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When we talk about gravitational potential energy in the context of launching satellites, we're looking at how much energy is needed to overcome Earth's gravity. This energy is dependent on both the mass of the object and its distance from Earth's center.
The formula for gravitational potential energy is \[-\frac{G M m}{r}\] where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the Earth.
• \(m\) is the mass of the satellite.
• \(r\) is the distance from the center of the Earth to the satellite.

At Earth's surface, \(r = R\) (Earth's radius), and the potential energy is \(-\frac{G M m}{R}\). When the satellite is at an altitude of \(2R\), the distance from the Earth's center is \(3R\), so its potential energy becomes \(-\frac{G M m}{3R}\).
This change demonstrates that as the satellite moves further from Earth, the gravitational potential energy becomes less negative, reflecting the energy needed to move the object from one point to another against the gravitational pull.

Kinetic energy is the energy an object has due to its motion. For a satellite in a stable circular orbit, this energy must be precisely balanced to keep the satellite moving smoothly.
In orbit, kinetic energy differs based on altitude. The formula used to express kinetic energy for a satellite is:\[\frac{1}{2} mv^2 \]For satellites, we often replace the velocity term with an expression based on gravitational elements. Thus, it becomes\[\frac{G M m}{2r}\] where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the Earth.
• \(m\) is the satellite's mass.
• \(r\) is the distance from the Earth's center.

When the satellite is at altitude \(2R\) (\(r = 3R\)), the kinetic energy becomes \(\frac{G M m}{6R}\).
This shows how the kinetic energy adapts to the orbital height and corresponds to the necessary velocity for maintaining that orbit. Without this proper kinetic energy, the satellite wouldn't be able to stay in its designated path.

Mechanical energy in this context is the sum of kinetic and potential energies of the satellite. To launch a satellite from Earth's surface into orbit, the change in mechanical energy must be carefully calculated.
The difference in mechanical energy gives us the minimum energy required to get our satellite into orbit.
On the Earth's surface, the energy is purely potential, calculated by:\[-\frac{G M m}{R}\]In orbit at altitude \(2R\), mechanical energy (both kinetic and potential) becomes:\[-\frac{G M m}{6R}\]Calculating the change gives us:\[\Delta E = \left(-\frac{G M m}{6R}\right) - \left(-\frac{G M m}{R}\right)\]which simplifies to:\[\frac{G M m}{2R}\]To relate this to familiar terms, we use gravitational acceleration \(g\), where \(g = \frac{G M}{R^2}\).
Thus, the change in mechanical energy becomes:\[\Delta E = \frac{1}{3} m g R\]Adding the energy needed to reach \(2R\) provides the total required energy:\[\Delta E_{total} = \frac{4}{3} m g R\]This calculation specifies the complete energy landscape necessary for successful satellite launch. It's crucial for matching the gravitational retentive forces and achieving a sustainable orbit.