How Does a Relay Satellite Move in an Elliptical Orbit? FRQ
Pause here: To build true exam stamina, attempt these questions on your own before checking the model answers.
FRQ: Angular Momentum in an Elliptical Orbit — The Vyom-1 Relay Satellite
Assessments aligned to 2026 AP Physics 1 standards
Translation Between Representations (TBR) | CHALLENGE | 12 points | 30 min
▤ Scenario
The relay satellite Vyom-1 orbits the planet Dhruva in an ellipse, shown in Figure 1. Three points of the orbit are marked: P, the point closest to Dhruva, where the satellite is a distance r_P = 2.0 × 10⁷ m from Dhruva’s centre; A, the point farthest from Dhruva, at r_A = 8.0 × 10⁷ m; and Q, an intermediate point between them. The satellite’s speed at P is v_P, its speed at A is v_A, and its orbital period is T. At P and at A, and only there, the satellite’s velocity is perpendicular to its position vector from Dhruva’s centre.
Figure 1 — The orbit of Vyom-1 (distances to scale, Dhruva not to scale), with P, Q and A marked.
The orbit lies in a single plane. The satellite travels counterclockwise, and the counterclockwise sense is taken as positive. Its thrusters are off throughout, atmospheric drag is negligible at every point of the orbit, and the gravitational forces exerted on the satellite by every body other than Dhruva are negligible. Treat the satellite as a particle of constant mass m. All quantities are measured in a reference frame in which Dhruva’s centre is at rest, which may be treated as inertial, and all distances are measured from Dhruva’s centre.
✎ Free Response Questions
Part A
Figure 2 shows the satellite at P, at Q and at A as dots on grids. Each dotted line runs from a dot toward Dhruva’s centre. The side of each grid square represents a force of magnitude GMm / r_A², where M is the mass of Dhruva.
On the dots in Figure 2 that represent the satellite at P, at Q and at A, draw and label the forces (not components) that are exerted on the satellite.
• 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 2.
Figure 2 — The satellite at P, at Q and at A; one grid square is a force of magnitude GMm/r_A².
Part B
Starting with conservation of the satellite’s angular momentum about Dhruva’s centre, derive an expression for the ratio K_P / K_A of the satellite’s kinetic energy K_P at P to its kinetic energy K_A at A. Express your final answer in terms of m, r_P, r_A, and physical constants, as appropriate. Begin your derivation by writing the fundamental physics principle or an equation from the reference information.
Part C
The satellite passes P at t = 0 and reaches A at t = T/2. Figure 3 shows axes for the satellite’s angular momentum L about Dhruva’s centre and for its speed v, each as a function of time t.
On the axes shown in Figure 3, do the following for t = 0 to t = T.
• Sketch a line or curve that represents L as a function of t.
• Sketch a line or curve that represents v as a function of t.
Figure 3 — Axes for part C: angular momentum L and speed v against time t.
Part D
Justify how your sketch of L in part C is or is not consistent with the forces you drew in part A.
❖ 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
Dhruva’s gravitational pull is the only force exerted on the satellite at each of the three points. The thrusters are off, drag is negligible, and so are the pulls of all other bodies, so each dot carries exactly one arrow, directed along the dotted line toward Dhruva’s centre (Figure 4).
By Newton’s law of universal gravitation, the force at a distance r is F_g = GMm / r². Writing F_A = GMm / r_A² for the force at A, one grid square, the force at a distance r is F_A multiplied by (r_A / r)². At P, (r_A / r_P)² = (8.0 × 10⁷ m / 2.0 × 10⁷ m)² = 16, so the arrow at P is sixteen grid squares long. The force varies as 1/r², so at a quarter of the distance it is sixteen times as large. Q is farther than P and closer than A, so its arrow is longer than the arrow at A and shorter than the arrow at P. As drawn in Figure 1, Q is 5.0 × 10⁷ m from Dhruva’s centre, where the force is (8.0 / 5.0)² ≈ 2.6 times F_A.
Figure 4 — One force at each point, toward Dhruva’s centre: sixteen squares at P, one at A.
Scoring (3 points):
▸ A1 — 1 point: For drawing, on each dot, exactly one arrow, which starts on the dot and is directed toward Dhruva’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 Dhruva’s centre (at Q, any arrow directed below the horizontal in Figure 2, into the orbit) 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 satellite’s position), an expression of the form GMm / r² with any distance (such as GMm / r_P², or 16GMm / r_A² at P; the magnitude a label states is not scored, since A3 scores the arrow’s length), or a label naming Dhruva, such as F_Dhruva on satellite; a magnitude may accompany any of these, as in F_g = 16F_A. 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, or a magnitude alone, such as a number of squares, F_A or 16F_A. The response is not required to name the object exerting the force.
▸ A3 — 1 point: For drawing, on each dot, an arrow directed toward Dhruva’s centre (at Q, any arrow directed below the horizontal in Figure 2) whose length is one grid square at A and sixteen grid squares at P, with the arrow at Q longer than the arrow at A and shorter than the arrow at P.
Scoring Note: A length within half a grid square of one square at A, and within one grid square of sixteen squares at P, meets A3. Where a dot also carries a velocity, acceleration or motion arrow directed toward Dhruva’s centre, A3 is 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 Dhruva’s centre on each dot, but scales the force as 1/r, drawing four squares at P, or draws the three arrows the same length because the satellite’s mass does not change — 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: a quarter of the distance gives sixteen times the force, not four.
⚠︎ Common error (partial credit): Draws the gravitational arrows correctly but adds a second arrow on a dot — a “thrust” or “force of motion” along the satellite’s path, a separate “centripetal force”, or an outward “centrifugal force” — earns 2 of 3 points, A2 and A3. No force is needed to keep the satellite moving; Dhruva’s pull, the only force, changes the direction and the speed of its motion. “Centripetal” describes the direction of a net force, not an extra force, and in an inertial frame nothing pushes the satellite outward.
⚠︎ Common error (partial credit): Draws one arrow on each dot, labelled as the gravitational force, one square long at A, sixteen at P and in between at Q, but draws the arrow at Q perpendicular to the satellite’s path (straight down in Figure 2), as though the pull were always at right angles to the motion — earns 2 of 3 points, A2 and A3. The pull points along the line to Dhruva’s centre at every point; only at P and at A is that line perpendicular to the path.
⚠︎ Common error (partial credit): Draws one arrow toward Dhruva’s centre on each dot, one square long at A, sixteen at P and in between at Q, but labels the arrows F_c or “centripetal force” — earns 2 of 3 points, A1 and A3. The force is gravitational; “centripetal” names no force of its own.
✗ Common error (no credit): Draws each arrow along the satellite’s path, in the direction of its motion, as the force that keeps it moving around the orbit, with no arrow toward Dhruva’s centre. At P and at A, A1 to A3 all need an arrow toward Dhruva’s centre, and there is none, so no point is earned.
Part B — Model Answer
The satellite’s angular momentum about Dhruva’s centre is conserved, so its value at P equals its value at A:
L_P = L_A
The magnitude of the satellite’s angular momentum about Dhruva’s centre is L = rmv sin θ, where θ is the angle between its position vector from Dhruva’s centre and its velocity. At P and at A, and only there, θ = 90° and sin θ = 1, so
r_P m v_P = r_A m v_A v_P / v_A = r_A / r_P
With K = ½mv² at each point, the ½ and the mass cancel in the ratio:
K_P / K_A = (½ m v_P²) / (½ m v_A²)
= (v_P / v_A)² = (r_A / r_P)²
Kinetic energy goes as the square of the speed, so the ratio of the kinetic energies is the square of r_A / r_P. With the given distances, K_P / K_A = (8.0 × 10⁷ m / 2.0 × 10⁷ m)² = 16: at P the satellite moves four times as fast as at A and has sixteen times the kinetic energy.
Scoring (3 points):
▸ B1 — 1 point: For a multistep derivation that includes conservation of angular momentum: an equation setting the satellite’s angular momentum about Dhruva’s centre at P equal to its angular momentum about Dhruva’s centre at A, in any form. This point is earned for the starting principle alone, even if the steps that follow are incorrect.
Scoring Note: A final expression preceded only by words, or only by unapplied equations from the reference information, is not a multistep derivation. B1 and B2 carry no consistency credit, and B1 to B3 do not depend on the response in part A.
▸ B2 — 1 point: For indicating that the satellite’s angular momentum at P and at A is rmv (L = rmv sin θ with sin θ = 1 there), for example by writing r_P m v_P = r_A m v_A, v_P / v_A = r_A / r_P, rmv sin θ at both points with the same angle, which then cancels, or L = Iω with I = mr² and ω = v/r. A reason is not required.
▸ B3 — 1 point: For a correct expression for the ratio, K_P / K_A = (r_A / r_P)² or an algebraic equivalent, consistent with the indicated angular-momentum relation. Where B2 was not earned, an expression that follows correctly from the response’s own incorrect expression for the angular momentum also earns this point, provided that expression includes the satellite’s distance from Dhruva’s centre, is used in the same form at P and at A, and gives a result in terms of r_P and r_A.
Scoring Note: A correct, isolated, final expression for K_P / K_A earns points B2 and B3. A final expression is isolated when everything before it is words or unapplied equations from the reference information. A number with the given distances substituted, such as 16, is not an expression in terms of r_P and r_A; evaluating a correct expression afterwards, as in (r_A / r_P)² = 16, does not lose B3.
⚠︎ Common error (partial credit): Writes r_P m v_P = r_A m v_A correctly but inverts the ratio when solving, reaching K_P / K_A = (r_P / r_A)² — earns 2 of 3 points, B1 and B2. That would make the satellite slower at P, the closer point, than at A.
⚠︎ Common error (partial credit): Reaches v_P / v_A = r_A / r_P but writes K_P / K_A = r_A / r_P, as though kinetic energy were proportional to speed — earns 2 of 3 points, B1 and B2. K = ½mv² goes as the square of the speed: a speed ratio of 4 is a kinetic-energy ratio of 16.
⚠︎ Common error (partial credit): Writes the angular momentum as m r² v, the rotational inertia mr² multiplied by the speed instead of by the angular speed v/r, reaching K_P / K_A = (r_A / r_P)⁴ — earns 2 of 3 points, B1 and B3. For this orbit that is 256, not 16.
⚠︎ Common error (partial credit): Writes r_P m v_P = r_A m v_A and then substitutes the given distances without ever writing K_P / K_A in terms of r_P and r_A, answering 16 — earns 2 of 3 points, B1 and B2. The number is right for this orbit, but part B asks for an expression in terms of r_P and r_A.
⚠︎ Common error (partial credit): Writes L_P = L_A but takes the angular momentum to be mv, with no distance in it, so that v_P = v_A and K_P / K_A = 1 — earns 1 of 3 points, B1. Angular momentum about Dhruva’s centre depends on the satellite’s distance from it: at P and at A it is rmv.
✗ Common error (no credit): Starts from K_P = K_A, “because energy is conserved”, with no angular momentum equation. The satellite’s kinetic energy is not constant; the mechanical energy of the satellite–Dhruva system is, as the sum K + U_G, and U_G changes with the distance, so no point is earned.
Part C — Model Answer
The satellite’s angular momentum about Dhruva’s centre does not change (the principle part B started from), so L is a horizontal line from t = 0 to t = T. It is positive, because the satellite travels counterclockwise, and it is not zero: at P, for example, L = r_P m v_P.
The speed is greatest at P, the closest point, and least at A, the farthest. From part B, v_P / v_A = r_A / r_P = 4, so the curve starts at 4v_A at t = 0, falls to v_A when the satellite reaches A at t = T/2, and rises back to 4v_A at t = T, when it is again at P. On the way out, the satellite moves away from Dhruva and part of the pull acts against its motion, so it slows all the way from P to A; on the way back it speeds up all the way. At P and at A the pull is perpendicular to the motion, so at those instants it neither slows the satellite down nor speeds it up: the speed stops changing for a moment, so the curve is level at t = 0 and t = T and levels off into a rounded minimum at t = T/2 rather than meeting it at a corner. With L = rmv sin θ fixed, the speed falls fourfold from P to A because the distance from Dhruva’s centre grows fourfold.
Figure 5 — L is constant; v (schematic) falls from 4v_A at P to v_A at A and back, never zero.
Figure 5’s shape between the scored instants is schematic; any smooth curve with the features scored below earns full credit.
Scoring (4 points):
▸ C1 — 1 point: For sketching L as a horizontal line at a positive, nonzero value from t = 0 to t = T.
Scoring Note: The height of the line is not scored. C1 does not depend on any other part.
▸ C2 — 1 point: For sketching v as a curve that decreases from t = 0 to t = T/2 and increases from t = T/2 to t = T, so that its greatest values are at t = 0 and t = T and its least value is at t = T/2.
▸ C3 — 1 point: For a v curve at v_A at t = T/2 and, at both t = 0 and t = T, either at 4v_A or at the speed at P that follows from the response’s own part B — v_A multiplied by its ratio v_P / v_A where it states one, otherwise v_A√(K_P / K_A) from its expression for K_P / K_A, each evaluated with the given distances. Where that speed is greater than 5v_A, a curve that runs off the top of the axes at t = 0 and at t = T meets the requirement at those two instants; the curve must still be at v_A at t = T/2. An expression that does not evaluate to a pure number gives no speed at P.
Scoring Note: Each value within 0.25 v_A (half a grid division) meets C3. C1, C2 and C4 do not depend on the response in part B.
▸ C4 — 1 point: For a v curve that is horizontal (has zero slope) at its least value, so that the minimum is rounded, not a corner. A curve that does not fall to a least value and rise again, such as a horizontal line, does not earn C4.
⚠︎ Common error (partial credit): Draws L correctly and v as two straight lines meeting in a point at t = T/2, with the correct values — earns 3 of 4 points, C1, C2 and C3. At A the pull is perpendicular to the motion, so the speed stops changing for an instant; a corner would mean the pull switched from slowing the satellite to speeding it up all at once.
⚠︎ Common error (partial credit): Draws L correctly and v falling smoothly to zero at t = T/2, as though the satellite stopped at its farthest point like a ball at the top of its flight — earns 3 of 4 points, C1, C2 and C4. A satellite at rest would have zero angular momentum about Dhruva’s centre, but L never changes and is not zero at P; at A the satellite is at its slowest, not at rest.
⚠︎ Common error (partial credit): Draws v correctly but draws L falling and rising with the speed, as though angular momentum depended on the speed alone — earns 3 of 4 points, C2, C3 and C4. L = rmv sin θ: where the satellite is slower, it is also farther from Dhruva’s centre, and L does not change.
⚠︎ Common error (partial credit): Draws v correctly but draws L as a line along the time axis, at zero, because Dhruva’s pull exerts no torque — earns 3 of 4 points, C2, C3 and C4. Zero net torque means that L does not change, not that L is zero.
⚠︎ Common error (partial credit): After a correct part B, draws both L and v as horizontal lines, as though the orbit were a circle — earns 1 of 4 points, C1. Only on a circular orbit does the speed stay constant; here the distance from Dhruva changes fourfold between P and A, and the speed changes with it.
Part D — Model Answer
The sketch of L in part C is consistent with the forces in part A.
In part A, the only force on the satellite, at P, at Q and at A, is directed along the line from the satellite toward Dhruva’s centre. Its line of action therefore passes through the centre, and its lever arm about the centre is zero: the angle between the satellite’s position vector r and the force is 180°, so
τ = rF sin 180° = 0
This holds at every point of the orbit, including Q, where the force is not perpendicular to the satellite’s velocity (Figure 6): the torque depends on the angle between the force and the position vector.
Figure 6 — At Q the pull lies along the line to Dhruva’s centre; v is not perpendicular to r.
The net torque on the satellite about Dhruva’s centre is the slope of its L–t graph, and the angular impulse ΔL = τΔt is zero over every interval of the orbit. A force whose line of action passes through Dhruva’s centre exerts no torque about it, so the satellite’s angular momentum about the centre cannot change. A horizontal L line, with zero slope everywhere, is exactly what the forces in part A require.
(Beyond what part D asks.) The point matters: about the middle of the long axis, the pull misses that point except at P and at A, so L about it changes.
Scoring (2 points):
▸ D1 — 1 point: For using the forces drawn in part A to determine the torque each exerts about Dhruva’s centre, stating whichever of the following applies to the response’s own forces: a force whose line of action passes through the centre, directed toward or away from it, exerts no torque about the centre (its lever arm is zero, the angle between the position vector and the force is 180° or 0°, or it has no component perpendicular to the position vector); a force whose line of action misses the centre, such as a force along the satellite’s path, exerts a torque about it. A response whose forces all pass through the centre need not discuss any other force. Where part A shows no forces, making this determination for the gravitational force exerted by Dhruva earns this point.
▸ D2 — 1 point: For stating the relation between the net torque about Dhruva’s centre and L that applies to the response’s own forces — zero net torque means that L does not change, or a nonzero net torque means that L changes (for example, the net torque is the slope of the L–t graph, or ΔL = τΔt) — together with the verdict it gives for the response’s own sketch of L in part C. Where every force in part A has a line of action through Dhruva’s centre, or part A shows no forces, a horizontal L line at any value is consistent and a line or curve along which L changes is not. Where part A shows a force whose line of action misses the centre, a horizontal L line is not consistent, and an L that changes is consistent only if it increases wherever that torque is counterclockwise (as for a force along the satellite’s motion) and decreases wherever it is clockwise. Where part C shows no sketch of L, the verdict is a statement that L must be constant, or must change, as the torque requires. A verdict with no statement linking the torque to the change in L does not earn D2; “there is no torque, so L is conserved” is such a link.
Scoring Note: D1 and D2 are judged against the response’s own forces in part A and its own sketch of L in part C.
⚠︎ Common error (partial credit): With the forces in part A drawn toward Dhruva’s centre and a horizontal L line in part C, states that the sketch is consistent because Dhruva’s pull is perpendicular to the satellite’s velocity, so it exerts no torque and L stays constant — earns 1 of 2 points, D2. The pull is perpendicular to the velocity only at P and at A; at Q it is not, yet the torque is still zero, because the torque depends on the angle between the force and the position vector.
⚠︎ Common error (partial credit): With the forces in part A drawn toward Dhruva’s centre and a horizontal L line in part C, states that the sketch is consistent because there is no torque on the satellite, the orbit being stable or the net force on it being zero, so L does not change — earns 1 of 2 points, D2. The net force is not zero (the satellite is accelerating); the torque is zero because of where the force points, along the line through Dhruva’s centre.
⚠︎ Common error (partial credit): States correctly that each force in part A passes through Dhruva’s centre and so exerts no torque about it, but draws no conclusion about L or about the sketch in part C — earns 1 of 2 points, D1.
✗ Common error (no credit): Having drawn the pull at Q perpendicular to the path in part A, states that the horizontal L line in part C is consistent because Dhruva’s pull is perpendicular to the satellite’s velocity, so it exerts no torque. The response’s own force at Q misses Dhruva’s centre and exerts a torque about it, so neither its torque nor its verdict follows from its own forces, and no point is earned.
✗ Common error (no credit): States that the sketch is consistent because angular momentum is always conserved, with nothing said about torque. Angular momentum is constant only when the net external torque about the chosen point is zero; here it is the direction of Dhruva’s pull, through the centre, that makes the torque zero, so no point is earned.
✗ Common error (no credit): Sketches L falling from P to A and calls it consistent because the satellite slows down, so its angular momentum must decrease, with no torque considered. L = rmv sin θ does not change: the slower satellite is also farther away, so no point is earned.
