional to 1/r; the speed itself falls only as 1/√r.

⚠︎ Common error (partial credit): Draws a straight line from v_R at r = R down to zero at r = 5R, as if the speed dropped by the same amount for each metre of radius — earns 1 of 3 points, C1. v = v_R√(R/r) falls less and less steeply as r grows and is never zero at a finite radius.

✗ Common error (no credit): Draws a curve that rises from v_R at r = R, because a larger orbit is longer and so must be covered faster. A larger orbit is covered more slowly: the weaker pull at larger r holds only a slower orbiter in a circle, and the period grows because the path is longer and the speed is lower, so no point is earned.

Part D — Model Answer

Graph. Since v = √(GM/r), the speed at every radius is proportional to √M. Quadrupling M multiplies √M by 2, so every speed doubles: the whole curve is stretched vertically by a factor of 2, it begins at 2v_R at r = R, and its shape, decreasing and concave up, is unchanged. On the axes of Figure 3, the new curve would begin above the top of the grid and come down into it near r = 1.6R.

Period. From part B, T = 2π√(r³/GM), so the period at a given radius is proportional to 1/√M. Quadrupling M therefore halves the orbital period at every radius. This agrees with the speed result: the orbiter covers the same circumference 2πr at twice the speed, so it takes half the time.

Both the speed and the period depend on the square root of the central mass, so four times the mass doubles the speed and halves the period.

Figure 6 — Around mass 4M (solid), every orbital speed is twice that around mass M (dashed).

Scoring (3 points):

▸ D1 — 1 point: For predicting that the orbital speed at every radius doubles, so that every point of the curve moves to twice its height (a vertical stretch by a factor of 2; the curve is still decreasing and concave up); or, where the response’s own part B work gives the speed a different dependence on M, the factor that follows correctly from it (from v = 2πr/T with its expression for T, or, where part B has no expression for T, from its expression for v).

Scoring Note: No description of the graph beyond the speed factor is required, and “the new curve is steeper” is correct.

▸ D2 — 1 point: For justifying the speed factor by how the speed depends on M: that v is proportional to √M, from v = √(GM/r); that the gravitational force at each radius, and with it v²/r, is four times as large, so v² quadruples and v doubles; or that T is proportional to 1/√M, so the period at each radius halves and v = 2πr/T doubles; or that dependence, taken as in D1, in the response’s own part B work.

▸ D3 — 1 point: For predicting that the orbital period at a given radius is halved, with a reason: that T = 2π√(r³/GM) is proportional to 1/√M, or that the orbiter covers the same circumference 2πr at twice the speed; or the factor that follows correctly from the dependence on M of the response’s own part B expression for T (or, where part B has no expression for T, of T = 2πr/v with its expression for v), with that dependence as the reason.

Scoring Note: A factor must carry its direction, such as “multiplied by 2” or “increases by a factor of 2” for the speed and “multiplied by ½” or “decreases by a factor of 2” for the period; “changes by a factor of 2” alone does not meet D1 or D3.

⚠︎ Common error (partial credit): Predicts from part B that the period is halved, because T is proportional to 1/√M, but has the speeds rise by a factor of 4, overlooking the square root in v = √(GM/r) — earns 1 of 3 points, D3. The two predictions contradict each other: half the period around the same circle means twice the speed, not four times.

⚠︎ Common error (partial credit): Predicts that every speed doubles and, as a separate claim not tied to the doubled speed, that the period halves, justifying both only by “more mass means a stronger pull, so the orbiter goes faster”, with no relationship quoted — earns 1 of 3 points, D1. A stronger pull explains why the speed rises, not why it doubles rather than quadruples.

✗ Common error (no credit): With a correct expression for T in part B, predicts that every speed quadruples and the period is quartered because the mass is four times as large. Both quantities depend on the square root of M, so no point is earned.

✗ Common error (no credit): With a correct expression for T in part B, predicts no change because the orbiter’s mass is unchanged. The orbiter’s mass cancels; it is the central mass M that sets both v and T, so no point is earned.

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