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. 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. Simplifying, we obtain 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 is independent of the mass of the planet. Let's do an example. How fast is the Earth moving around the Sun? The mass of the Sun is 1.991 times 10 to the thirtieth kilograms, the average Earth-Sun distance is: 1.496 times 10 to the 11th meters, and the gravitational constant is equal to 6.67 times 10 to the negative 11th Newton meters squared per kilogram squared. ... In much the same way, we can find the centripetal acceleration for a planet in circular orbit. 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. Let's compare the centripetal acceleration of the Earth around the Sun to that of Venus. ... 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 find the orbital period T of a planet, 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. By squaring the velocity, and setting it equal to the value obtained previously, we can solve for the period T of the planet. Notice that the period of the planet is independent of the mass of the planet. Also notice that the square of the period is equal to a constant, Ks times the cube of the average distance of the planet to the Sun. What is Earth's orbital period? ...

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