explain what the centripetal force is equal to the gravitational force in orbits

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Welcome to our exploration of orbital motion. When we look at satellites orbiting Earth, we see objects that are actually in continuous free fall. However, instead of falling straight down, they maintain a circular path around our planet. This happens because they need a constant inward force, called centripetal force, to keep them moving in a circle rather than flying off in a straight line. Now let's understand centripetal force mathematically. For any object moving in a circular path, we start with the relationship between velocity and period: v equals 2 pi r over T. From this, we can derive the centripetal acceleration as v squared over r, which always points toward the center of the circle. Using Newton's second law, the centripetal force equals mass times acceleration, giving us F equals m v squared over r. This force can also be expressed as m omega squared r, or m times 4 pi squared r over T squared. Notice that centripetal force always points toward the center, just like the acceleration. The gravitational force that keeps objects in orbit follows Newton's law of universal gravitation. The force equals G times the product of the two masses, divided by the square of the distance between their centers. Here, G is the gravitational constant, M is the mass of the central body like Earth, m is the mass of the orbiting object, and r is the distance between their centers. This gravitational force has several key properties: it decreases with the square of distance, it always acts along the line connecting the two masses, and it represents a mutual attraction between both objects according to Newton's third law. Now we reach the crucial principle of orbital mechanics: the force balance in orbits. For a stable circular orbit, the centripetal force required to keep the satellite moving in a circle must exactly equal the gravitational force pulling it toward Earth. This gives us the equation: m v squared over r equals G M m over r squared. When we solve this equation for orbital velocity, we cancel the mass m from both sides, multiply both sides by r, and take the square root to get v equals the square root of G M over r. The key insight here is that gravitational force IS the centripetal force - these are not two separate forces acting on the satellite, but rather gravitational attraction provides exactly the inward force needed for circular motion. Using the force balance equation, we can derive key relationships for orbital motion. The orbital velocity equals the square root of G M over r, while the orbital period follows Kepler's third law: T equals 2 pi times the square root of r cubed over G M. These equations reveal important insights: higher orbits have slower velocities but longer periods, and the square of the period is proportional to the cube of the orbital radius. Watch how as we increase the orbital radius, the satellite moves more slowly but takes much longer to complete one orbit.