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The gravitational force is fundamental in understanding how objects, like satellites, remain in orbit around a planet. This force results from the attraction between two masses: the satellite and the planet. Newton's law of universal gravitation defines it with the formula:

• \[ F = \frac{G \cdot M \cdot m}{(R + h)^2} \]

where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the satellite, and \( R + h \) is the distance from the planet's center to the satellite.
The gravitational force provides the necessary centripetal force for maintaining a circular orbit. However, in problems like these, instead of using the constant \( G \) and the mass \( M \), we simplify calculations using the planet's surface gravity \( g \), which represents the acceleration due to gravity at the planet's surface. Thus, within a planetary system, the force felt by the satellite can be expressed simply as \( F = m \cdot g \).
This means that the satellite's experience of gravity depends on the planet's gravitational pull, which decreases with distance as the satellite moves further from the surface. This relationship is key in figuring out how fast a satellite must move to maintain its orbit.

In a circular orbit, a satellite moves around a planet at a constant speed along a circular path. The path's radius is determined by the planet's radius plus the satellite's altitude above the planet's surface. The formula \( r = R + h \) helps us calculate this value:

• \( R \) is the radius of the planet.
• \( h \) is the height above the surface where the satellite orbits.

For instance, for Venus, you would add its radius of 3761 miles to the satellite's altitude of 100 miles, giving a total orbit radius \( r \) of 3861 miles.
This distance often needs to be converted into a single unit for calculations, like feet, to match gravitational force calculations that are given in \( \text{ft/s}^2 \). In this exercise, multiplying by 5280 converts miles to feet, resulting in an orbit radius of 20308080 feet. Understanding the concept of orbit radius is pivotal because it directly influences how strong the gravitational force is at various distances from the planet.

Orbital velocity is the speed required for a satellite to orbit a planet in a stable circular path. This velocity ensures that the gravitational pull of the planet provides just enough centripetal force to keep the satellite from crashing back into the planet or flying off into space. The formula used to determine orbital velocity is:

• \( v = \sqrt{g \times r} \)

Where \( g \) is the gravitational acceleration at the planet's surface, and \( r \) is the radius of the orbit (sum of planetary radius and altitude of orbit). In this step-by-step solution, substituting values for Venus provides its orbital velocity as approximately 24594 feet per second.
It's important to note that this velocity is specific to the altitude of 100 miles and the gravitational characteristics of Venus. If the altitude or the planet changes, the values involved, including \( g \) and \( r \), must be adjusted accordingly to find the new correct value for orbital velocity.
For any satellite to maintain this perfect balance, understanding and calculating these velocities is crucial. It ensures continued orbit without energy loss or deviation from the path, embodying the harmony needed for any successful satellite mission.