Around 1500, Copernicus proposed that planets move in circular orbits around the sun. Copernicus used his heliocentric model to estimate the periods of planets and the radii of their orbits. We can determine the velocity and period of a planet in orbit by using Newton's inverse square laws. The gravitational force on a planet of mass Mp moving around the Sun of mass Ms in a circular orbit of radius r is equal to the gravitational constant, capital G, times the product of MS and MP, divided by r squared. The centripetal force that keeps the planet moving in a circular orbit is equal to: the mass of the planet M p times the square of the velocity of the planet, divided by the distance r between the planet and the Sun. You have already learned that centripetal force is the force needed to keep an object moving in a circle. In this case, the force is the force of gravity. If we assume a planet moves in a circular orbit around the Sun, we can set the gravitational force of the Sun equal to the centripetal force needed to keep it moving in a circle. We can simplify this equation by: canceling the mass of the planet and one r from both sides of the equation and then solving for v. In this way we find that the velocity of the planet is equal to the square root of the Gravitational Constant times the mass of the sun, divided by the radius of the orbit. Notice that the speed of the planet does not depend on the mass of the planet. Let's work through an example: Given the mass of the sun, the average earth-sun distance, and the gravitational constant, G, we will determine how fast the Earth moves around the sun? ... Similarly, we can find the centripetal acceleration of a planet in a circular orbit. Remember that a mass accelerates in response to unbalanced forces. Thus we set the mass of the planet multiplied by its centripetal acceleration equal to the gravitational force that causes the acceleration. The centripetal acceleration of a planet is independent of the mass of the planet, and it is equal to the product of the gravitational constant and the mass of the Sun, divided by the square of the average distance between the planet and the Sun. Let's compare the centripetal acceleration of the Earth around the Sun to that of Venus. The average Earth-Sun distance is: 1.496 times 10 to the 11th meters, and the average Venus-Sun distance is 1.08 times 10 to the 11th meters. Venus' centripetal acceleration around the Sun is? ... Because the centripetal acceleration of a planet is inversely proportional to the square of the distance between the Sun and the planet, the centripetal acceleration of Venus around the Sun is almost twice Earth's centripetal acceleration around the Sun! We can also find the orbital period of a planet-denoted by a capital T- by using the expression we derived for the square of the velocity. Recall that the orbital velocity of a planet is equal to 2 pi times the distance of the planet to the Sun, r divided by the orbital period T of the planet. Squaring both sides gives us an expression for v squared. Earlier we obtained another expression for v squared. By setting these expressions equal to each other, we reach an expression for the period, T, of the planet. Notice that the period of the planet is independent of the mass of the planet. Part of this expression can be expressed as a single constant K-sub-s. Thus the square of the period is equal to the constant K-sub-s times the cube of the average distance from the planet to the sun. What is the orbital period of planet X, whose orbit has a radius four times that of the Earth? ... The ratio of the orbital period squared to the orbit radius cubed is the same for both Earth and Planet X. Therefore the orbital period of planet X is eight times that of the Earth's orbital period.

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