What is the relationship between a satellite's orbital radius and its velocity?

As a satellite's orbital radius increases, its velocity decreases. This is because the gravitational force weakens with distance, so less velocity is needed to maintain orbit. The product of the orbital radius and velocity remains constant for a given central body, a principle known as the conservation of angular momentum.

How does the concept of escape velocity relate to satellite orbits?

Escape velocity is the minimum speed an object needs to break free from a body's gravitational field without further propulsion. For satellites, any velocity below escape velocity results in an orbit, while reaching escape velocity allows the object to leave Earth's influence entirely. The escape velocity decreases with increasing distance from Earth.

How does atmospheric drag affect the orbit of low Earth orbit (LEO) satellites?

Atmospheric drag causes LEO satellites to lose energy over time, gradually lowering their orbits. This effect is more pronounced for satellites in very low orbits and can eventually cause them to re-enter the atmosphere if their orbits are not periodically boosted.

How does the concept of angular momentum apply to satellite orbits?

Angular momentum is conserved in satellite orbits. As a satellite moves closer to Earth, its velocity increases to maintain constant angular momentum. This principle explains why satellites move faster at perigee and slower at apogee in elliptical orbits.

What is the significance of the Lagrange points in satellite orbits?

Lagrange points are positions in space where the gravitational forces of two large bodies (like Earth and the Moon) balance the centrifugal force felt by a much smaller third body (like a satellite). These points are useful for positioning satellites with minimal station-keeping requirements.

Why do satellites in lower orbits have shorter periods than those in higher orbits?

Satellites in lower orbits experience stronger gravitational force from Earth, requiring higher velocities to maintain their orbits. These higher velocities result in shorter orbital periods. Conversely, satellites in higher orbits experience weaker gravity and move more slowly, leading to longer periods.

What is the significance of the "critical altitude" for satellite orbits?

The critical altitude is where the orbital period of a satellite equals the rotational period of the body it's orbiting. For Earth, this is the altitude of geosynchronous orbits. Satellites below this altitude move faster than Earth's rotation, while those above it move slower.

Why do satellites in higher orbits require less frequent station-keeping maneuvers?

Satellites in higher orbits experience less atmospheric drag and weaker perturbations from Earth's non-spherical shape. They also have longer orbital periods, so any deviations accumulate more slowly. As a result, they require less frequent adjustments to maintain their intended orbits.

How does Earth's oblateness affect satellite orbits?

Earth's oblateness (its slight flattening at the poles) causes perturbations in satellite orbits. It leads to precession of the orbit's line of nodes and rotation of the line of apsides. These effects are more pronounced for satellites in lower orbits and can be used advantageously for certain missions, like sun-synchronous orbits.

What is meant by "orbital resonance," and how can it affect satellite orbits?

Orbital resonance occurs when two orbiting bodies exert regular, periodic gravitational influences on each other. For satellites, resonances with Earth's rotation can lead to stable or unstable orbits. For example, the geosynchronous orbit is in a 1:1 resonance with Earth's rotation, while certain other resonances are avoided to prevent orbital instabilities.

How does the energy of a satellite change as it moves to a higher orbit?

As a satellite moves to a higher orbit, its total energy increases. This is because it gains potential energy by moving farther from Earth, which outweighs the decrease in kinetic energy due to lower velocity. The increase in energy must be supplied externally, such as by firing thrusters.

How does the shape of an orbit affect a satellite's energy?

In elliptical orbits, a satellite's energy remains constant, but its distribution between kinetic and potential energy changes. At the closest point (perigee), kinetic energy is maximum and potential energy is minimum. At the farthest point (apogee), the opposite is true. Circular orbits maintain a constant distribution of energy.

What is the "Hohmann transfer orbit," and how does it relate to satellite energy?

The Hohmann transfer orbit is the most energy-efficient way to move a satellite between two circular orbits in the same plane. It's an elliptical orbit that touches both the initial and final circular orbits. The satellite performs two burns: one to enter the transfer orbit and another to circularize at the new altitude.

What is meant by the term "orbital energy," and how is it calculated?

Orbital energy is the total mechanical energy of a satellite in orbit, which is the sum of its kinetic and gravitational potential energy. It's calculated as E = -GMm/2r, where G is the gravitational constant, M is Earth's mass, m is the satellite's mass, and r is the orbital radius. The negative sign indicates it's a bound system.

How does the inclination of a satellite's orbit affect its coverage area on Earth?

The inclination determines the range of latitudes the satellite passes over. A satellite with 0° inclination (equatorial orbit) only covers equatorial regions, while one with 90° inclination (polar orbit) covers all latitudes. Higher inclinations provide broader coverage but require more energy to achieve.

What determines the time period of a satellite orbiting Earth?

The time period of a satellite depends primarily on its distance from the center of Earth. The farther the satellite is from Earth, the longer its orbital period. This relationship is described by Kepler's Third Law of Planetary Motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.

How does the mass of a satellite affect its orbital period?

Surprisingly, the mass of the satellite does not affect its orbital period. The period depends only on the orbit's radius and the mass of the body being orbited (in this case, Earth). This is because the gravitational force increases proportionally with the satellite's mass, but so does its inertia, resulting in no net effect on the period.

What is the difference between geosynchronous and geostationary orbits?

While both have the same orbital period as Earth's rotation, a geosynchronous orbit can have any inclination, whereas a geostationary orbit must be directly above the equator (0° inclination). Geostationary satellites appear fixed in the sky, while geosynchronous satellites may appear to move in a figure-8 pattern.

What is the geosynchronous orbit, and why is it special?

The geosynchronous orbit is a circular orbit above Earth's equator with a period of exactly one sidereal day (23 hours, 56 minutes, 4 seconds). It's special because satellites in this orbit appear stationary from Earth's surface, making it ideal for communication satellites and weather monitoring.

What is the relationship between a satellite's orbital period and its altitude?

The orbital period increases with altitude, but not linearly. The relationship follows Kepler's Third Law: T² ∝ r³, where T is the orbital period and r is the orbital radius. This means that doubling the altitude (measured from Earth's center) increases the period by a factor of 2√2, or about 2.83.

What is the "critical inclination" for satellite orbits, and why is it significant?

The critical inclination (about 63.4° or 116.6°) is an orbital inclination where the precession of the line of apsides due to Earth's oblateness is zero. Orbits at this inclination maintain a stable orientation of their major axis relative to Earth. This property is useful for certain types of communication and reconnaissance satellites.

How does the concept of specific orbital energy apply to satellites?

Specific orbital energy is the total energy per unit mass of a satellite. It's independent of the satellite's mass and depends only on the orbit's characteristics. For circular orbits, it's given by ε = -μ/2r, where μ is Earth's standard gravitational parameter and r is the orbital radius. This concept is useful for comparing different orbits.

How does solar radiation pressure affect satellite orbits and energy?

Solar radiation pressure is the force exerted on a satellite by sunlight. While small, it can significantly affect satellite orbits over time, especially for satellites with large surface areas relative to their mass (like solar sails). It can cause orbital perturbations and changes in the satellite's energy, requiring correction maneuvers.

How does the concept of "orbital eccentricity" relate to a satellite's energy distribution?

Orbital eccentricity measures how much an orbit deviates from circular. In circular orbits (e=0), the energy distribution remains constant. As eccentricity increases, the variation in energy distribution grows. In highly eccentric orbits, the satellite's speed and altitude change dramatically throughout its orbit, leading to large fluctuations in the kinetic-to-potential energy ratio.

What is the significance of the "sphere of influence" concept in satellite orbits?

The sphere of influence is the region around a celestial body where its gravitational influence dominates over other bodies. For Earth satellites, it defines the approximate boundary where Earth's gravity is more significant than the Sun's. Understanding this concept is crucial for planning interplanetary missions and determining when to switch reference frames in orbital calculations.

How does the "Oberth effect" relate to satellite energy and orbital maneuvers?

The Oberth effect states that a rocket burn is most efficient when the vehicle's velocity is greatest. For satellites, this means that propulsive maneuvers are most effective at the lowest point of an orbit (perigee). This effect is used to design efficient orbital transfers and is particularly important for interplanetary missions.

What is meant by "orbital decay," and how does it affect a satellite's energy?

Orbital decay is the gradual decrease in a satellite's orbital altitude due to atmospheric drag or other perturbations. As the orbit decays, the satellite loses energy, converting orbital energy into heat. This process accelerates as the satellite enters denser parts of the atmosphere, eventually leading to re-entry if not corrected.

How does the concept of "specific angular momentum" apply to satellite orbits?

Specific angular momentum is the angular momentum per unit mass of a satellite. It remains constant throughout the orbit due to the central nature of the gravitational force. This conservation principle explains why satellites move faster when closer to Earth and slower when farther away, maintaining a constant r × v (radius times velocity) product.

What is a "sun-synchronous orbit," and how is it achieved?

A sun-synchronous orbit is one where the satellite passes over any given point on Earth's surface at the same local solar time. This is achieved by carefully choosing the orbit's altitude and inclination so that the precession rate of the orbit matches Earth's revolution around the Sun. These orbits are useful for Earth observation satellites that need consistent lighting conditions.

How does the "nodal regression" of satellite orbits relate to Earth's oblateness?

Nodal regression is the gradual rotation of a satellite's orbital plane caused by Earth's oblateness. It causes the line of nodes (the intersection of the orbital plane with Earth's equatorial plane) to rotate. The rate of regression depends on the orbit's inclination and altitude. This effect is used to design sun-synchronous orbits and can affect the long-term evolution of satellite constellations.

What is a "Molniya orbit," and how does its energy distribution differ from circular orbits?

A Molniya orbit is a highly elliptical orbit with a period of half a sidereal day and an inclination of 63.4° (the critical inclination). Its energy distribution varies greatly throughout the orbit, with high velocity and kinetic energy at perigee, and low velocity but high potential energy at apogee. This orbit type is used for communication satellites serving high latitudes.

How does the "third-body effect" influence satellite orbits and energy?

The third-body effect refers to the gravitational influence of bodies other than Earth (like the Moon or Sun) on satellite orbits. These perturbations can cause long-term changes in orbital elements and energy. For some orbits, like geostationary, these effects must be actively compensated for through station-keeping maneuvers.

Time Period And Energy Of A Satellite

Physics
Gravitation
Time Period And Energy Of A Satellite

Time Period And Energy Of A Satellite

Vishal kumarUpdated on 02 Jul 2025, 07:57 PM IST

Satellites, whether natural like the Moon or man-made, orbit Earth due to a delicate balance of gravitational pull and velocity. The time period of a satellite, which is the time it takes to complete one full orbit, is a fundamental concept in understanding its motion. This period is influenced by the satellite's altitude and the mass of the Earth. The energy of a satellite, comprising both kinetic and potential energy, determines its speed and orbit stability. In real life, these principles are crucial for the operation of GPS satellites that guide our navigation systems, communications satellites that enable global connectivity, and weather satellites that monitor Earth's climate. Understanding the time period and energy of satellites helps engineers design efficient and reliable systems that are integral to modern technology, ensuring everything from accurate weather forecasts to uninterrupted phone calls.

This Story also Contains

The Time Period of the Satellite
Height of Satellite
The Energy of Satellite
Binding Energy (B.E.)
Solved Examples Based on Time Period And Energy of a Satellite

Time Period And Energy Of A Satellite

The Time Period of the Satellite

The time period of a satellite is the duration it takes for the satellite to complete one full orbit around a celestial body, such as Earth. This time period is determined by the satellite's altitude and the mass of the celestial body it orbits. According to Kepler's laws of planetary motion, the time period is directly related to the size of the orbit—the higher the satellite, the longer the time period.

The time period (T) of the satellite is given by
$\begin{array}{ll} T = \frac{2 π r}{v} = 2 π r \sqrt{\frac{r}{G M}} & [As v = \sqrt{\frac{G M}{r}}] \\ T = 2 π \sqrt{\frac{r^{3}}{G M}} = 2 π \sqrt{\frac{r^{3}}{g R^{2}}} & [As G M = g R^{2}] \\ T = 2 π \sqrt{\frac{(R + h)^{3}}{g R^{2}}} = 2 π \sqrt{\frac{R}{g}} {(1 + \frac{h}{R})}^{3 / 2} & [As r = R + h] \end{array}$
Where
$r =$ radius of orbit
$T \to$ Time period
$M \to$ Mass of planet
If the satellite is very close to the Earth's surface,
i.e., $h ≪≪ R$ ,

$\begin{aligned} T = 2 π \sqrt{\frac{R}{g}} ≅ 84.6 minutes \\ then \\ or T ≃ 1.4 h r \end{aligned}$

The time period of a satellite in terms of density
$T = \sqrt{\frac{3 π}{G ρ}}$

$ρ \to$ Density of planet
$T \to$ Time period
$G \to$ Gravitational constant

$ρ = 5478.4 Kg / m_{for earth}^{3}$

For a satellite, the time interval between the two consecutive appearances overhead

If a satellite in the equilateral planes moves from west to east Angular velocity of the satellite with respect to an observer on earth will be $(ω_{S} - ω_{E})$
$ω_{S} \to$ Satellite angular velocity
$ω_{E} \to$ earth angular velocity
So $T = \frac{2 π}{ω_{S} - ω_{E}} = \frac{T_{S} T_{E}}{T_{E} - T_{S}}$
if $ω_{S} = ω_{E}, T = \infty$

means the satellite will appear stationary relative to Earth.

Height of Satellite

As we know, time period of satellite $T = 2 π \sqrt{\frac{r^{3}}{G M}} = 2 π \sqrt{\frac{(R + h)^{3}}{g R^{2}}}$ By squaring and rearranging both sides $\frac{q R^{2} T^{2}}{4 π^{2}} = (R + h)^{3}$ $\Rightarrow h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{1 / 3} - R$

Putting the value of the period in the above formula we can calculate the height of the satellite from the surface of the earth.

The Energy of Satellite

When a satellite revolves around a planet in its orbit, it possesses both kinetic energy (due to orbital motion) and potential energy (due to its position against the gravitational pull of Earth). These energies are given by

Potential energy: $U = m V = \frac{- G M m}{G M m} = \frac{- L^{2}}{m r^{2}}$
Kinetic energy : $K = \frac{1}{2} m v^{2} = \frac{G M m}{2 r} = \frac{m L^{2}}{2 m r^{2}}$
Total energy : $E = U + K = \frac{- G M m}{r} + \frac{(2 M m}{2 r} = \frac{- G M m}{2 r} = \frac{- L^{2}}{2 m r^{2}}$
Where

$\begin{aligned} M \to mass of planet \\ m \to mass of satellite \end{aligned}$
And

$\begin{aligned} K = - E \\ U = 2 E \\ U = - 2 K \end{aligned}$

Energy Graph of Satellite

Where

$E \to$ Energy of satellite
$K \to$ Kinetic energy
$U \to$ Potential energy

Energy Distribution in an Elliptical Orbit

In this Total Energy
$E = - \frac{G M m}{2 a} = const$
Where $a =$ semi-major axis

Binding Energy (B.E.)

The minimum energy required to remove the satellite from its orbit to infinity is called Binding Energy.

And It is given by
$B \cdot E = \frac{G M m}{2 r}$

where
B.E Binding energy
$M \to$ mass of planet
$m \to$ mass of satellite

Work done in changing the orbit

When the satellite is transferred to a higher orbit i.e. $(r_{2} > r_{1})$ as shown in the figure.

$\begin{aligned} W & = E_{2} - E_{1} \\ W & = \frac{G M m}{2} [\frac{1}{r_{1}} - \frac{1}{r_{2}}] \end{aligned}$
Where
$W \to$ work done
$r_{1} \to$ radius of 1st orbit
$r_{2} \to$ radius of $2 n d$ orbit

Solved Examples Based on Time Period And Energy of a Satellite

Example 1: A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has a radius R (R<<L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the period of the star is T and its distance from the Galaxies axis is r, then :

1) $T^{2} α r^{3}$
2) $T α r^{2}$
3) $T α r$
4) $T α \sqrt{r}$

Solution:

The period of satellite
$\begin{aligned} T = 2 π \sqrt{\frac{r^{3}}{G M}} \\ r = radius of orbit \\ T \to Period \\ M \to Mass of planet \\ T = \frac{Circum ference of orbit}{orbital velocity} \\ F = \frac{2 G M}{L r} m, F = (\frac{k}{r}) m \end{aligned}$
Here K is some constant

$\begin{aligned} (\frac{m v^{2}}{r}) = \frac{k m}{r} \Rightarrow v = constant \\ T = \frac{2 π r}{v} \\ ∴ T \propto r \end{aligned}$

Hence, the answer is the option (3).

Example 2: The relative uncertainty in the period of a satellite orbiting around the earth is $10^{- 2}$ . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is

1) $10^{- 2}$
2) $2 \times 10^{- 2}$
3) $3 \times 10^{- 2}$
4) $6 \times 10^{- 2}$

Solution:

Period of satellite

$\begin{aligned} T = 2 π \sqrt{\frac{r^{3}}{G M}} \\ r = radius of orbit \\ T \to \underset{―}{Time period} \\ M \to Mass of planet \\ wherein \\ T = \frac{Circum ference of orbit}{orbital velocity} \\ v = \sqrt{(\frac{Q M}{R})} \Rightarrow T = \frac{2 π R}{v} = \frac{2 π}{\sqrt{(4 m)}} \cdot R^{3 / 2} \\ M = \frac{4 π^{2}}{4} \cdot \frac{R^{3}}{T^{2}} \\ \frac{Δ M}{M} = 3 \cdot \frac{Δ R}{R} + 2 \frac{Δ T}{T} = 3.0 + 2 \times 10^{- 2} \\ \frac{Δ M}{M} = 2 \times 10^{- 2} \end{aligned}$

Hence, the answer is the option (2).

Example 3: A spaceship orbits around a planet at a height of 20 Km from its surface. Assuming that only the gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? (Given : Mass of planet $= 8 \times 10^{22} Kg$ , Radius of planet $= 2 \times 10^{6} m$ Gravitational Constant $G = 6.67 \times 10^{- 11} {Nm}^{2} {Kg}^{- 2}$ )

1) 11

2) 17

3) 13

4) 9

Solution:

( $T =$ time, $r =$ radius, $G =$ gravitational force, $m =$ mass of planet $)$
using r=R+h

$\begin{aligned} T = 2 π \sqrt{\frac{r^{3}}{G M}} = 2 π \sqrt{\frac{{(2.02 \times 10^{6})}^{3}}{6.67 * 10^{- 11} \times 8 \times 10^{22}}} = 7800 \sec . \\ Now, No of revolution = \frac{24 \times 3600}{7800} = 11.07 = 11 \end{aligned}$

Hence, the answer is the Option (1).

Example 4: If the satellite whose mass m is revolving in a circular orbit of radius r around the earth (mass of earth =M). The time of the revolution of satellites is:

1) $T α \frac{r^{3}}{\sqrt{G M}}$
2) $T \propto \frac{r^{\frac{3}{2}}}{\sqrt{G M}}$
3) $T α \frac{\sqrt{G M}}{r 3}$
4) $T α \frac{\sqrt{G M}}{r^{\frac{3}{2}}}$

Solution:

$\begin{aligned} T & = \frac{2 π r}{v_{0}} = \frac{2 π r}{\sqrt{\frac{G M}{r}}} \\ T & = \frac{2 π r}{\sqrt{\frac{G M}{r}}} \\ T & \propto \frac{r^{\frac{3}{2}}}{\sqrt{G M}} \end{aligned}$

Hence, the answer is the option (2).

Example 5: If the density of the planet is $ρ$ and a satellite is revolving in a circular orbit of radius r then the time of revolution varies with density $ρ$ as :

1) $T \propto \sqrt{ρ}$
2) $T \propto \frac{1}{ρ}$
3) $T \propto ρ$
4) $T \propto \frac{1}{\sqrt{ρ}}$

Solution

The time period of a satellite in terms of density
$T = \sqrt{\frac{3 π}{G ρ}}$

$ρ \to$ Density of planet
$T \to$ Time period
$G \to$ Gravitational constant
wherein

$ρ = 5478.4 Kg / m^{3} for earth$
As we known

$\begin{aligned} T = \frac{2 π r^{\frac{3}{2}}}{\sqrt{G M}} \\ M = ρ * \frac{4}{3} π r^{3} \\ T = \frac{2 π r^{\frac{3}{2}}}{\sqrt{G ρ * \frac{4}{3} π R^{3}}} \\ so, \\ T = \sqrt{\frac{3 π}{G ρ}} \end{aligned}$

$T \propto \frac{1}{\sqrt{ρ}}$

Hence, the answer is the option (4).

Summary

The time period of a satellite depends on its orbit radius and the mass of the planet, following Kepler's laws. It is influenced by gravitational forces and can be calculated using specific formulas that account for satellite altitude and Earth's mass. The satellite's energy consists of kinetic and potential energy, which together determine orbit stability. Understanding these factors helps in applications like GPS, communication, and weather satellite

Frequently Asked Questions (FAQs)

Q: What is the "Hill sphere," and how does it relate to a satellite's orbital stability?

The Hill sphere is the region around a celestial body where its gravitational influence dominates over other nearby massive bodies. For Earth satellites, it defines the approximate volume where stable orbits are possible. Beyond the Hill sphere, a satellite's orbit becomes unst

Q: How does the "Kozai mechanism" affect the long-term evolution of satellite orbits and energy?

The Kozai mechanism is a gravitational effect that causes coupled oscillations in the eccentricity and inclination of a satellite's orbit due to the gravitational influence of a third body (like the Moon for Earth satellites). While conserving the orbit's semi-major axis and energy, it can lead to significant long-term changes in the orbit's shape and orientation.

Q: What is meant by "orbital injection," and how does it relate to a satellite's initial energy?

Orbital injection is the process of placing a satellite into its intended orbit. It involves providing the satellite with the correct velocity and position to achieve the desired orbital parameters. The energy required for orbital injection depends on the target orbit's altitude and shape. Higher or more eccentric orbits generally require more energy for injection.

Q: How does the concept of "gravitational assist" relate to changing a satellite's energy?

Gravitational assist, or gravity assist, is a technique where a satellite uses the gravity of a planet or moon to change its trajectory and speed. This maneuver can increase or decrease the satellite's kinetic energy and alter its direction without using propellant. It's commonly used for interplanetary missions to save fuel and achieve velocities that would be impractical with propulsion alone.

Q: How does the concept of "orbital velocity" relate to a satellite's energy?

Orbital velocity is directly related to a satellite's kinetic energy. For a circular orbit, the velocity is given by v = √(GM/r), where G is the gravitational constant, M is Earth's mass, and r is the orbital radius. As the orbit's radius increases, the velocity decreases, but the total energy (kinetic + potential) increases due to the greater increase in potential energy.

Q: What is the significance of the "semi-major axis" in determining a satellite's orbital energy?

The semi-major axis is a key parameter in orbital mechanics, directly related to a satellite's total energy. For elliptical orbits, it's half the longest diameter of the ellipse. The specific orbital energy is given by ε = -μ/(2a), where μ is Earth's standard gravitational parameter and a is the semi-major axis. This relationship shows that larger orbits have higher (less negative) total energy.

Q: How does the concept of "orbital resonance" affect the long-term stability and energy of satellite orbits?

Orbital resonance occurs when the orbital periods of a satellite and another body (or Earth's rotation) have a simple integer ratio. Some resonances can lead to stable configurations, while others can cause orbital instabilities. Resonances can either help maintain a satellite's orbit with minimal energy input or require additional energy for station-keeping to avoid undesirable effects.

Q: What is the "frozen orbit" concept, and how does it relate to satellite energy conservation?

A frozen orbit is designed to minimize variations in orbital elements over time due to perturbations like Earth's oblateness. By carefully selecting the eccentricity, inclination, and argument of perigee, these orbits maintain nearly constant average altitude and ground track. This stability helps conserve satellite energy by reducing the need for frequent orbital corrections.

Q: How does atmospheric drag affect the energy of satellites in very low Earth orbits?

Atmospheric drag causes satellites in very low Earth orbits to lose energy continuously. This energy loss manifests as a decrease in orbital altitude and an increase in orbital velocity (as the satellite falls into a lower orbit). Without intervention, this leads to a spiral decay of the orbit and eventually atmospheric re-entry.

Q: What is the significance of "Lagrange points" in terms of satellite energy?

Lagrange points are locations in space where a small object can maintain a stable position relative to two larger bodies. At these points, the gravitational and centrifugal forces balance, creating a local energy minimum. Satellites placed at Lagrange points require minimal energy for station-keeping, making them ideal for certain space observation missions.

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