# Video: Determining the Centripetal Acceleration of an Orbital Body

What is the magnitude of the acceleration of Venus toward the Sun, assuming a circular orbit with a radius of 1.082 ร 10ยนยน m, and an orbital period of 0.6152 years (Use a value of exactly 365 for the number of days in a year.)

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### Video Transcript

What is the magnitude of the acceleration of Venus toward the Sun, assuming a circular orbit with a radius of 1.082 times 10 to eleventh meters, and an orbital period of 0.6152 years. Use a value of exactly 365 for the number of days in a year.

We can call this acceleration magnitude of Venus toward the sun ๐. And if we make a sketch of the planet Venus orbiting circularly around the sun, we know that the distance between the center of the sun and the center of Venus is given as ๐ 1.082 times 10 to eleventh meters. And that the planet Venus makes it once all the way around this orbit in a period weโve called capital ๐ of 0.6152 years.

Since the planet Venus is moving in a circular orbit, that means it will accelerate centripetally towards the center of the circle. We recall that an objectโs centripetal acceleration is equal to its linear speed squared divided by the radius of the circle it moves in. Recalling further that an objectโs speed is equal to the distance it travels divided by the time it takes to travel that distance, we can say that the linear speed of Venus is equal to the distance it travels the circumference of the circle two times ๐ times its radius divided by the period ๐.

This means that the centripetal acceleration of Venus is equal to two ๐๐ over ๐ quantity squared all divided by the radius ๐. This simplifies to four ๐ squared ๐ over ๐ squared. Since weโre given both the radius ๐ and the period ๐, we have all the information we need. But before we plug in and solve for ๐, we like to convert the period ๐ from units of years to units of seconds.

To make that conversion, weโll take ๐, which is given in years, multiply it by the number of days in a year multiply that by the number of hours in a day and multiply that by the number of seconds in an hour. This then will give us a time value in units of seconds.

With the radial distance ๐ plugged into our expression, weโre already to calculate ๐. When we do, we find a value of 1.135 times 10 to the negative two meters per second squared. Thatโs the magnitude of the center-seeking acceleration of the planet Venus as it moves in its circular orbit.