Energy of Elliptical Orbits: Videos & Practice Problems

In elliptical orbits, kinetic and potential energies vary, unlike in circular orbits where they remain constant. At periapsis, the spacecraft's speed and kinetic energy are at their maximum, while potential energy is at its minimum. Conversely, at apoapsis, speed and kinetic energy are at their minimum, and potential energy is at its maximum. The relationship between velocities at two points can be expressed as mi>v1r1=v2r2. Transitioning between circular and elliptical orbits involves energy conservation, requiring work to change the orbit's shape.

Speed and Energy of Elliptical Orbits

In the study of elliptical orbits, it's essential to understand that the kinetic and potential energies are not constant, unlike in circular orbits. In an elliptical orbit, the distance from the central body, denoted as \( r \), varies throughout the orbit, leading to changes in both kinetic energy (\( K \)) and potential energy (\( U \)). At the farthest point in the orbit, known as the apoapsis, the spacecraft has minimum velocity and thus minimum kinetic energy. Conversely, at the closest point, called the periapsis, the spacecraft reaches maximum velocity and maximum kinetic energy. This relationship is governed by the principle of conservation of energy, which states that as one form of energy increases, the other must decrease to maintain a constant total energy.

To quantify the relationship between velocity and distance at two points in an elliptical orbit, the equation \( v_1 r_1 = v_2 r_2 \) can be utilized. This equation arises from the conservation of angular momentum, where angular momentum (\( L \)) is defined as \( L = mvr \). In this context, \( m \) represents mass, \( v \) is velocity, and \( r \) is the distance from the central body. Since the mass remains constant, it cancels out, simplifying the relationship to the aforementioned equation. This allows for the comparison of velocities at different points in the orbit, enabling the calculation of an unknown variable when three of the four variables are provided.

For example, if a spacecraft's velocity at periapsis (\( v_1 \)) and the corresponding distance (\( r_1 \)) are known, along with the distance at apoapsis (\( r_2 \)), the velocity at apoapsis (\( v_2 \)) can be calculated. This is done by rearranging the equation to \( v_2 = \frac{v_1 r_1}{r_2} \). It is important to ensure that the units used for velocity and distance are consistent, even if they are not in SI units, to achieve accurate results.

In summary, understanding the dynamics of elliptical orbits involves recognizing the interplay between kinetic and potential energy, as well as applying the conservation of angular momentum to relate velocities and distances at different points in the orbit.

Pluto travels in a fairly elliptical orbit around the Sun. At its closest distance of $4.43\times 10^9$ km, its orbital speed is 6.12 km/s. At its farthest, its orbital speed reduces to just 3.71 km/s. How far is it from the Sun at this point (in km)?

• A. $2.69\times 10^{12}$ km
• B. $7.31\times 10^{12}$ km
• C. $2.69\times 10^9$ km
• D. $7.31\times 10^9$ km

Answer: (D)

Comet Traveling Around the Sun

In this problem, we analyze the orbit of a comet that travels around the sun with an orbital period of 2000 years. The closest approach to the sun, known as periapsis (denoted as \( r_p \)), is given as \( 3 \times 10^8 \) kilometers. To find the farthest distance from the sun, or apoapsis (denoted as \( r_a \)), we first need to determine the semi-major axis (\( a \)) of the comet's elliptical orbit.

Using Kepler's Third Law, which states that the square of the orbital period (\( T \)) is proportional to the cube of the semi-major axis, we can express this relationship mathematically as:

\[ T^2 = \frac{4\pi^2 a^3}{G M} \]

Here, \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)) and \( M \) is the mass of the sun (\( 2 \times 10^{30} \, \text{kg} \)). To use this equation, we must convert the orbital period from years to seconds:

\[ T = 2000 \, \text{years} \times 365 \, \text{days/year} \times 86400 \, \text{seconds/day} = 6.31 \times 10^{10} \, \text{s} \]

Substituting \( T \) into Kepler's equation and rearranging gives us:

\[ a^3 = \frac{G M T^2}{4\pi^2} \]

Calculating \( a \) yields:

\[ a = \sqrt[3]{\frac{(6.67 \times 10^{-11})(2 \times 10^{30})(6.31 \times 10^{10})^2}{4\pi^2}} \approx 2.38 \times 10^{13} \, \text{m} \]

Next, we can find the apoapsis distance using the relationship between the semi-major axis and the periapsis:

\[ a = \frac{r_a + r_p}{2} \Rightarrow r_a = 2a - r_p \]

Substituting the values gives:

\[ r_a = 2(2.38 \times 10^{13}) - (3 \times 10^{11}) \approx 4.73 \times 10^{13} \, \text{m} \]

For part b, we need to find the ratio of the comet's speed at periapsis (\( v_p \)) to its speed at apoapsis (\( v_a \)). Using the conservation of angular momentum, we can express this as:

\[ \frac{v_p}{v_a} = \frac{r_a}{r_p} \]

Substituting the known distances:

\[ \frac{v_p}{v_a} = \frac{4.73 \times 10^{13}}{3 \times 10^{11}} \approx 158 \]

This indicates that the comet travels approximately 158 times faster at its closest point to the sun compared to its farthest point. This analysis highlights the significant variation in speed due to the elliptical nature of the orbit.

Energy Conservation in Changing Elliptical Orbits

Understanding the transition from circular to elliptical orbits involves the principle of energy conservation. When changing orbits, work must be done, whether moving from a circular orbit to an elliptical one or vice versa. The total mechanical energy in an orbit is the sum of kinetic and potential energy, which can be expressed as a single equation for circular orbits:

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

where \( G \) is the gravitational constant, \( M \) is the mass of the central body, \( m \) is the mass of the orbiting object, and \( r \) is the radius of the circular orbit. However, in elliptical orbits, the distance \( r \) varies as the object moves from periapsis (closest point) to apoapsis (farthest point). Therefore, the semi-major axis \( a \) becomes the key variable, leading to the modified energy equation for elliptical orbits:

\[ E = -\frac{GMm}{2a} \]

To calculate the work done in transitioning from a circular orbit to an elliptical orbit, we can set up the energy conservation equation:

\[ E_{\text{initial}} + W = E_{\text{final}} \]

Substituting the energy expressions gives:

\[ -\frac{GMm}{2r} + W = -\frac{GMm}{2a} \]

Rearranging this yields:

\[ W = \frac{GMm}{2r} - \frac{GMm}{2a} \]

In this context, \( W \) represents the work done, which is equivalent to the energy required to change the orbit. The semi-major axis \( a \) is calculated using the relationship between the periapsis and apoapsis distances:

\[ a = \frac{r_a + r_p}{2} \]

where \( r_a \) is the apoapsis distance and \( r_p \) is the periapsis distance. When transitioning from a circular orbit to an elliptical orbit, the periapsis distance becomes equal to the radius of the circular orbit. This relationship is crucial for determining the semi-major axis accurately.

For example, if a spacecraft is initially in a circular orbit with a radius \( r \) and transitions to an elliptical orbit with a known apoapsis distance, the periapsis can be calculated, allowing for the determination of the semi-major axis. The work done can then be computed using the derived equations.

When transitioning from an elliptical orbit back to a circular orbit, the process is reversed. If the new circular orbit is larger, the apoapsis distance becomes the radius of the new circular orbit, and if it is smaller, the periapsis distance will become the new radius. In both cases, adjustments in velocity (delta-v) are necessary to achieve the desired orbit.

In summary, the transition between circular and elliptical orbits relies on understanding energy conservation, the significance of the semi-major axis, and the relationships between periapsis and apoapsis distances. This knowledge is essential for calculating the work required for orbital maneuvers.

A comet is in a highly elliptical orbit around the Sun with a period of 2400 years. If this comet has a mass of approximately $5\times 10^{14}$ kg, what is the total energy of its orbit?

• A. $1.24\times 10^{21}$ J
• B. $-1.24\times 10^{21}$ J
• C. $2.48\times 10^{21}$ J
• D. $-1.24\times 10^{20}$ J

Answer: (B)

Changing from a Elliptical to Circular Orbit

In this scenario, we analyze the energy required for a spacecraft to transition from an elliptical orbit to a circular orbit. The spacecraft has a mass of 2,000 kilograms, and we need to determine the energy needed to change its orbit, specifically focusing on the perigee and apogee of the elliptical orbit.

To visualize the problem, we can represent the Earth at the center, with the elliptical orbit defined by its perigee (the closest point to Earth) and apogee (the farthest point). The key point is that the apogee of the elliptical orbit corresponds to the radius of the final circular orbit. This transition involves using the principle of energy conservation, which states that the total mechanical energy (kinetic plus potential) remains constant unless work is done on the system.

The energy conservation equation can be expressed as:

ΔE = E_final - E_initial

Where:

• ΔE is the work done to change the orbit.
• E_final is the energy of the circular orbit.
• E_initial is the energy of the elliptical orbit.

The energy for an elliptical orbit can be calculated using the formula:

E_{elliptical} = -\frac{GMm}{2a}

And for a circular orbit, it is given by:

E_{circular} = -\frac{GMm}{2r}

Here, G is the gravitational constant, M is the mass of the Earth, m is the mass of the spacecraft, a is the semi-major axis of the elliptical orbit, and r is the radius of the circular orbit.

To find the work done, we rearrange the energy conservation equation:

W = E_{circular} - E_{elliptical} = -\frac{GMm}{2r} + \frac{GMm}{2a} = \frac{GMm}{2} \left( \frac{1}{a} - \frac{1}{r} \right)

Next, we need to calculate the semi-major axis (a) of the elliptical orbit using the formula:

a = \frac{r_a + r_p}{2}

Where \( r_a \) is the apogee and \( r_p \) is the perigee. Given the values of \( r_a = 8.87 \times 10^6 \) m and \( r_p = 6.87 \times 10^6 \) m, we find:

a = \frac{8.87 \times 10^6 + 6.87 \times 10^6}{2} = 7.87 \times 10^6 \text{ m}

Substituting the values into the work equation, we can calculate the work done, which results in:

W = 5.7 \times 10^9 \text{ J}

In the second part of the problem, we need to determine the change in velocity (Δv) required to achieve this energy change. The relationship between work and change in kinetic energy is given by:

W = \Delta E = \frac{1}{2} m (\Delta v)^2

Rearranging this gives us:

\Delta v = \sqrt{\frac{2W}{m}}

Substituting the work done and the mass of the spacecraft:

\Delta v = \sqrt{\frac{2 \times 5.7 \times 10^9}{2000}} \approx 2410 \text{ m/s}

This indicates that the spacecraft must increase its velocity by approximately 2410 meters per second to successfully circularize its orbit. The positive work done aligns with the expectation that energy is required to move from a smaller orbit to a larger one.

Here’s what students ask on this topic:

What is the difference between kinetic and potential energy in elliptical orbits compared to circular orbits?

In elliptical orbits, unlike circular orbits, kinetic and potential energies are not constant. As the distance from the focus changes, kinetic energy is maximum at the closest approach (perigee) and minimum at the farthest point (apogee). Conversely, potential energy is minimum at perigee and maximum at apogee. This variation occurs because the distance from the central body changes, affecting the gravitational potential energy. Despite these changes, the total energy of the system remains conserved. In circular orbits, both kinetic and potential energies remain constant because the distance from the central body does not change. Understanding these differences is crucial for analyzing the dynamics of celestial bodies in elliptical orbits.

How can you compare velocities at different points in an elliptical orbit?

To compare velocities at different points in an elliptical orbit, you can use the conservation of angular momentum. This principle leads to the equation $V_1{R}_1=V_2{R}_2$, where $V_1$ and $R_1$ are the velocity and distance at one point, and $V_2$ and $R_2$ are the velocity and distance at another point. This equation allows you to solve for an unknown variable if three of the four are known. It is derived from the fact that angular momentum, given by $\mathrm{mvr}$, is conserved in an orbit. This method is particularly useful in problems involving celestial mechanics and satellite dynamics.

What is the role of the semi-major axis in energy calculations for elliptical orbits?

The semi-major axis, denoted as $a$, plays a crucial role in energy calculations for elliptical orbits. It replaces the radius $R$ used in circular orbit energy equations. The total energy of an elliptical orbit is given by the equation $-\frac{GMm}{2a}$, where $G$ is the gravitational constant, $M$ is the mass of the central body, and $m$ is the mass of the orbiting body. The semi-major axis is the average of the apogee and perigee distances and is essential for determining the energy required for orbit transitions. It is distinct from the apogee and is a fundamental parameter in Kepler's laws of planetary motion.

How is work calculated when transitioning from a circular to an elliptical orbit?

When transitioning from a circular to an elliptical orbit, work is calculated using energy conservation principles. The initial energy of the circular orbit is given by $-\frac{GMm}{2r}$, where $r$ is the radius of the circular orbit. The final energy of the elliptical orbit is $-\frac{GMm}{2a}$, where $a$ is the semi-major axis. The work done is the difference between the final and initial energies, expressed as $\frac{GMm}{2}(\frac{1}{r}-\frac{1}{a})$. This calculation helps determine the energy required to change the orbit, which is equivalent to the work done on the system.

What happens to the radius of a circular orbit when transitioning to an elliptical orbit?

When transitioning from a circular to an elliptical orbit, the radius of the original circular orbit can become either the perigee or apogee of the new elliptical orbit, depending on the size change. If the new elliptical orbit is larger, the original radius becomes the perigee, the closest point to the central body. Conversely, if the new orbit is smaller, the original radius becomes the apogee, the farthest point. This transition involves work and energy changes, calculated using energy conservation principles. Understanding this concept is essential for analyzing orbit transitions in celestial mechanics and satellite dynamics, as it affects the energy required for such changes.