What is the kinetic energy of a satellite and Why is satellite energy negative?

This lesson answers all questions around motion of satellites in orbits. Learn how energy conservation governs satellite motion in orbit, from circular to elliptical paths. This lesson covers insights into velocity, energy, and orbital mechanics.

What You’ll Learn

1. Calculate kinetic, potential, and total mechanical energy of a satellite in orbit
2. Derive the orbital velocity and period of a satellite
3. Understand why satellite energy is negative and mass-independent
4. Analyze how orbital radius affects speed and energy
5. Apply energy principles to explain weightlessness in space
6. Extend concepts from circular to elliptical satellite orbits

Key Concepts Covered

• Satellite orbits
• Gravitational potential energy
• Kinetic energy of a satellite
• Mechanical energy in orbits
• Orbital velocity and period
• Circular and elliptical orbits
• Kepler’s Third Law
• Apparent weightlessness

Understanding satellite energy dynamics is very relevant in real-world applications like launching spacecraft, predicting satellite motion, and explaining phenomena such as weightlessness. These concepts are essential in AP Physics 1 & C, IB Physics, and competitive exams like JEE and NEET.

Prerequisite or Follow-Up Lessons

• Newton’s Law of Gravitation
• Centripetal Force and Acceleration

Full Lesson Transcript: Satellite Energy and Orbits

Gravitational Potential Energy in Orbit

A satellite orbiting Earth experiences gravitational potential energy (U), given by:

U = –GMm / r

Here, G is the gravitational constant, M is Earth’s mass, m is the satellite’s mass, and r is the distance from Earth’s center. The negative sign reflects the convention that U = 0 at infinite separation.

Kinetic Energy in a Circular Orbit

To derive kinetic energy, apply Newton’s second law using gravitational force as the centripetal force:

F = ma = mv² / r = GMm / r²

Solving for v and substituting into K = ½ mv², we find:

K = GMm / 2r

And since U = –GMm / r, it follows that:

K = –U / 2

So if U = –1000 J, then K = 500 J.

Orbital Velocity

Using the above:

v = √(GM / r)

This reveals two important insights:

• Orbital speed is fixed once radius r is chosen.
• Satellite mass does not affect orbital velocity.

Thus, a 100 kg or 500 kg satellite at the same r move at identical speeds.

Apparent Weightlessness

Astronauts inside a space shuttle experience weightlessness because both shuttle and astronaut are in free fall, orbiting Earth at the same speed. This eliminates contact forces and creates the sensation of floating.

Orbital Period (T)

Equating velocity with circumference over time:

v = 2πr / T → √(GM / r) = 2πr / T

Solving for T:

T = 2π √(r³ / GM)

This is a specific form of Kepler's Third Law, stating that T² ∝ r³.

Total Mechanical Energy

Total energy (E) is:

E = K + U = GMm / 2r – GMm / r = –GMm / 2r

Which also means:

E = –K

So for U = –1000 J, K = 500 J, and E = –500 J.

Elliptical Orbits

This total energy formula also holds for elliptical orbits by replacing r with the semi-major axis a:

E = –GMm / 2a

Notably, E is independent of eccentricity (e); only a matters. So orbits with equal a have identical total energy.

Energy Variation with Orbital Radius

As r increases:

• Kinetic energy (K) decreases
• Orbital velocity decreases
• Potential energy (U) becomes less negative
• Total energy (E) approaches zero

At geosynchronous orbit (r ≈ 42,000 km), all energies are smaller in magnitude than in low Earth orbit (r ≈ 7,000 km).

In summary, a satellite’s motion, energy, and period are fully determined by its distance from Earth. Once r or a is fixed, all other quantities are set—showing the elegant predictability of orbital mechanics.

Frequently Asked Questions (FAQ) on Satellites, Orbits and Energy

Q1. Why does a satellite's speed change in an elliptical orbit, but its total mechanical energy remains constant?

A satellite's speed varies in an elliptical orbit because its distance from Earth changes, affecting its kinetic energy (KE) and gravitational potential energy (U). However, these changes are inversely related: as KE increases, U decreases, and vice-versa. The total mechanical energy (E), which is KE + U, remains constant due to the conservation of energy in a conservative gravitational field.

Q2. How are kinetic energy and potential energy related for a satellite in a circular orbit?

For a satellite in a circular orbit, its kinetic energy (K) is exactly half the magnitude of its potential energy (U), but positive: K = -U/2. This specific relationship, derived from balancing gravitational force and centripetal force, is key to understanding stable circular orbits.

Q3. What is the total mechanical energy of a satellite, and why is it negative?

The total mechanical energy (E) of a satellite is the sum of its kinetic and potential energies, given by E = -GMm / 2r for a circular orbit (or 2a for an elliptical orbit, where a is the semi-major axis). It is always negative, indicating a bound system because the satellite does not have enough energy to escape Earth's gravitational pull.

Q4. Does a satellite's mass affect its orbital velocity or period?

No, a satellite's mass does not affect its orbital velocity. The orbital velocity formula, v = sqrt(GM/r), shows that velocity depends only on the gravitational constant (G), the central body's mass (M), and the orbital radius (r). This highlights the universality of orbital motion, meaning two satellites with different masses but the same orbital radius will move at the same speed.

Q5. Why do astronauts float in a space shuttle or the ISS?

Astronauts float because they, along with the space shuttle or International Space Station (ISS), are constantly in free fall around Earth. They are continuously falling towards Earth while simultaneously moving forward at a very high orbital speed, creating a sensation of apparent weightlessness or microgravity, not an absence of gravity.

Q6. How is Kepler's Third Law related to a satellite's orbital period?

Kepler's Third Law states that the square of a satellite's orbital period (T) is directly proportional to the cube of its orbital radius (r) for circular orbits. The derived equation for the time period is T = 2π * sqrt(r^3 / GM), showcasing the relationship between a satellite's distance and the time it takes to complete an orbit.

Q7. How do kinetic, potential, and total energy change as a satellite's orbital radius increases?

As a satellite's orbital radius (r) increases: its kinetic energy decreases (the satellite slows down), its gravitational potential energy becomes less negative (moves towards zero), and its total mechanical energy also increases (approaches zero). This reflects that less energy is required to maintain a larger, slower orbit.

Q8. Can the orbital speed of a satellite at a given altitude be anything we want?

No. At a fixed altitude, there is only one specific speed at which an object can stay in a stable circular orbit. If a satellite moves faster than this speed, it will rise into an elliptical orbit, increasing its total mechanical energy. Conversely, if its speed is reduced, it will spiral into a lower orbit.

Q9. Why do satellites not just fall to Earth?

Satellites don't fall to Earth because they are constantly falling toward the Earth due to gravity, but they also possess a high sideways (tangential) velocity. This combination means the surface of the Earth curves away beneath them at the same rate they fall, keeping them perpetually in free fall and thus staying in orbit.

Q10. Do satellites need fuel to stay in orbit once they are there?

No, satellites in a stable orbit do not need fuel to keep moving. Once in orbit, their motion is maintained by inertia and the balance with gravity. They only need fuel for adjustments (like changing orbit or attitude) or to counteract atmospheric drag if they are in a very low orbit.