6.6 Motion of Orbiting Satellites

1. AEach exerts an equal force on the other, but the planet's acceleration is tiny. Correct
   The gravitational forces the satellite and planet exert on each other are a Newton's third-law pair, equal in magnitude. By Newton's second law the planet's acceleration is that force divided by its mass, about 10²⁰ times smaller than the satellite's, so its motion is negligible.
2. BThe planet pulls on the satellite much harder than the satellite pulls on it.
   A student who thinks the more massive object exerts the larger force picks this. The two forces are a third-law pair and have equal magnitudes; the difference is in the accelerations they produce.
3. CThe satellite's mass is far too small for it to exert a gravitational force at all.
   A student who thinks only massive bodies exert gravitational forces picks this. The satellite pulls on the planet with a force equal in magnitude to the planet's pull on it, G Mm/r².
4. DIn orbit the satellite is weightless, so the two of them exert no forces on each other.
   A student who thinks there is no gravity in orbit picks this. Gravity is the force that keeps the satellite in orbit; the satellite is only apparently weightless, because gravity is the only force on it.

CED 6.6.A.1 · Read this in Fix

1. Athe system's total mechanical energy
   A student who equates the system's total energy with the satellite's kinetic energy picks this. With only gravitational forces between the two objects, the total mechanical energy is the same at P and A; the larger K at P is balanced by a lower U_g.
2. Bthe satellite's angular momentum
   A student who thinks angular momentum increases with speed picks this. The satellite's angular momentum about the planet is constant; at P its larger speed is matched by its smaller distance, so rv is the same at P and A.
3. Cthe system's potential energy U_g
   A student who compares the sizes of the negative values picks this, since |U_g| is larger at P. U_g = −GMm/r is more negative, so smaller, at P; it is greater at A, farther from the planet.
4. Dthe kinetic energy of the satellite Correct
   P is much closer to the planet than A. The system's mechanical energy is constant, and U_g = −GMm/r is lower (more negative) at P, so K is greater at P: the satellite moves fastest at its closest point.

CED 6.6.A.2.ii · Read this in Fix

1. A16.0K₀
   A student who thinks a satellite farther out moves faster, as a point on a rotating rigid object does (v = rω), takes the speed to be four times as large at 4r and the kinetic energy (1/2)mv² to be 16 times as large. In a larger circular orbit the satellite moves more slowly: K = GMm/(2r) falls as r increases.
2. B0.06K₀
   A student who takes the kinetic energy to vary as 1/r², like the gravitational force, picks this: K₀/16 ≈ 0.06K₀. From K = −(1/2)U = GMm/(2r), the kinetic energy varies as 1/r.
3. C0.50K₀
   A student who finds that the speed halves (v = √(GM/r)) and takes kinetic energy as proportional to speed picks this. K = (1/2)mv², so halving v divides K by 4.
4. D0.25K₀ Correct
   In a circular orbit K = −(1/2)U = GMm/(2r), which is proportional to 1/r, so four times the radius gives one-quarter of the kinetic energy.

Working For a circular orbit, K = −(1/2)U = GMm/(2r), so K ∝ 1/r. At radius 4r, K = K₀/4 = 0.25K₀. (Equivalently v = √(GM/r) halves, and K ∝ v² falls by 4.)

CED 6.6.A.3 · Read this in Fix

1. AAt launch, its kinetic energy equals the magnitude of the system's U_g. Correct
   At escape speed the mechanical energy of the probe–planet system is zero: K + U_g = 0. Since U_g = −GMm/R is negative, the probe's kinetic energy at launch equals |U_g|, GMm/R.
2. BBeyond some distance from the planet, the planet's gravity stops acting on it.
   A student who thinks gravity has a limited range picks this. The planet's gravitational force on the probe weakens as 1/r² but acts at every distance, so the probe keeps slowing down.
3. CIts launch speed equals the speed of a circular orbit just above the surface.
   A student who confuses escape with going into orbit picks this. A circular orbit at the surface needs √(GM/R), giving negative total energy; escape needs zero total energy, √(2GM/R), √2 times as fast.
4. DA probe of greater mass would need a greater launch speed to escape.
   A student who thinks a heavier probe is harder to launch away picks this. The probe's mass multiplies both K and U_g, so it cancels in (1/2)mv² − GMm/R = 0: the escape speed is the same for any probe.

CED 6.6.A.4 · Read this in Fix

1. AIts speed does not change as it goes around the orbit, so its angular momentum cannot change.
   A student who thinks a satellite's speed is constant in any orbit picks this. In an elliptical orbit the speed does change; angular momentum is constant because gravity exerts no torque about the center, and that is why the satellite moves faster when it is closer.
2. BGravity is perpendicular to its velocity at each point, so gravity does no work on the satellite.
   A student who thinks gravity is perpendicular to the velocity in every orbit picks this. That holds only in a circular orbit; in an ellipse gravity does work and changes the speed. In any case, zero work concerns energy, not angular momentum.
3. CWhile it is in orbit it is weightless, so no net force acts to change its motion.
   A student who thinks there is no gravity in orbit picks this. Gravity acts on the satellite throughout and constantly changes its velocity; angular momentum is constant because that force exerts no torque about the planet's center.
4. DThe force on it points along the line to the planet's center, so it exerts no torque about that center. Correct
   Torque about the center is τ = rF sin θ, where θ is the angle between the position vector from the center and the force. Gravity acts along that line (θ = 180°), so the torque is zero at every point, and with zero net torque the angular momentum about the center is constant.

CED 6.6.A.2 · Read this in Fix

1. Athe satellite's kinetic energy
   A student who thinks a force acting on a moving object always changes its kinetic energy picks this. In a circular orbit gravity is perpendicular to the velocity, does no work, and the speed, so K = (1/2)mv², is the same at X and Y.
2. Bthe system's potential energy U_g
   A student who links potential energy to height on the page picks this, since X is drawn above the planet. U_g = −GMm/r depends only on the distance r from the planet's center, which is the same at every point of a circular orbit.
3. Cthe velocity of the satellite Correct
   At X the velocity points to the left and at Y up the page: the speed is the same, but velocity is a vector, so it differs. Every other listed quantity depends only on the speed or on the distance from the planet's center, which are the same at X and Y.
4. Dthe satellite's angular momentum
   A student who thinks angular momentum changes when the velocity changes direction picks this. About the planet's center L = rmv sin θ has the same magnitude at X and Y, and its direction, perpendicular to the orbit's plane, does not change.

CED 6.6.A.2.i · Read this in Fix

1. A8.2 × 10³ m/s
   A student who thinks a satellite keeps the same speed all around any orbit picks this. In an elliptical orbit the product rv is constant at perigee and apogee, so at twice the distance the speed is halved.
2. B4.1 × 10³ m/s Correct
   The satellite's angular momentum about Earth's center is constant. At perigee and apogee the velocity is perpendicular to the radius, so v_p r_p = v_a r_a and v_a = (8.2 × 10³)(8.0 × 10⁶)/(1.6 × 10⁷) = 4.1 × 10³ m/s.
3. C5.8 × 10³ m/s
   A student who uses the circular-orbit relation v = √(GM/r) at each point picks this: v_a = v_p√(r_p/r_a) = 5.8 × 10³ m/s. That relation holds only for circular orbits; on an ellipse angular momentum conservation gives v_a = v_p r_p/r_a.
4. D1.6 × 10⁴ m/s
   A student who thinks a satellite moves faster farther out, as points on a rigid rotating object do, picks this: v_p r_a/r_p. A satellite's angular momentum is constant, so it moves slower where it is farther away.

Working Gravity exerts no torque about Earth's center, so the satellite's angular momentum is constant. At perigee and apogee the velocity is perpendicular to the line from Earth's center, so L = mvr there: m v_p r_p = m v_a r_a. v_a = v_p r_p/r_a = (8.2 × 10³ m/s)(8.0 × 10⁶ m)/(1.6 × 10⁷ m) = 4.1 × 10³ m/s.

CED 6.6.A.2.ii · Read this in Fix

1. A−1.0 × 10¹² J
   A student who uses the altitude as r picks this: −GMm/(1.6 × 10⁶ m). In U_g = −GMm/r, r is the distance between the centers, Earth's radius plus the altitude, 8.0 × 10⁶ m.
2. B−2.0 × 10¹¹ J Correct
   The distance between the centers is r = 6.4 × 10⁶ + 1.6 × 10⁶ = 8.0 × 10⁶ m, and U_g = −GMm/r = −(6.67 × 10⁻¹¹)(6.0 × 10²⁴)(4000)/(8.0 × 10⁶) = −2.0 × 10¹¹ J, negative because the zero is at infinite separation.
3. C+2.0 × 10¹¹ J
   A student who writes the gravitational potential energy as +GMm/r picks this. With the zero at infinite separation, U_g is negative at every finite distance: work must be done on the system to separate the satellite and Earth.
4. D+6.4 × 10¹⁰ J
   A student who uses mgΔy over the whole altitude, with the surface value g ≈ 10 m/s², picks this: (4000)(10)(1.6 × 10⁶) = 6.4 × 10¹⁰ J. The field GM/r² falls as 1/r², so at r = 8.0 × 10⁶ m it is only (6.4/8.0)² ≈ 0.64 of its surface value and g is not constant over that distance; mgΔy also measures U from the surface, not from infinity. Use U_g = −GMm/r.

Working r = 6.4 × 10⁶ m + 1.6 × 10⁶ m = 8.0 × 10⁶ m from Earth's center. U_g = −GMm/r = −(6.67 × 10⁻¹¹ N·m²/kg²)(6.0 × 10²⁴ kg)(4000 kg)/(8.0 × 10⁶ m) = −2.0 × 10¹¹ J.

CED 6.6.A.2.iii · Read this in Fix

1. AGraph 1
   A student who writes the potential energy as +GMm/r picks the positive curve falling toward zero. With zero at infinity, U = −GMm/r is negative at every finite distance.
2. BGraph 2
   A student who uses U = mgΔy at every distance picks the straight line rising from zero at the surface. That treats the field as constant and puts the zero at the surface; for a satellite system U = −GMm/r, which levels off toward zero.
3. CGraph 3 Correct
   U = −GMm/r is negative for every finite r, has its lowest value, −GMm/R, at the surface, and rises toward zero as r increases, more and more slowly because its slope GMm/r² decreases.
4. DGraph 4
   A student who keeps the zero of potential energy at the planet's surface picks the curve rising from zero at R. That curve, GMm(1/R − 1/r), has the right shape but the wrong zero: the stem defines U = 0 at infinite separation.

Working U = −GMm/r: negative at every finite r, equal to −GMm/R at the surface, and increasing toward zero (from below) as r increases, with slope GMm/r² that decreases with r.

CED 6.6.A.2.iii · Read this in Fix

1. AGMm/(3r) Correct
   In a circular orbit of radius r′ the total energy is E = −GMm/(2r′). ΔE = −GMm/(6r) − (−GMm/(2r)) = GMm/(3r). The potential energy rises by 2GMm/(3r), but the kinetic energy falls by GMm/(3r).
2. B2GMm/(3r)
   A student who counts only the change in gravitational potential energy picks this: ΔU = GMm/r − GMm/(3r). The satellite's kinetic energy is lower in the larger orbit, by GMm/(3r), so the energy that must be added is less than ΔU.
3. C4GMm/(9r²)
   A student who writes the orbital energy with r² in the denominator, like the force, E = −GMm/(2r²), gets ΔE = GMm/(2r²) − GMm/(18r²) = 4GMm/(9r²). That expression has units of force, not energy; the total energy of a circular orbit is E = −GMm/(2r), which varies as 1/r.
4. D2GMm/r
   A student who treats the gravitational force as constant at its value at radius r, GMm/r², over the 2r move picks this. The force falls as 1/r², so the work against it is much less; use energies with U = −GMm/r.

Working For a circular orbit of radius r′, E = (1/2)U = −GMm/(2r′). ΔE = E(3r) − E(r) = −GMm/(6r) + GMm/(2r) = GMm/(3r). (ΔU = +2GMm/(3r), while ΔK = −GMm/(3r): the satellite moves more slowly in the higher orbit.)

CED 6.6.A.3 · Read this in Fix

1. A−6.9 × 10¹⁰ J
   A student who leaves the satellite's kinetic energy out of the system's energy picks U = −GMm/r = −6.9 × 10¹⁰ J. The satellite has K = +3.4 × 10¹⁰ J, so E = K + U = −3.4 × 10¹⁰ J.
2. B−4.0 × 10¹¹ J
   A student who uses the altitude, 6.0 × 10⁵ m, as r picks this: −GMm/(2 × 6.0 × 10⁵ m). In the gravitational equations r is the distance from Earth's center, 6.4 × 10⁶ m + 6.0 × 10⁵ m = 7.0 × 10⁶ m.
3. C−3.4 × 10¹⁰ J Correct
   The distance from Earth's center is r = 6.4 × 10⁶ + 6.0 × 10⁵ = 7.0 × 10⁶ m. For a circular orbit the kinetic energy is K = −(1/2)U, so E_total = (1/2)U = −GMm/(2r) = −(6.67 × 10⁻¹¹)(6.0 × 10²⁴)(1200)/(1.4 × 10⁷) = −3.4 × 10¹⁰ J.
4. D+1.0 × 10¹¹ J
   A student who takes the gravitational potential energy as positive, +GMm/r, adds it to K: 3.4 × 10¹⁰ + 6.9 × 10¹⁰ ≈ 1.0 × 10¹¹ J. With zero at infinity U_g is negative, and E = −3.4 × 10¹⁰ J.

Working r = 6.4 × 10⁶ m + 6.0 × 10⁵ m = 7.0 × 10⁶ m from Earth's center. E_total = (1/2)U = −GMm/(2r) = −(6.67 × 10⁻¹¹)(6.0 × 10²⁴)(1200)/(2 × 7.0 × 10⁶) J = −3.4 × 10¹⁰ J. (U = −6.9 × 10¹⁰ J and K = +3.4 × 10¹⁰ J.)

CED 6.6.A.3 · Read this in Fix

1. AGraph 1
   A student who thinks the planet's gravity stops acting beyond some distance picks the graph in which v levels off at a positive value. Gravity acts at every distance, so the probe keeps slowing, though ever more gently.
2. BGraph 2 Correct
   With zero mechanical energy, v = √(2GM/r) at every distance: always positive, decreasing as r increases. The decrease gets slower and slower because gravity weakens as 1/r², and v approaches zero only as the probe gets infinitely far away.
3. CGraph 3
   A student who thinks a probe whose velocity reaches zero stays at rest picks the graph that drops to zero and stays there. Zero velocity does not mean zero acceleration; besides, at escape speed v = √(2GM/r) is never zero at a finite distance.
4. DGraph 4
   A student who treats gravity as constant at its surface value picks the straight line: constant deceleration would stop the probe and bring it back. Gravity weakens with distance, so a probe at escape speed never stops at a finite distance.

Working With E = 0, (1/2)mv² = GMm/r, so v = √(2GM/r) > 0 at every finite r: the probe never stops or turns back. Gravity, GMm/r², weakens as r grows, so v decreases ever more slowly, approaching zero only as r → ∞.

CED 6.6.A.4.i · Read this in Fix

1. A7.1 × 10³ m/s
   A student who uses the altitude, 1.7 × 10⁶ m, as r picks this. In v_esc = √(2GM/r), r is the distance from the center of Mars, 5.1 × 10⁶ m.
2. B5.0 × 10³ m/s
   A student who uses the escape speed from the surface of Mars, √(2GM/R), picks this. The escape speed depends on the starting distance from the center; at 5.1 × 10⁶ m the system's U_g is less negative, so less speed is needed.
3. C4.1 × 10³ m/s Correct
   The spacecraft is r = 5.1 × 10⁶ m from the center. Zero mechanical energy there requires (1/2)v² = GM/r, so v_esc = √(2GM/r) = √(2(6.67 × 10⁻¹¹)(6.4 × 10²³)/(5.1 × 10⁶)) = 4.1 × 10³ m/s.
4. D2.9 × 10³ m/s
   A student who equates escape speed with circular-orbit speed picks √(GM/r) = 2.9 × 10³ m/s. That speed gives E = −GMm/(2r) < 0; escape needs E = 0, √2 times faster.

Working r = 3.4 × 10⁶ m + 1.7 × 10⁶ m = 5.1 × 10⁶ m. Setting E = 0: (1/2)mv² − GMm/r = 0, v_esc = √(2GM/r) = √(2(6.67 × 10⁻¹¹)(6.4 × 10²³)/(5.1 × 10⁶)) m/s = √(1.67 × 10⁷) m/s = 4.1 × 10³ m/s.

CED 6.6.A.4.ii · Read this in Fix

1. A√(2GMr_a/(r_p(r_p + r_a))) Correct
   Conservation of angular momentum gives v_p r_p = v_a r_a, and conservation of energy gives (1/2)v_p² − GM/r_p = (1/2)v_a² − GM/r_a. Eliminating v_a: v_p² = 2GMr_a/(r_p(r_p + r_a)). The factor r_a/r_p > 1 makes v_p greater than v_a, as it must be.
2. B√(2GMr_p/(r_a² + r_arp))
   A student who expects the satellite to move faster at the farther point assigns the smaller of the two speeds to the closest point. This expression is v_a; since v_p r_p = v_a r_a with r_p < r_a, the speed at the closest point is the larger one.
3. C√(GM(2r_a − r_p)/(r_p r_a))
   A student who uses energy conservation but takes the speed at the farthest point to be the circular-orbit speed √(GM/r_a) gets this. The satellite is not in a circular orbit at r_a; its speed there is fixed by angular momentum, v_a = v_p r_p/r_a.
4. D√(2GMr_a²/(r_p²(r_p + r_a)))
   A student who treats the gravitational field as constant at its value at the closest point, GM/r_p², and uses (1/2)v_p² − (1/2)v_a² = (GM/r_p²)(r_a − r_p) with v_a = v_p r_p/r_a, gets this. Over a change of distance comparable to r_p the field changes greatly, so U_g = −GMm/r must be used.

Working Angular momentum (velocity ⟂ radius at both points): v_p r_p = v_a r_a, so v_a = v_p r_p/r_a. Energy: (1/2)v_p² − GM/r_p = (1/2)v_a² − GM/r_a. Substituting: (1/2)v_p²(1 − r_p²/r_a²) = GM(1/r_p − 1/r_a) = GM(r_a − r_p)/(r_p r_a). With 1 − r_p²/r_a² = (r_a − r_p)(r_a + r_p)/r_a²: v_p² = 2GM r_a/(r_p(r_p + r_a)). (Checked with sympy.)

CED 6.6.A.2.ii · Read this in Fix

1. A0.25v₀
   A student who takes kinetic energy as proportional to speed, so that v (rather than v²) varies as 1/R, picks this. From (1/2)mv² = GMm/R, v² ∝ 1/R, so four times the radius halves the escape speed.
2. B0.50v₀ Correct
   Setting the mechanical energy to zero at the surface gives v_esc = √(2GM/R). With the same mass and four times the radius, the escape speed is multiplied by √(1/4) = 1/2.
3. C1.00v₀
   A student who thinks escape speed depends only on the planet's mass picks this. v_esc = √(2GM/R) also depends on the distance from the center; the surface of Y is four times farther out, where U_g is less negative.
4. D2.00v₀
   A student who thinks a bigger planet is harder to escape picks this, letting v_esc grow as √R. At fixed mass a larger radius makes the escape speed smaller: v_esc ∝ 1/√R.

Working v_esc = √(2GM/R). With M unchanged and R → 4R: v = v₀√(1/4) = 0.50v₀.

CED 6.6.A.4.ii · Read this in Fix

1. AmgR
   A student who uses mgΔy with the surface value of g, height R, picks this. That treats the field as constant out to 2R and puts the zero at the surface; with U = 0 at infinity, U_g = −GMm/r = −mgR/2.
2. B−mgR
   A student who uses the height above the surface, R, as r picks this: −GMm/R = −mgR. The distance in U_g = −GMm/r is measured from the planet's center, 2R.
3. C−mg/4
   A student who writes the potential energy with r² in the denominator, like the force, picks this: −GMm/(2R)² = −mg/4. That expression is a force, not an energy; U_g = −GMm/r.
4. D−mgR/2 Correct
   At the surface g = GM/R², so GM = gR². The satellite is 2R from the planet's center, so U_g = −GMm/(2R) = −mgR/2.

Working At the surface, g = GM/R², so GM = gR². The satellite is r = R + R = 2R from the center. U_g = −GMm/r = −gR²m/(2R) = −mgR/2.

CED 6.6.A.2.iii · Read this in Fix

1. A0.82√(gR)
   A student who uses the height above the surface, 3R, as r in v = √(2GM/r) picks this: √(2gR/3). The distance in the gravitational equations is measured from the planet's center, 4R here.
2. B0.50√(gR)
   A student who uses the circular-orbit speed √(GM/r) at r = 4R picks this: √(gR/4). That speed holds only for a circular orbit; the probe's speed follows from its zero total energy, v = √(2GM/r).
3. C0.35√(gR)
   A student who sees that the kinetic energy at r = 4R is one-fourth of its launch value, and takes the speed to fall to one-fourth as well, picks this: √(2gR)/4. Kinetic energy is proportional to v², so the speed falls only to one-half of √(2gR).
4. D0.71√(gR) Correct
   With the escape speed, the system's mechanical energy is zero throughout, so (1/2)mv² = GMm/r and v = √(2GM/r). With GM = gR² and r = 4R (the distance from the center at a height 3R), v = √(gR/2) ≈ 0.71√(gR).

Working At the surface, g = GM/R², so GM = gR². Launching at exactly the escape speed makes the system's total mechanical energy zero, and it stays zero: (1/2)mv² − GMm/r = 0 at every point, so v = √(2GM/r). At a height 3R, r = R + 3R = 4R: v = √(2gR²/(4R)) = √(gR/2) ≈ 0.71√(gR). Distractors: r taken as the height 3R: √(2gR/3) ≈ 0.82√(gR); circular-orbit speed √(GM/r) at r = 4R: √(gR/4) = 0.50√(gR); kinetic energy falls to 1/4 of its launch value, and the speed is taken to fall by the same factor: √(2gR)/4 ≈ 0.35√(gR).

CED 6.6.A.4 · Read this in Fix

1. Av − GM/r
   A student who takes kinetic energy to be proportional to speed, writing K = (1/2)mv, gets (1/2)mv − GMm/r = (1/2)mv_Q − GMm/(2r) and so v_Q = v − GM/r. That expression does not even have units of speed; kinetic energy is (1/2)mv², which gives v_Q² = v² − GM/r.
2. Bv/2
   A student who takes the satellite's angular momentum to be mvr at every point, so that vr is constant, picks this: v_Q(2r) = vr. Angular momentum is mvr sin θ; at P and Q the velocity is not perpendicular to the line from the planet's center, so the speeds are not in the inverse ratio of the distances.
3. C√(v² − GM/r) Correct
   The system's mechanical energy is the same at P and Q: (1/2)mv² − GMm/r = (1/2)mv_Q² − GMm/(2r). Solving, v_Q² = v² − GM/r. The satellite moves away from the planet, so it slows down.
4. D√(GM/(2r))
   A student who uses the circular-orbit speed √(GM/r) at Q, with the distance 2r, picks this. That speed holds only for a circular orbit; in an elliptical orbit the speed at Q follows from the conservation of mechanical energy.

Working Only gravity acts, so the mechanical energy of the satellite–planet system is constant: (1/2)mv² − GMm/r = (1/2)mv_Q² − GMm/(2r). So v_Q² = v² − 2GM/r + GM/r = v² − GM/r, and v_Q = √(v² − GM/r). The satellite's angular momentum is also constant, but mvr sin θ involves angles that are not given, so it cannot give v_Q here. Distractors: K taken as (1/2)mv: v_Q = v − GM/r; vr taken as constant: v(r) = v_Q(2r), v_Q = v/2; circular-orbit speed at 2r: √(GM/(2r)).

CED 6.6.A.2 · Read this in Fix