Energy of Circular Orbits: Videos & Practice Problems

Understanding the energy of orbits involves the conservation of energy when a satellite changes its orbital distance or velocity. The kinetic energy (KE) is given by $\frac{1}{2}m{v}^2$ , while gravitational potential energy (PE) is $-\frac{GmM}{r}$ . To change orbits, positive work is required to increase altitude, resulting in decreased velocity, while negative work is needed to decrease altitude, increasing velocity. The relationship between orbital radius and velocity is crucial for solving these problems.

Energy of Circular Orbits

Understanding the energy of orbits is crucial when analyzing the motion of satellites. When a satellite transitions from one orbital distance (denoted as \( r_1 \)) to another (denoted as \( r_2 \)), it experiences changes in both kinetic energy and gravitational potential energy. The relationship between the orbital radius and the satellite's velocity is governed by the equation for orbital speed, which can be expressed as:

\[ v_{\text{sat}} = \sqrt{\frac{GM}{r}} \]

where \( G \) is the gravitational constant and \( M \) is the mass of the celestial body being orbited. The kinetic energy (\( KE \)) of the satellite is given by:

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

and the gravitational potential energy (\( PE \)) is described by:

\[ PE = -\frac{GMm}{r} \]

To analyze the work done when changing orbits, we apply the principle of energy conservation. The total mechanical energy in orbit is the sum of kinetic and potential energy. When moving from an initial orbit to a final orbit, the work done (\( W \)) can be expressed as:

\[ W = \left( \frac{1}{2} mv_f^2 - \frac{GMm}{r_f} \right) - \left( \frac{1}{2} mv_i^2 - \frac{GMm}{r_i} \right) \]

In this equation, \( v_i \) and \( v_f \) are the initial and final velocities, while \( r_i \) and \( r_f \) are the initial and final distances from the center of the mass being orbited. If the velocities are unknown, we can substitute them using the orbital speed equation, allowing us to express everything in terms of the known distances.

For example, if a spacecraft of mass \( m \) is moving from a lower orbit to a higher orbit, the work done is positive, indicating that energy is added to the system. Conversely, if the spacecraft is moving to a lower orbit, the work done is negative, as energy is removed from the system. This results in an increase in velocity as the radius decreases.

In summary, the energy of orbits can be analyzed through the conservation of energy, where the total energy is a combination of kinetic and potential energy. The equations governing these energies allow for the calculation of work done during orbital transitions, highlighting the relationship between orbital radius and velocity changes.

The 12,000-kg Lunar Command Module is in a circular orbit above the Moon's surface. If it spends ¼ of its fuel energy ($-1.74\times 10^9$ J) bringing it to a circular orbit just above the surface, how high was its original orbit?

• A. $1.9\times 10^5$ m
• B. $1.6\times 10^6$ m
• C. $1.8\times 10^6$ m
• D. $3.7\times 10^6$ m

Answer: (A)

a) How much work do you have to do on a 100-kg payload to move it from Earth's surface to a height of 1000 km?
b) How much additional work must you do to put this payload into orbit at this altitude?

• A. (a) $W_{AB}=3.37\times10^8$ J;(b) $W_{BC}=1.99\times 10^9$ J
• B. (a) $W_{AB}=8.48\times10^8$ J;(b) $W_{BC}=2.7\times 10^9$ J
• C. (a) $W_{AB}=3.37\times10^8$ J;(b) $W_{BC}=5.97\times 10^9$ J
• D. (a) $W_{AB}=8.48\times10^8$ J;(b) $W_{BC}=8.1\times10^9$ J

Answer: (B)

Here’s what students ask on this topic:

What is the relationship between orbital radius and velocity in circular orbits?

The relationship between orbital radius (r) and velocity (v) in circular orbits is given by the equation:

$\sqrt{\frac{GM}{r}}$

where G is the gravitational constant and M is the mass of the central object (e.g., Earth). This equation shows that the orbital velocity is inversely proportional to the square root of the orbital radius. As the orbital radius increases, the velocity decreases, and vice versa.

How do you calculate the kinetic energy of a satellite in a circular orbit?

The kinetic energy (KE) of a satellite in a circular orbit is given by the equation:

$\frac{1}{2}m{v}^2$

where m is the mass of the satellite and v is its orbital velocity. The orbital velocity can be found using the relationship:

$\sqrt{\frac{GM}{r}}$

Substituting this into the kinetic energy equation, we get:

$\frac{1}{2}m\frac{GM}{r}$

What is the total energy of a satellite in a circular orbit?

The total energy (E) of a satellite in a circular orbit is the sum of its kinetic energy (KE) and gravitational potential energy (PE). The kinetic energy is given by:

$\frac{1}{2}m{v}^2$

and the gravitational potential energy is:

$\frac{-GmM}{r}$

The total energy is:

$\frac{-GmM}{2r}$

This shows that the total energy of a satellite in a circular orbit is negative, indicating a bound system.

How does work affect the change in orbital distance of a satellite?

To change the orbital distance of a satellite, work must be done on the system. Positive work is required to increase the orbital distance, which results in a decrease in orbital velocity. Conversely, negative work is needed to decrease the orbital distance, leading to an increase in orbital velocity. The work done changes the total energy of the satellite, which is the sum of its kinetic and potential energies. The relationship between work (W), initial and final kinetic energies (KE_i and KE_f), and initial and final potential energies (PE_i and PE_f) is given by:

$(\frac{1}{2}m{v}^2-\frac{G}{m}$

What happens to the velocity of a satellite when its orbital radius increases?

When the orbital radius of a satellite increases, its orbital velocity decreases. This is due to the inverse relationship between orbital radius (r) and velocity (v) given by the equation:

$\sqrt{\frac{GM}{r}}$

As the radius increases, the gravitational force acting on the satellite decreases, requiring a lower velocity to maintain a stable orbit. This decrease in velocity is a result of the conservation of energy, where the increase in gravitational potential energy is balanced by a decrease in kinetic energy.