AP Physics C Mechanics Quiz: Motion Of Orbiting Satellites

20 questions · exam conditions

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Motion Of Orbiting SatellitesQuestion 1 of 20

In circular orbit, gravity supplies the centripetal force: $\frac{GMm}{r^2}=\frac{mv^2}{r}$ . This yields $v=\sqrt{\mu/r}$ and, combining with $v=2\pi r/T$ , gives Kepler's 3rd law $T^2\propto r^3$ . How does the gravitational force act as the centripetal force for an orbiting satellite?

It points radially outward, balancing inertia to keep speed constant.

It provides a tangential force that continuously increases orbital speed.

It points toward Earth's center, supplying

a_c=v^2/r

for curved motion.

It is replaced by magnetic forces once the satellite reaches orbit.

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Motion Of Orbiting Satellites

Practice Motion Of Orbiting Satellites in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Motion Of Orbiting Satellites, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

It points radially outward, balancing inertia to keep speed constant.
It provides a tangential force that continuously increases orbital speed.
It points toward Earth's center, supplying $a_c=v^2/r$ for curved motion. (correct answer)
It is replaced by magnetic forces once the satellite reaches orbit.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). In circular orbital motion, gravitational force serves as the centripetal force, always pointing toward the center of the circular path and providing the acceleration needed to continuously change the velocity's direction. The gravitational force F = GMm/r² must equal the required centripetal force mv²/r for circular motion, leading to the orbital velocity relationship v = √(GM/r). Choice C is correct because it accurately describes gravity pointing toward Earth's center and providing the centripetal acceleration ac = v²/r necessary for curved motion. Choice D is incorrect as it suggests magnetic forces replace gravity in orbit, a fundamental misconception - gravity is the only significant force acting on satellites. To help students: Use free body diagrams showing only gravitational force on the satellite, emphasize that this single force causes the acceleration that curves the path, and demonstrate how setting Fgrav = Fcentripetal yields orbital relationships. Address the misconception that satellites are beyond gravity's reach.

Question 2

How does conservation of angular momentum affect satellite speed if orbital radius decreases due to a brief inward impulse?

Speed decreases so that $L=mrv$ remains constant.
Speed increases so that $L=mrv$ remains constant. (correct answer)
Speed is unchanged because gravity cancels angular momentum.
Speed increases because potential energy increases as $r$ decreases.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Angular momentum L = mrv is conserved for a satellite when no external torques act, which is true for central forces like gravity. When orbital radius r decreases due to an inward impulse, conservation of angular momentum requires that velocity v must increase to keep L constant. Choice B is correct because it accurately applies conservation of angular momentum: as r decreases, v must increase proportionally so that the product mrv remains constant. Choice A is incorrect because it states speed decreases, which would violate angular momentum conservation when radius decreases. To help students: Emphasize that gravity exerts no torque about the central body, practice applying L = mrv for different orbital scenarios, and use the ice skater analogy where pulling arms inward increases rotation speed. Have students calculate specific velocity changes for given radius changes.

Question 3

A satellite in an elliptical orbit follows Kepler's laws: (1) orbits are ellipses with Earth at a focus, (2) equal areas in equal times, and (3) $T^2\propto a^3$ . The equal-areas law is a consequence of constant angular momentum $L=mrv_\perp$ when gravity (a central force) exerts zero torque about Earth's center. How does conservation of angular momentum affect the orbit of a satellite?

It forces the semi-major axis to stay constant even if energy changes.
It implies faster motion near perigee and slower motion near apogee. (correct answer)
It makes orbital speed constant everywhere along an ellipse.
It requires an external torque from Earth's rotation to maintain orbit.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Kepler's second law (equal areas in equal times) is a direct consequence of angular momentum conservation, as gravitational force exerts zero torque about the central body. In an elliptical orbit, the perpendicular component of velocity v⊥ varies inversely with distance r to maintain constant angular momentum L = mrv⊥. Choice B is correct because conservation of angular momentum requires the satellite to move faster when closer to Earth (perigee) and slower when farther away (apogee), as rv⊥ must remain constant. Choice C is incorrect as it claims constant speed throughout the ellipse, which would violate both energy conservation and angular momentum conservation in a varying gravitational field. To help students: Use visual demonstrations of equal area sweeping, derive Kepler's second law from L = constant, and have students calculate velocities at different points in elliptical orbits. Emphasize the distinction between constant angular momentum and varying linear speed.

Question 4

A satellite in an elliptical Earth orbit experiences gravitational force $F_g=\frac{GMm}{r^2}$ directed toward Earth's center, providing centripetal acceleration for the instantaneous curved motion. Kepler's 2nd law implies equal areas in equal times, consistent with conservation of angular momentum $L=mrv_\perp$ . If no external torque acts, $L$ is constant as $r$ changes. How does conservation of angular momentum affect the orbit of a satellite?

As $r$ decreases, $v_\perp$ increases to keep $mrv_\perp$ constant. (correct answer)
As $r$ decreases, $v_\perp$ decreases to keep $mrv_\perp$ constant.
Angular momentum increases because gravity does positive torque about Earth.
Angular momentum is irrelevant; only kinetic energy determines orbital shape.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Angular momentum conservation is a fundamental principle in orbital mechanics, stating that L = mrv⊥ remains constant when no external torques act on the system. Since gravitational force points radially toward Earth's center, it exerts zero torque about that center, making angular momentum conserved throughout the orbit. Choice A is correct because as r decreases, v⊥ must increase proportionally to maintain constant L = mrv⊥, explaining why satellites move faster at perihelion (closest approach) in elliptical orbits. Choice B is incorrect as it contradicts the conservation principle - if r decreases and v⊥ also decreased, angular momentum would not be conserved. To help students: Use the ice skater analogy (pulling arms in while spinning), derive why central forces produce zero torque, and practice applying L = mrv⊥ = constant to various orbital scenarios. Emphasize that this principle leads directly to Kepler's second law of equal areas.

Question 5

Consider raising a satellite from a circular orbit at $r_1$ to a larger circular orbit at $r_2$ . Using $E=-\frac{\mu m}{2r}$ and $v=\sqrt{\mu/r}$ (from gravity providing centripetal force), Kepler's 3rd law predicts the higher orbit has a longer period. What energy transformations occur as a satellite moves from a low Earth orbit to a high Earth orbit?

K increases because speed increases; U increases, so E is unchanged.
K decreases, U increases (less negative), so E increases (less negative). (correct answer)
K increases, U decreases (more negative), so E increases.
K decreases and U decreases, so E becomes more negative.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). When raising a satellite from a lower to higher circular orbit, both kinetic and potential energies change in specific ways determined by the orbital mechanics relationships. Since v = √(μ/r), as r increases, velocity decreases, causing kinetic energy K = ½mv² to decrease. Choice B is correct because K decreases while U = -GMm/r becomes less negative (increases toward zero), and the total energy E = -GMm/(2r) also becomes less negative, representing an overall energy increase. Choice C is incorrect as it claims kinetic energy increases when moving to a higher orbit, contradicting the v ∝ 1/√r relationship - a common misconception among students. To help students: Use energy bar charts showing K, U, and E at different orbital radii, emphasize that work must be done against gravity to raise the orbit, and practice numerical examples. Reinforce that satellites in higher orbits move more slowly but have greater total mechanical energy.

Question 6

For a satellite moving from a lower circular orbit of radius $r_1$ to a higher circular orbit of radius $r_2>r_1$ , the orbital speed is $v=\sqrt{\mu/r}$ and total mechanical energy is $E=-\frac{\mu m}{2r}$ . Gravitational force remains the centripetal force, and Kepler's 3rd law predicts a longer period at larger $r$ . What energy transformations occur as a satellite moves from a low Earth orbit to a high Earth orbit?

K decreases, U increases (less negative), and total E increases (less negative). (correct answer)
K increases, U decreases (more negative), and total E decreases.
Both K and U increase, so total E becomes more negative.
K stays constant while U increases, because orbital speed is unchanged.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). The total mechanical energy of an orbiting satellite is E = K + U = ½mv² - GMm/r, which simplifies to E = -GMm/(2r) for circular orbits. When moving to a higher orbit, the orbital radius increases, causing specific energy changes in both kinetic and potential components. Choice A is correct because as r increases, v = √(μ/r) decreases, so K = ½mv² decreases; meanwhile, U = -GMm/r becomes less negative (increases), and the total energy E = -GMm/(2r) also becomes less negative (increases). Choice B is incorrect as it reverses the energy changes - students often confuse 'less negative' with 'decreasing'. To help students: Use energy diagrams showing how U and E vary with r, emphasize that 'less negative' means 'increasing' for negative quantities, and practice calculating specific values. Reinforce that energy must be added to raise an orbit, making total energy less negative.

Question 7

Using Kepler's third law, how does orbital period $T$ scale with orbital radius $r$ for satellites around Earth?

$T\propto r$ because gravity is constant near Earth.
$T\propto r^{3/2}$ because $T^2\propto r^3$ . (correct answer)
$T\propto r^2$ because $v\propto r$ in orbit.
$T\propto r^{1/2}$ because $v\propto r^{1/2}$ in orbit.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Kepler's third law states that the square of the orbital period is proportional to the cube of the orbital radius, or T² ∝ r³. This relationship can be derived from Newton's laws by equating gravitational and centripetal forces and using the relationship between period and velocity. Choice B is correct because it accurately states that T ∝ r^(3/2), which follows from taking the square root of both sides of T² ∝ r³. Choice C is incorrect because it suggests T ∝ r², which would imply T² ∝ r⁴, violating Kepler's third law. To help students: Derive Kepler's third law from first principles using F = ma, practice applying the T² ∝ r³ relationship to compare orbital periods, and use log-log plots to verify the 3/2 power relationship. Emphasize that this law applies to all satellites orbiting the same central body.

Question 8

For a circular orbit, which expression correctly gives total mechanical energy $E$ in terms of $r$ and Earth's mass $M$ ?

$E=-\dfrac{GMm}{2r}$ (correct answer)
$E=-\dfrac{GMm}{r}$
$E=+\dfrac{GMm}{2r}$
$E=+\dfrac{GMm}{r}$

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). The total mechanical energy of a satellite in circular orbit is the sum of kinetic energy K = ½mv² and gravitational potential energy U = -GMm/r. For circular orbits, using v² = GM/r from the centripetal force condition, we find K = GMm/2r and U = -GMm/r, giving total energy E = K + U = -GMm/2r. Choice A is correct because it accurately gives the total mechanical energy as E = -GMm/2r, which is negative indicating a bound orbit. Choice B is incorrect because it gives only the potential energy, omitting the kinetic energy contribution. To help students: Derive the total energy step-by-step from K and U, emphasize that E is always negative for bound orbits, and show that |E| = K for circular orbits (virial theorem). Practice calculating escape velocity where E = 0.

Question 9

A geostationary satellite orbits Earth at radius $r_{geo}$ . A spy satellite is placed in a circular orbit at radius $r = \frac{r_{geo}}{4}$ . How many times does the spy satellite orbit Earth while the geostationary satellite completes one orbit?

$4$
$8$ (correct answer)
$16$
$64$

Explanation: Using Kepler's third law:

\frac{T_{spy}^2}{T_{geo}^2} = \frac{r_{spy}^3}{r_{geo}^3} = \frac{(r_{geo}/4)^3}{r_{geo}^3} = \frac{1}{64}

. Therefore

T_{spy} = \frac{T_{geo}}{8}

. In one geostationary period, the spy satellite completes

\frac{T_{geo}}{T_{spy}} = 8

orbits. Choice A incorrectly uses linear scaling. Choice C incorrectly uses quadratic scaling. Choice D uses the cube of the radius ratio directly.

Question 10

A satellite in an elliptical orbit around Earth has speeds $v_p = 8.0 \times 10^3$ m/s at periapsis (closest approach) and $v_a = 3.0 \times 10^3$ m/s at apoapsis (farthest point). If the periapsis distance is $r_p = 7.0 \times 10^6$ m, what is the apoapsis distance $r_a$ ?

$1.9 \times 10^7$ m (correct answer)
$2.8 \times 10^7$ m
$5.6 \times 10^7$ m
$1.5 \times 10^8$ m

Explanation: Conservation of angular momentum gives

v_p r_p = v_a r_a

, so

r_a = \frac{v_p r_p}{v_a} = \frac{(8.0 \times 10^3)(7.0 \times 10^6)}{3.0 \times 10^3} = \frac{56 \times 10^9}{3.0 \times 10^3} = 1.87 \times 10^7

\approx 1.9 \times 10^7

m. Choice B uses incorrect velocity ratio. Choice C assumes energy conservation alone without angular momentum. Choice D incorrectly applies inverse square scaling.

Question 11

A satellite orbits Earth in a circular orbit at altitude $h$ above Earth's surface. If the satellite's orbital radius is doubled while maintaining a circular orbit, which of the following correctly describes the changes in the satellite's orbital speed and orbital period?

Speed decreases by a factor of $\sqrt{2}$ ; period increases by a factor of $2\sqrt{2}$ (correct answer)
Speed decreases by a factor of $2$ ; period increases by a factor of $4$
Speed decreases by a factor of $\sqrt{2}$ ; period increases by a factor of $4$
Speed decreases by a factor of $2$ ; period increases by a factor of $2\sqrt{2}$

Explanation: For circular orbits, orbital speed

v = \sqrt{\frac{GM}{r}}

and period

T = 2\pi\sqrt{\frac{r^3}{GM}}

. When radius doubles: speed becomes

v' = \sqrt{\frac{GM}{2r}} = \frac{v}{\sqrt{2}}

(decreases by

\sqrt{2}

), and period becomes

T' = 2\pi\sqrt{\frac{(2r)^3}{GM}} = 2\pi\sqrt{\frac{8r^3}{GM}} = 2\sqrt{2}T

(increases by

2\sqrt{2}

). Choice B incorrectly applies inverse-square scaling to speed. Choice C uses correct speed scaling but incorrect period scaling. Choice D combines incorrect scalings.

Question 12

A satellite in a circular orbit at radius $r$ around Earth experiences a gravitational force $F$ . If this satellite is moved to a new circular orbit where it experiences half the gravitational force, what is the ratio of its new orbital period to its original period?

$\sqrt{2}$
$2$
$2\sqrt{2}$ (correct answer)
$4$

Explanation: If gravitational force is halved:

\frac{F'}{F} = \frac{GMm/r'^2}{GMm/r^2} = \frac{r^2}{r'^2} = \frac{1}{2}

, so

r' = r\sqrt{2}

. Using Kepler's third law:

\frac{T'^2}{T^2} = \frac{r'^3}{r^3} = \frac{(r\sqrt{2})^3}{r^3} = (\sqrt{2})^3 = 2\sqrt{2}

. Therefore

\frac{T'}{T} = \sqrt{2\sqrt{2}} = (2\sqrt{2})^{1/2} = 2^{3/4} \approx 2.38

. The closest answer is

2\sqrt{2} \approx 2.83

. Actually, let me recalculate:

(\sqrt{2})^3 = 2^{3/2} = 2\sqrt{2}

, so

T'/T = \sqrt{2\sqrt{2}} = (2^{3/2})^{1/2} = 2^{3/4}

. Wait, this equals

(2\sqrt{2})^{1/2}

, which is approximately 1.68, not 2.83. Let me restart: If

r' = r\sqrt{2}

, then

T'/T = (r'/r)^{3/2} = (\sqrt{2})^{3/2} = 2^{3/4} = \sqrt{2\sqrt{2}} \approx 1.68

. This doesn't match any given choice exactly, suggesting an error in the problem setup.

Question 13

A satellite moves in an elliptical orbit around Earth. At point P, the satellite is at distance $r_P = 8.0 \times 10^6$ m from Earth's center and has speed $v_P = 6.0 \times 10^3$ m/s. At point Q, the satellite is at distance $r_Q = 1.2 \times 10^7$ m from Earth's center. What is the satellite's speed at point Q?

$2.0 \times 10^3$ m/s
$9.0 \times 10^3$ m/s
$4.8 \times 10^3$ m/s
$4.0 \times 10^3$ m/s (correct answer)

Explanation: When dealing with satellites in elliptical orbits, you need to apply conservation of energy and angular momentum. These fundamental principles remain constant throughout the orbital motion, even as the satellite's speed and distance from Earth change. For this problem, conservation of angular momentum gives us

L = mvr = \text{constant}

. At points P and Q:

m v_P r_P = m v_Q r_Q

. The mass cancels out, so:

v_Q = v_P \frac{r_P}{r_Q}

. Substituting the given values:

v_Q = (6.0 \times 10^3) \times \frac{8.0 \times 10^6}{1.2 \times 10^7} = (6.0 \times 10^3) \times \frac{8.0}{12.0} = (6.0 \times 10^3) \times \frac{2}{3} = 4.0 \times 10^3 \text{ m/s}

This confirms answer D is correct. Looking at the wrong answers: A (

2.0 \times 10^3

m/s) would result from incorrectly using the ratio

\frac{r_P}{2r_Q}

or making an arithmetic error. B (

9.0 \times 10^3

m/s) comes from mistakenly using

v_Q = v_P \frac{r_Q}{r_P}

, which inverts the correct relationship and violates conservation of angular momentum. C (

4.8 \times 10^3

m/s) likely results from calculation errors in the fraction simplification. Study tip: Remember that in elliptical orbits, satellites move faster when closer to the central body (higher gravitational potential energy converts to kinetic energy). Angular momentum conservation gives you

vr = \text{constant}

, so larger radius means smaller speed.

Question 14

A satellite in elliptical orbit obeys Kepler's second law; what happens to its speed as it approaches perigee?

Speed decreases because angular momentum must decrease near Earth.
Speed increases because angular momentum conservation requires larger $v$ at smaller $r$ . (correct answer)
Speed stays constant because gravity does no work in orbit.
Speed increases because gravitational potential energy increases at smaller $r$ .

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Kepler's second law states that a satellite sweeps out equal areas in equal times, which is a consequence of angular momentum conservation. As a satellite in elliptical orbit approaches perigee (closest point to Earth), its distance r from Earth decreases while angular momentum L = mrv remains constant. Choice B is correct because it accurately applies angular momentum conservation: as r decreases at perigee, v must increase to maintain constant L = mrv. Choice D is incorrect because it misstates the energy relationship - gravitational potential energy actually becomes more negative (decreases) at smaller r. To help students: Emphasize the connection between Kepler's second law and angular momentum conservation, use area-sweeping diagrams to visualize the law, and practice calculating speeds at apogee and perigee using L conservation. Show energy conservation as a separate constraint.

Question 15

For a circular geostationary orbit, how does gravitational force provide the centripetal force needed to keep the satellite moving?

It is replaced by magnetic force, which supplies $mv^2/r$ .
It equals $GMm/r^2$ and must match $mv^2/r$ toward Earth's center. (correct answer)
It equals $GMm/r$ and must match $mv^2/r^2$ toward Earth's center.
It acts tangentially, increasing speed to maintain circular motion.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). For circular orbits, gravitational force provides the centripetal force needed to maintain the satellite's circular path around Earth. The gravitational force on a satellite of mass m at distance r from Earth's center is GMm/r², where G is the gravitational constant and M is Earth's mass. Choice B is correct because it accurately states that gravitational force equals GMm/r² and this must equal the required centripetal force mv²/r directed toward Earth's center. Choice A is incorrect because magnetic forces play no role in satellite orbits - gravity alone provides the centripetal force. To help students: Emphasize that for any circular orbit, the gravitational force IS the centripetal force, practice setting GMm/r² = mv²/r to derive orbital velocity, and use free-body diagrams showing only gravity acting radially inward. Have students verify dimensional consistency and practice with numerical examples.

Question 16

In a circular orbit, what energy exchange occurs if a satellite is boosted from LEO to a higher circular orbit?

Kinetic energy increases and gravitational potential energy decreases.
Kinetic energy decreases while gravitational potential energy increases (less negative). (correct answer)
Both kinetic and gravitational potential energies decrease.
Both kinetic and gravitational potential energies remain constant.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). When a satellite moves to a higher orbit, both its kinetic and potential energies change, but total mechanical energy increases due to work done by the boosting force. In a higher orbit, the satellite moves more slowly (v = √(GM/r) decreases with increasing r), so kinetic energy K = ½mv² decreases. Choice B is correct because it accurately states that kinetic energy decreases while gravitational potential energy U = -GMm/r becomes less negative (increases) as r increases. Choice A is incorrect because it reverses the energy changes - kinetic energy actually decreases in higher orbits. To help students: Emphasize that orbital speed decreases with altitude, practice calculating both K and U for different orbits, and show that total energy E = -GMm/2r becomes less negative (increases) for higher orbits. Use energy bar charts to visualize the trade-off between kinetic and potential energy.

Question 17

For a geostationary satellite, what orbital radius $r$ satisfies $T=24\,\text{h}$ using $T=2\pi\sqrt{r^3/(GM)}$ ?

$r=\left(\dfrac{GM\,T^2}{2\pi}\right)^{1/3}$
$r=\left(\dfrac{GM\,T^2}{4\pi^2}\right)^{1/3}$ (correct answer)
$r=\left(\dfrac{4\pi^2GM}{T^2}\right)^{1/3}$
$r=\left(\dfrac{GM\,T}{4\pi^2}\right)^{1/3}$

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). For circular orbits, the period T and radius r are related by T = 2π√(r³/GM), which can be rearranged to solve for r when T is known. Squaring both sides gives T² = 4π²r³/GM, and solving for r yields r = (GMT²/4π²)^(1/3). Choice B is correct because it properly rearranges the period formula to isolate r, giving r = (GMT²/4π²)^(1/3) for a geostationary satellite with T = 24 hours. Choice A is incorrect because it omits the factor of 2 in the denominator, using 2π instead of 4π². To help students: Practice algebraic manipulation of the period formula, emphasize keeping track of all factors including 2π vs 4π², and work through the calculation with actual values for Earth to find r ≈ 42,000 km. Show how this places geostationary satellites about 36,000 km above Earth's surface.

Question 18

A satellite is launched into a circular orbit; which condition on initial tangential speed ensures stable orbit at radius $r$ ?

$v=\sqrt{GM/r}$ so that $GMm/r^2=mv^2/r$ . (correct answer)
$v=\sqrt{GM/r^2}$ so that $GMm/r=mv^2/r$ .
$v=\sqrt{GM/r}$ so that $GMm/r=mv^2/r^2$ .
$v=\sqrt{2GM/r}$ so that total energy is zero at radius $r$ .

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). For a stable circular orbit, the gravitational force must exactly provide the centripetal force needed for circular motion at radius r. Setting GMm/r² = mv²/r and solving for v gives the required orbital velocity v = √(GM/r). Choice A is correct because it gives the proper orbital velocity v = √(GM/r) and shows the correct force balance equation GMm/r² = mv²/r that this velocity satisfies. Choice D is incorrect because v = √(2GM/r) is the escape velocity, not the circular orbital velocity - this would give the satellite too much speed for a bound orbit. To help students: Derive the orbital velocity from F = ma step-by-step, emphasize the difference between orbital and escape velocities, and practice checking solutions by verifying force balance. Show that any other initial speed results in elliptical or hyperbolic trajectories.

Question 19

What is the velocity required for a stable circular orbit at altitude $h=300\,\text{km}$ above Earth, using $v=\sqrt{GM/r}$ ?

$v=\sqrt{GM/(R_E+h)}$ (correct answer)
$v=\sqrt{G(M+m)/(R_E+h)}$
$v=\sqrt{GM/(R_E+h)^2}$
$v=\sqrt{GM/(R_E-h)}$

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). The orbital velocity formula v = √(GM/r) derives from equating gravitational force to centripetal force, where r is the distance from Earth's center. For a satellite at altitude h above Earth's surface, the orbital radius r = RE + h, where RE is Earth's radius. Choice A is correct because it properly substitutes r = RE + h into the orbital velocity formula, giving v = √(GM/(RE+h)). Choice C is incorrect because it incorrectly squares the denominator, suggesting a misunderstanding of how radius enters the formula. To help students: Emphasize that r in orbital equations always means distance from Earth's center, not altitude above surface, practice converting between altitude and orbital radius, and work through the derivation starting from F = ma. Use diagrams showing Earth's radius and satellite altitude clearly labeled.

Question 20

A satellite orbits Earth in a circular orbit with period $T$ . If Earth's mass were suddenly doubled while keeping the orbital radius constant, what would be the new orbital period?

$\frac{T}{2}$
$2T$
$T\sqrt{2}$
$\frac{T}{\sqrt{2}}$ (correct answer)

Explanation: When you encounter orbital mechanics problems, focus on the relationship between gravitational force, centripetal force, and orbital parameters. For circular orbits, the gravitational force provides the centripetal force needed to maintain the orbit. Start with the fundamental equation:

\frac{GMm}{r^2} = \frac{mv^2}{r}

, where

G

is the gravitational constant,

M

is Earth's mass,

m

is the satellite's mass,

r

is the orbital radius, and

v

is the orbital velocity. Simplifying gives

v = \sqrt{\frac{GM}{r}}

. Since the orbital period is

T = \frac{2\pi r}{v}

, substituting the velocity expression yields:

T = 2\pi r \sqrt{\frac{r}{GM}} = 2\pi\sqrt{\frac{r^3}{GM}}

This shows that

T \propto \frac{1}{\sqrt{M}}

when radius is constant. If Earth's mass doubles (

M \rightarrow 2M

), the new period becomes:

T_{new} = T \sqrt{\frac{M}{2M}} = \frac{T}{\sqrt{2}}

Answer D is correct because the period is inversely proportional to the square root of the central mass. Answer A (

\frac{T}{2}

) incorrectly assumes the period is inversely proportional to mass itself, not its square root. Answer B (

2T

) mistakenly suggests the period increases with mass, which would violate the physics—stronger gravity pulls harder, speeding up the orbit. Answer C (

T\sqrt{2}

) also incorrectly shows an increase in period. Remember: In orbital problems, always derive relationships from first principles using

F_{gravity} = F_{centripetal}

. Kepler's laws confirm that orbital period decreases as the central mass increases.