College Physics Quiz: Motion Of Orbiting Satellites

20 questions · exam conditions

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Motion Of Orbiting SatellitesQuestion 1 of 20

A satellite orbits Earth at a distance of $2R$ from Earth's center, where $R$ is Earth's radius. If the satellite's orbital radius is increased to $3R$ , what happens to its orbital speed and orbital period?

Speed decreases by a factor of

\sqrt{\frac{3}{2}}

; period increases by a factor of

\left(\frac{3}{2}\right)^{3/2}

Speed decreases by a factor of

\sqrt{\frac{2}{3}}

; period increases by a factor of

\left(\frac{2}{3}\right)^{3/2}

Speed increases by a factor of

\sqrt{\frac{3}{2}}

; period decreases by a factor of

\left(\frac{3}{2}\right)^{3/2}

Speed decreases by a factor of

\frac{3}{2}

; period increases by a factor of

\left(\frac{3}{2}\right)^2

Speed increases by a factor of

\frac{2}{3}

; period decreases by a factor of

\left(\frac{2}{3}\right)^2

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College Physics Quiz

College Physics Quiz: Motion Of Orbiting Satellites

Practice Motion Of Orbiting Satellites in College Physics 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 College Physics.

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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

Speed decreases by a factor of $\sqrt{\frac{3}{2}}$ ; period increases by a factor of $\left(\frac{3}{2}\right)^{3/2}$ (correct answer)
Speed decreases by a factor of $\sqrt{\frac{2}{3}}$ ; period increases by a factor of $\left(\frac{2}{3}\right)^{3/2}$
Speed increases by a factor of $\sqrt{\frac{3}{2}}$ ; period decreases by a factor of $\left(\frac{3}{2}\right)^{3/2}$
Speed decreases by a factor of $\frac{3}{2}$ ; period increases by a factor of $\left(\frac{3}{2}\right)^2$
Speed increases by a factor of $\frac{2}{3}$ ; period decreases by a factor of $\left(\frac{2}{3}\right)^2$

Explanation: When you encounter orbital mechanics problems, focus on two key relationships: orbital speed depends on

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

and Kepler's third law states that

T^2 \propto r^3

. For orbital speed, as radius increases from

2R

3R

, you have:

\frac{v_2}{v_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{2R}{3R}} = \sqrt{\frac{2}{3}}

This means the new speed is

v_2 = v_1 \sqrt{\frac{2}{3}}

, so speed decreases by a factor of

\sqrt{\frac{2}{3}}

. For the period, using

T^2 \propto r^3

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \frac{(3R)^3}{(2R)^3} = \left(\frac{3}{2}\right)^3

Taking the square root:

\frac{T_2}{T_1} = \left(\frac{3}{2}\right)^{3/2}

So the period increases by a factor of

\left(\frac{3}{2}\right)^{3/2}

. Answer A correctly states both relationships. Answer B has the wrong speed factor - it uses

\sqrt{\frac{2}{3}}

instead of the reciprocal, and the wrong period factor with

\left(\frac{2}{3}\right)^{3/2}

instead of

\left(\frac{3}{2}\right)^{3/2}

. Answer C incorrectly suggests speed increases when it actually decreases for larger orbits, and gets the period direction wrong. Answer D uses incorrect power relationships - it applies

\frac{3}{2}

directly to speed instead of its square root, and

\left(\frac{3}{2}\right)^2

to period instead of

\left(\frac{3}{2}\right)^{3/2}

. Remember: larger orbits mean slower speeds and longer periods, with speed scaling as

r^{-1/2}

and period as

r^{3/2}

Question 2

A satellite in a circular orbit around Earth has angular momentum $L$ . If Earth's mass were suddenly doubled while keeping the satellite's orbital radius constant, what would happen to the satellite's motion?

The satellite would maintain circular motion with angular momentum $L\sqrt{2}$ and increased orbital speed
The satellite would maintain circular motion with the same angular momentum $L$ but increased orbital speed
The satellite would spiral inward because the gravitational force becomes too strong for circular motion (correct answer)
The satellite would escape because the increased gravitational force exceeds the centripetal requirement
The satellite would maintain circular motion with angular momentum $2L$ and the same orbital speed

Explanation: When analyzing satellite motion problems, you need to consider the delicate balance between gravitational force and the requirements for circular motion. A satellite maintains circular orbit when gravitational force provides exactly the centripetal force needed. Initially, the satellite orbits with gravitational force

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

providing centripetal force

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

. Setting these equal gives the orbital speed:

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

. When Earth's mass suddenly doubles to

2M

, the gravitational force immediately doubles to

\frac{2GMm}{r^2}

, but the satellite's speed remains

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

(it can't instantaneously change). For circular motion with doubled mass, the required speed would be

v_{new} = \sqrt{\frac{2GM}{r}} = v\sqrt{2}

. Since the actual speed is less than required, the gravitational force now exceeds what's needed for circular motion at radius

r

. The satellite will spiral inward toward Earth. Option A incorrectly assumes the satellite adjusts to maintain circular motion - orbital mechanics don't work this way. Option B makes the same error while incorrectly claiming angular momentum stays constant (

L = mvr

changes when

v

changes). Option D confuses the scenario - stronger gravity pulls objects inward, not outward toward escape. Remember: in orbital mechanics problems, sudden changes in gravitational parameters create imbalances. The object's motion can't instantly adjust, so you must analyze whether the new forces support the existing trajectory or cause deviation.

Question 3

Two satellites orbit Earth in circular orbits with the same orbital period. Satellite A has mass $m$ and satellite B has mass $2m$ . What is the ratio of their orbital radii $r_A/r_B$ ?

$1$ (correct answer)
$\frac{1}{2}$
$2$
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$

Explanation: When you encounter orbital mechanics problems, remember that Kepler's Third Law governs the relationship between orbital period and radius, while the mass of the orbiting object plays no role in determining its orbit. For any satellite in circular orbit, gravitational force provides the centripetal force:

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

, where

M

is Earth's mass and

m

is the satellite's mass. Notice that

m

cancels out completely, leaving

\frac{GM}{r} = v^2

. Since orbital velocity is

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

for period

T

, substituting gives us

\frac{GM}{r} = \frac{4\pi^2 r^2}{T^2}

. Rearranging:

T^2 = \frac{4\pi^2 r^3}{GM}

. This is Kepler's Third Law: period depends only on orbital radius, not on the satellite's mass. Since both satellites have identical periods, they must have identical orbital radii. Therefore

r_A/r_B = 1

. Looking at the wrong answers: Choice B (

\frac{1}{2}

) and C (

2

) suggest the ratio equals the inverse or direct mass ratio, respectively—a common misconception that satellite mass affects orbital radius. Choice D (

\frac{1}{\sqrt{2}

) might tempt students who incorrectly apply the square root relationship from Kepler's law but confuse mass with period. Study tip: Remember that in orbital mechanics, the orbiting object's mass never determines its orbital characteristics—only the central body's mass matters. This counterintuitive fact means a feather and a hammer would orbit Earth identically at the same altitude.

Question 4

A satellite orbits Earth at distance $R$ from Earth's center with orbital speed $v_0$ . What would be the escape speed from this same orbital distance?

$\sqrt{2}v_0$ (correct answer)
$2v_0$
$v_0$
$\frac{v_0}{\sqrt{2}}$
$\frac{v_0}{2}$

Explanation: When you encounter orbital mechanics problems involving escape velocity, you're dealing with energy conservation principles. The key insight is understanding the relationship between circular orbital motion and the minimum speed needed to escape gravitational pull entirely. For a satellite in circular orbit at distance

R

, the gravitational force provides the centripetal force:

\frac{GMm}{R^2} = \frac{mv_0^2}{R}

. This gives us the orbital speed

v_0 = \sqrt{\frac{GM}{R}}

. Escape velocity is the minimum speed needed for an object to break free from gravitational attraction. Using energy conservation, an object escapes when its total energy (kinetic + potential) equals zero:

\frac{1}{2}mv_{escape}^2 - \frac{GMm}{R} = 0

. Solving for escape speed:

v_{escape} = \sqrt{\frac{2GM}{R}}

. Comparing these expressions, you can see that

v_{escape} = \sqrt{2} \cdot v_0

, making (A) $\sqrt{2}v_0$ correct. (B) $2v_0$ represents a common error where students might think escape speed is simply double the orbital speed, confusing the factor of 2 that appears under the square root. (C) $v_0$ would suggest escape speed equals orbital speed, which is impossible since orbital speed maintains a bound orbit. (D) $\frac{v_0}{\sqrt{2}}$ inverts the correct relationship and would represent a speed less than orbital velocity. Remember this universal relationship: escape speed from any distance is always

\sqrt{2}

times the circular orbital speed at that same distance. This

\sqrt{2}

factor comes directly from energy considerations and appears frequently in orbital mechanics problems.

Question 5

A satellite moves from a circular orbit of radius $r$ to a circular orbit of radius $4r$ . By what factor does its angular momentum change?

$2$ (correct answer)
$4$
$\sqrt{2}$
$\frac{1}{2}$
$\frac{1}{4}$

Explanation: When you encounter orbital mechanics problems involving changes in radius, you need to connect angular momentum to the orbital velocity, which depends on the gravitational force providing centripetal force. For a circular orbit, gravitational force equals centripetal force:

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

. Solving for velocity gives

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

, so orbital speed decreases as

\frac{1}{\sqrt{r}}

when radius increases. Angular momentum is

L = mvr

. Substituting our expression for velocity:

L = m \cdot \sqrt{\frac{GM}{r}} \cdot r = m\sqrt{GMr}

. This shows that angular momentum is proportional to

\sqrt{r}

. When the radius changes from

r

4r

, the angular momentum changes by a factor of

\sqrt{\frac{4r}{r}} = \sqrt{4} = 2

. Therefore, answer A (

2

) is correct. Let's examine the wrong answers: Answer B (

4

) would be correct if angular momentum were proportional to

r

itself, which ignores how velocity decreases with radius. Answer C (

\sqrt{2}

) might come from incorrectly using

2r

instead of

4r

, or from other algebraic errors. Answer D (

\frac{1}{2}

) represents the opposite relationship and suggests confusion about whether angular momentum increases or decreases with radius. Remember: In orbital problems, always account for how velocity changes with radius. The satellite moves slower in higher orbits, but the increased radius more than compensates, resulting in greater angular momentum overall.

Question 6

A geosynchronous satellite orbits Earth with period equal to Earth's rotation period (24 hours). If a satellite were placed in orbit with half this orbital radius, what would be its approximate orbital period?

$8.5$ hours (correct answer)
$12$ hours
$6$ hours
$17$ hours
$3$ hours

Explanation: When you encounter orbital mechanics problems, think about Kepler's Third Law, which relates orbital period to orbital radius. This fundamental relationship governs all satellite motion around Earth. Kepler's Third Law states that the square of the orbital period is proportional to the cube of the orbital radius:

T^2 \propto r^3

, or

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

. For our problem, the geosynchronous satellite has period

T_1 = 24

hours at radius

r_1

. The new satellite has radius

r_2 = \frac{r_1}{2}

. Substituting into Kepler's Third Law:

\frac{(24)^2}{T_2^2} = \frac{r_1^3}{(r_1/2)^3} = \frac{r_1^3}{r_1^3/8} = 8

Solving for

T_2

T_2^2 = \frac{(24)^2}{8} = \frac{576}{8} = 72

Therefore:

T_2 = \sqrt{72} \approx 8.5

hours Choice A (8.5 hours) is correct. Choice B (12 hours) assumes a simple inverse relationship where halving the radius halves the period—this ignores the cubic dependence on radius. Choice C (6 hours) might come from incorrectly applying

T \propto \frac{1}{r^2}

, giving

24/4 = 6

. Choice D (17 hours) doesn't correspond to any reasonable orbital mechanics relationship. Study tip: Remember that orbital period decreases as satellites get closer to Earth, but not linearly. The

T^2 \propto r^3

relationship means closer orbits are disproportionately faster. Always use Kepler's Third Law for period-radius problems—never assume simple proportionality.

Question 7

Two identical satellites are initially in the same circular orbit around Earth. One satellite fires its engines briefly to increase its speed by 10% while maintaining its position in orbit. Immediately after this speed increase, what happens to this satellite?

It continues in the same circular orbit but with higher angular momentum
It moves to a larger circular orbit with the new speed as its orbital speed
It enters an elliptical orbit with its current position as the perigee (correct answer)
It enters an elliptical orbit with its current position as the apogee
It escapes Earth's gravitational field since the speed increase exceeds the escape requirement

Explanation: When a satellite suddenly increases its speed while in orbit, you're dealing with orbital mechanics and the relationship between velocity and orbital paths. The key insight is that circular orbits require a very specific velocity at each radius - any deviation from this velocity will create an elliptical orbit. In a stable circular orbit, the satellite's velocity is perfectly balanced with Earth's gravitational pull. When the satellite fires its engines and increases speed by 10%, it now has more velocity than required for a circular orbit at that distance. This excess energy means the satellite will move outward from Earth, but gravity will eventually slow it down and pull it back. The satellite enters an elliptical orbit where its current position becomes the closest point to Earth (perigee). From there, it will travel outward to a maximum distance (apogee) before falling back toward Earth. Answer C correctly identifies this scenario. Looking at the wrong answers: A is incorrect because the satellite cannot maintain the same circular orbit with higher speed - circular orbits require specific velocities. B is wrong because the satellite doesn't instantly jump to a new circular orbit; orbital mechanics doesn't work that way. D confuses perigee and apogee - the current position becomes the closest point (perigee), not the farthest point (apogee). Remember this pattern: increasing orbital velocity always moves you away from the central body initially, making your current position the perigee of the new elliptical orbit. Decreasing velocity would make it the apogee.

Question 8

A satellite orbits Earth with orbital radius $r$ and period $T$ . If this satellite were moved to orbit a planet with twice Earth's mass and half Earth's radius, what would be its period if it maintains the same orbital radius $r$ ?

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

Explanation: When you encounter orbital mechanics problems, the key relationship is Kepler's Third Law, which connects orbital period to the gravitational parameters of the system. For circular orbits, the period is given by

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

, where

r

is the orbital radius,

G

is the gravitational constant, and

M

is the mass of the central body. To find the new period, you need to compare the gravitational parameter

GM

between Earth and the new planet. The new planet has twice Earth's mass, so

M_{new} = 2M_{Earth}

. This means

GM_{new} = 2GM_{Earth}

. Notice that the planet's radius doesn't affect orbital mechanics for satellites orbiting at the same distance

r

from the center. Using the period formula:

T_{new} = 2\pi\sqrt{\frac{r^3}{GM_{new}}} = 2\pi\sqrt{\frac{r^3}{2GM_{Earth}}}

. Since the original period was

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

, we get

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

, which is answer B. Looking at the wrong answers: A)

\frac{T}{2}

incorrectly assumes a direct proportional relationship rather than the square root relationship. C)

\frac{T}{2\sqrt{2}}

mistakenly incorporates the planet's radius, which is irrelevant for orbital mechanics at radius

r

. D)

T\sqrt{2}

gets the direction wrong—more massive planets should decrease orbital periods, not increase them. Remember: in orbital problems, only the central body's mass matters for satellites at a given orbital radius. The square root relationship in Kepler's Third Law is crucial.

Question 9

A satellite orbits Earth at distance $R$ from Earth's center. At this distance, what is the relationship between the satellite's gravitational potential energy $U$ and kinetic energy $K$ for circular motion?

$U = -2K$ (correct answer)
$U = -K$
$U = K$
$U = 2K$
$U = -\frac{K}{2}$

Explanation: When you encounter orbital mechanics problems, you're dealing with the balance between gravitational forces and circular motion. For any satellite in a stable circular orbit, two key energy relationships govern its motion. Start with the force balance: gravitational force provides the centripetal force needed for circular motion. This gives us

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

, which simplifies to

v^2 = \frac{GM}{R}

. The kinetic energy is

K = \frac{1}{2}mv^2 = \frac{GMm}{2R}

, while gravitational potential energy is

U = -\frac{GMm}{R}

(negative because it's bound to Earth). Comparing these expressions:

U = -\frac{GMm}{R} = -2 \cdot \frac{GMm}{2R} = -2K

. This confirms answer choice A is correct. Let's examine why the other options fail. Choice B (

U = -K

) would violate the virial theorem for gravitational systems and lead to incorrect orbital velocities. Choice C (

U = K

) ignores the negative sign of gravitational potential energy entirely—gravitational PE is always negative for bound systems. Choice D (

U = 2K

) not only has the wrong magnitude but also the wrong sign, suggesting the satellite somehow has positive potential energy. Remember the "factor of 2" rule for circular orbits: gravitational potential energy is always twice the kinetic energy in magnitude but opposite in sign. This relationship (

U = -2K

) appears frequently in orbital mechanics problems and is a direct consequence of the virial theorem for inverse-square force laws.

Question 10

Consider a satellite in an elliptical orbit around Earth. The satellite's distance from Earth's center varies between $r_{min}$ and $r_{max}$ . At which point in its orbit does the satellite have the greatest angular momentum magnitude?

At $r_{min}$ (perigee)
At $r_{max}$ (apogee)
At the point where $r = \sqrt{r_{min} \cdot r_{max}}$
Angular momentum is constant throughout the orbit (correct answer)
At the point where the satellite's velocity is perpendicular to the radius vector

Explanation: When analyzing orbital motion, you need to recognize that satellites follow conservation laws that govern their behavior throughout the entire orbit. The key insight here is understanding angular momentum conservation in central force fields. Angular momentum is defined as

L = mvr\sin\theta

, where for orbital motion, the velocity is always perpendicular to the radius vector, so

L = mvr

. In any central force system like gravitational orbits, angular momentum is conserved because there are no external torques acting on the satellite. This means

L

remains constant at every point in the orbit, regardless of the satellite's distance from Earth. While the satellite's speed and distance from Earth change continuously throughout the elliptical orbit, these changes occur in precisely the right proportions to keep

mvr

constant. At perigee (choice A), the satellite moves fastest but is closest to Earth. At apogee (choice B), it moves slowest but is farthest away. These effects exactly compensate each other. Choice C suggests a specific geometric mean distance where angular momentum peaks, but this misunderstands the fundamental physics. There's no special point where angular momentum changes—it's the same everywhere. Choices A and B reflect the common misconception that extremes in orbital parameters (closest/farthest points) correspond to extremes in angular momentum. Remember this key principle: in orbital mechanics, angular momentum conservation is absolute. When you see questions about elliptical orbits, always consider what quantities are conserved (angular momentum, total energy) versus what varies (kinetic energy, potential energy, speed, radius).

Question 11

A satellite is in a circular orbit around Earth at distance $2R$ from Earth's center, where $R$ is Earth's radius. The satellite's engines fire to decrease its speed by exactly 10%. Immediately after the engines stop firing, what type of orbit does the satellite follow?

A circular orbit of smaller radius than $2R$
A circular orbit of larger radius than $2R$
An elliptical orbit with apogee at $2R$ (correct answer)
An elliptical orbit with perigee at $2R$
The satellite crashes into Earth because its speed is too low

Explanation: When a satellite's speed changes while in orbit, you need to understand how this affects orbital mechanics. The key principle is that at any point in an orbit, the satellite's speed and position determine the entire future orbital path. Initially, the satellite is in a circular orbit at distance

2R

with the exact speed needed for that circular motion:

v_{circular} = \sqrt{\frac{GM}{2R}}

. When the engines reduce this speed by 10%, the satellite suddenly has less speed than required for circular motion at that distance. With reduced speed, the satellite cannot maintain its circular orbit. However, its current position (

2R

from Earth's center) becomes a special point in its new orbit. Since the satellite now moves too slowly to "keep up" with circular motion, it will fall inward toward Earth under gravity's pull. The point where the speed change occurred becomes the apogee (farthest point) of the new elliptical orbit, because the satellite will never return to a point farther from Earth than where it started. Looking at the wrong answers: (A) and (B) suggest circular orbits, but you cannot instantly change from one circular orbit to another—orbital mechanics doesn't work that way. The satellite must follow a continuous path determined by its new speed and current position. (D) incorrectly identifies

2R

as the perigee (closest point), but since the satellite will fall inward from this point,

2R

must be the farthest point it reaches. Study tip: Remember that when you decrease orbital speed, the current position becomes the apogee of the new elliptical orbit; when you increase speed, it becomes the perigee.

Question 12

A satellite moves in an elliptical orbit around Earth. At its closest approach (perigee), it is at distance $r_1$ from Earth's center and has speed $v_1$ . At its farthest point (apogee), it is at distance $r_2$ from Earth's center. What is the speed $v_2$ at apogee?

$v_1\frac{r_1}{r_2}$ (correct answer)
$v_1\frac{r_2}{r_1}$
$v_1\sqrt{\frac{r_1}{r_2}}$
$v_1\sqrt{\frac{r_2}{r_1}}$
$v_1\left(\frac{r_1}{r_2}\right)^2$

Explanation: When you encounter orbital mechanics problems involving elliptical orbits, the key insight is that two fundamental conservation laws apply: conservation of energy and conservation of angular momentum. These allow you to relate conditions at different points in the orbit. At any two points in an elliptical orbit, angular momentum is conserved. Angular momentum

L = mvr

for circular motion, so at perigee and apogee:

m v_1 r_1 = m v_2 r_2

. The masses cancel, giving us

v_1 r_1 = v_2 r_2

. Solving for

v_2

v_2 = v_1 \frac{r_1}{r_2}

. This confirms answer A is correct. Notice that since

r_2 > r_1

(apogee is farther than perigee), we get

v_2 < v_1

, which makes physical sense—the satellite moves slower when it's farther from Earth. Answer B gives

v_1\frac{r_2}{r_1}

, which would make the satellite faster at apogee—physically impossible since it's moving against Earth's gravitational pull. Answer C,

v_1\sqrt{\frac{r_1}{r_2}}

, incorrectly applies an energy-like relationship (square roots suggest kinetic energy considerations) without proper derivation. Answer D,

v_1\sqrt{\frac{r_2}{r_1}}

, makes the same error but in the wrong direction. Remember this pattern: for orbital problems, angular momentum conservation gives you direct proportional relationships (

v \propto 1/r

), while energy methods typically involve square roots. When you see elliptical orbits, start with

v_1 r_1 = v_2 r_2

—it's usually your fastest path to the answer.

Question 13

A satellite in a circular orbit around Earth has total mechanical energy $E$ . What is the satellite's kinetic energy in terms of $E$ ?

$-E$ (correct answer)
$E$
$\frac{E}{2}$
$-\frac{E}{2}$
$2E$

Explanation: When analyzing orbital mechanics problems, you need to understand the relationship between kinetic energy, potential energy, and total mechanical energy for circular orbits. For a satellite in circular orbit, the gravitational force provides the centripetal force:

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

. This gives us the orbital velocity and kinetic energy:

K = \frac{1}{2}mv^2 = \frac{GMm}{2r}

. The gravitational potential energy is

U = -\frac{GMm}{r}

. Notice that

U = -2K

, so the total mechanical energy becomes:

E = K + U = K + (-2K) = -K

. Therefore,

K = -E

. Looking at the answer choices: Choice A (

-E

) is correct because the kinetic energy equals the negative of the total mechanical energy. Choice B (

E

) incorrectly assumes kinetic energy equals total energy, ignoring potential energy entirely. Choice C (

\frac{E}{2}

) might seem reasonable if you mistakenly think kinetic energy is half the total energy, but this ignores that total energy is negative. Choice D (

-\frac{E}{2}

) could result from incorrectly applying the virial theorem without accounting for the sign of total energy. Remember this key relationship for circular orbits:

K = -E

and

U = 2E

. The total energy is always negative (bound orbit), the kinetic energy is always positive, and potential energy is always negative with twice the magnitude of total energy. This pattern appears frequently in orbital mechanics problems.

Question 14

A satellite in a circular orbit has orbital energy $E_0$ . If the satellite's mass is tripled while keeping it in the same orbit, what is the new total orbital energy?

$3E_0$ (correct answer)
$E_0$
$\frac{E_0}{3}$
$9E_0$
$\sqrt{3}E_0$

Explanation: When you encounter orbital mechanics problems, focus on how orbital energy depends on the masses involved and the orbital parameters. For a satellite in a circular orbit, the total orbital energy consists of kinetic energy and gravitational potential energy. The key insight is that orbital energy is directly proportional to the satellite's mass. For a circular orbit, the total energy is

E = -\frac{GMm}{2r}

, where

G

is the gravitational constant,

M

is the central body's mass,

m

is the satellite's mass, and

r

is the orbital radius. Since we're keeping the satellite in the same orbit (same

r

) around the same central body (same

M

), the energy is directly proportional to the satellite mass

m

. When you triple the satellite's mass, you triple the orbital energy. If the original energy was

E_0

, the new energy becomes

3E_0

, making (A) correct. (B)

E_0

would suggest that orbital energy is independent of satellite mass, which contradicts the energy formula. (C)

\frac{E_0}{3}

represents an inverse relationship that doesn't exist in orbital mechanics—this might tempt students who confuse how mass affects different orbital quantities. (D)

9E_0

suggests the energy depends on the square of the mass, which would be incorrect and likely comes from confusion with other physics relationships where quantities scale as

m^2

. Remember: orbital energy scales linearly with satellite mass when orbital parameters remain constant. Don't confuse this with other orbital relationships that may have different mass dependencies.

Question 15

A satellite orbits Earth in a circular orbit of radius $r$ . What is the minimum energy required to move the satellite to a circular orbit of radius $2r$ ?

$\frac{GMm}{4r}$ (correct answer)
$\frac{GMm}{2r}$
$\frac{GMm}{r}$
$\frac{3GMm}{4r}$
$\frac{GMm}{8r}$

Explanation: When you encounter orbital mechanics problems involving energy changes, focus on the total mechanical energy of the system. For a satellite in circular orbit, the total energy is

E = -\frac{GMm}{2r}

, where the negative sign indicates the satellite is bound to Earth. To find the minimum energy required to change orbits, calculate the difference in total energies between the final and initial states. Initially, at radius

r

E_1 = -\frac{GMm}{2r}

. In the final orbit at radius

2r

E_2 = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}

. The energy difference is:

\Delta E = E_2 - E_1 = -\frac{GMm}{4r} - (-\frac{GMm}{2r}) = -\frac{GMm}{4r} + \frac{GMm}{2r} = \frac{GMm}{4r}

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

\frac{GMm}{2r}

) represents the magnitude of the initial orbital energy, not the energy difference. C (

\frac{GMm}{r}

) is twice the initial kinetic energy, which students might incorrectly calculate if they forget about potential energy changes. D (

\frac{3GMm}{4r}

) could result from incorrectly adding energies instead of finding their difference. Remember this key insight: when a satellite moves to a higher orbit, it actually moves slower and has less kinetic energy, but its potential energy increases by a larger amount. The energy you must supply equals the difference in total mechanical energies between the orbits.

Question 16

Two satellites of equal mass orbit Earth in circular orbits. Satellite A orbits at altitude $h$ above Earth's surface, while satellite B orbits at altitude $2h$ . What is the ratio of the kinetic energy of satellite A to the kinetic energy of satellite B?

$\frac{R + 2h}{R + h}$ (correct answer)
$\frac{R + h}{R + 2h}$
$\sqrt{\frac{R + 2h}{R + h}}$
$\left(\frac{R + 2h}{R + h}\right)^2$
$\frac{2(R + h)}{R + 2h}$

Explanation: When analyzing orbital mechanics problems involving energy, remember that for circular orbits, both kinetic and potential energy depend on the orbital radius, and there's a specific relationship between them. For a satellite in circular orbit, the gravitational force provides the centripetal force:

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

, where

r

is the distance from Earth's center. Solving for velocity gives

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

. Since kinetic energy is

KE = \frac{1}{2}mv^2

, we get

KE = \frac{GMm}{2r}

. For satellite A at altitude

h

r_A = R + h

, so

KE_A = \frac{GMm}{2(R + h)}

For satellite B at altitude

2h

r_B = R + 2h

, so

KE_B = \frac{GMm}{2(R + 2h)}

The ratio is:

\frac{KE_A}{KE_B} = \frac{\frac{GMm}{2(R + h)}}{\frac{GMm}{2(R + 2h)}} = \frac{R + 2h}{R + h}

This matches answer A. Answer B gives the inverse ratio - a common mistake when confusing which satellite has higher energy. Answer C provides the square root of the correct ratio, which you might get if you incorrectly related kinetic energy directly to orbital velocity instead of velocity squared. Answer D gives the square of the correct ratio, possibly from incorrectly squaring the entire energy relationship. Remember: kinetic energy in circular orbits is inversely proportional to orbital radius. Closer satellites move faster and have more kinetic energy, even though their total energy is more negative.

Question 17

A satellite moves from a circular orbit of radius $r$ to a circular orbit of radius $4r$ around the same planet. What is the ratio of its final kinetic energy to its initial kinetic energy?

$KE_f/KE_i = 1/4$ (correct answer)
$KE_f/KE_i = 1/16$
$KE_f/KE_i = 4$
$KE_f/KE_i = 1/2$

Explanation: For circular orbits,

KE = \frac{1}{2}mv^2 = \frac{1}{2}m\frac{GM}{r} = \frac{GMm}{2r}

. Since

KE \propto \frac{1}{r}

, when radius increases by factor 4, kinetic energy decreases by factor 4:

\frac{KE_f}{KE_i} = \frac{r_i}{r_f} = \frac{r}{4r} = \frac{1}{4}

. Choice B incorrectly uses

r^2

dependence. Choice C inverts the relationship. Choice D uses incorrect proportionality.

Question 18

A satellite orbits Earth at an altitude where the orbital period is 12 hours. If the satellite is moved to a new orbit where the orbital radius is doubled, what will be the new orbital period?

24 hours
33.9 hours (correct answer)
48 hours
6 hours

Explanation: Using Kepler's third law,

T^2 \propto r^3

, so

\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}

. With

r_2 = 2r_1

and

T_1 = 12

hours:

T_2 = 12 \times (2)^{3/2} = 12 \times 2\sqrt{2} = 12 \times 2.828 = 33.9

hours. Choice A incorrectly uses linear relationship. Choice C incorrectly squares the period ratio. Choice D incorrectly inverts the relationship.

Question 19

A satellite in an elliptical orbit around Earth has its closest approach at distance $r_1$ from Earth's center and its farthest point at distance $r_2 = 3r_1$ . If the satellite's speed at closest approach is $v_1$ , what is its speed $v_2$ at the farthest point?

$v_2 = v_1\sqrt{3}/3$
$v_2 = v_1/\sqrt{3}$
$v_2 = v_1/3$ (correct answer)
$v_2 = v_1/9$

Explanation: When you encounter orbital mechanics problems involving elliptical orbits, you need to apply conservation laws—specifically conservation of angular momentum and energy. For any object in orbit, angular momentum

L = mvr

remains constant throughout the orbit. At the closest approach (periapsis) and farthest point (apoapsis), the velocity is perpendicular to the radius vector, so:

m v_1 r_1 = m v_2 r_2

Since

r_2 = 3r_1

, we can substitute:

v_1 r_1 = v_2 (3r_1)

v_1 = 3v_2

v_2 = \frac{v_1}{3}

This confirms answer C is correct. Let's examine why the other options are wrong: Option A (

v_2 = v_1\sqrt{3}/3

) gives

v_2 = v_1/\sqrt{3}

, which would result from incorrectly taking the square root of the radius ratio—a common mistake when students confuse this with energy relationships. Option B (

v_2 = v_1/\sqrt{3}

) also involves an incorrect square root relationship, possibly from misapplying Kepler's laws or confusing orbital period relationships with velocity ratios. Option D (

v_2 = v_1/9

) would come from incorrectly squaring the radius ratio, treating this like an inverse-square law problem rather than the linear relationship that angular momentum conservation actually provides. Study tip: In orbital problems, always start with conservation laws. Angular momentum conservation gives you velocity ratios directly from radius ratios:

v_1 r_1 = v_2 r_2

. Don't overthink it with square roots or squares unless energy considerations specifically require them.

Question 20

A satellite orbits Earth with total energy $E = -2.0 \times 10^{10}$ J. What is the satellite's kinetic energy?

$KE = 1.0 \times 10^{10}$ J
$KE = -1.0 \times 10^{10}$ J
$KE = 4.0 \times 10^{10}$ J
$KE = 2.0 \times 10^{10}$ J (correct answer)

Explanation: When you encounter orbital mechanics problems involving total energy, remember that satellites in stable orbits follow a specific energy relationship governed by gravitational forces. For any satellite in a circular or elliptical orbit, the total energy

E

relates to kinetic energy

KE

and potential energy

PE

through the virial theorem. In gravitational systems, the kinetic energy is always exactly half the magnitude of the potential energy:

KE = -\frac{1}{2}PE

. Since total energy equals

E = KE + PE

, we can substitute to get

E = KE + (-2KE) = -KE

. Therefore,

KE = -E

. Given that

E = -2.0 \times 10^{10}

J, the kinetic energy is

KE = -(-2.0 \times 10^{10}) = 2.0 \times 10^{10}

J. This confirms answer D is correct. Let's examine why the other options are wrong: Option A (

1.0 \times 10^{10}

J) incorrectly assumes kinetic energy is half the total energy magnitude, confusing the relationship between kinetic and potential energy. Option B (

-1.0 \times 10^{10}

J) is impossible since kinetic energy must always be positive—it represents motion energy. Option C (

4.0 \times 10^{10}

J) incorrectly doubles the total energy magnitude, perhaps confusing the factor of 2 in the virial relationship. Remember this key orbital relationship: for bound gravitational systems, kinetic energy always equals the negative of total energy (

KE = -E

). This makes orbital energy problems much more straightforward once you recognize the pattern.