# Centripetal Acceleration help

• Coldchillin
In summary, a circular Earth satellite with a radius of 597 km and a period of 96.42 min has a speed of 648.39 m/s and a centripetal acceleration of 0.704 m/s^2. The difference between speed and magnitude is unclear. Further assistance is needed.

#### Coldchillin

"An Earth satellite moves in a circular orbit 597 km above Earth's surface with a period of 96.42 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?"

Ok, so first I converted everything to standard units. I made a circle diagram with radius 597,000m and it's period is 5785.2 seconds. So to find the acceleration I had to find the velocity, which I had as v=(2*(Pi)*(597,000))/5785.2s and got v=648.39 m/s. Then to find the acceleration I used a=(648.39)^2/597,000 and got a=.704 m/s^2, which apparently is wrong. Can someone help me out? Also, what is the difference between the speed and magnitude? Thank you!

Since the satellite is orbiting above the Earth the radius of its orbit must include the radius of the earth!

Hello, thank you for reaching out for help with your calculation. It seems like you have the right approach, but there may be a small error in your calculations. First, let's address the difference between speed and magnitude. Speed is the rate at which an object is moving, while magnitude is the size or amount of a quantity. In this case, the speed refers to the velocity of the satellite, while the magnitude of the centripetal acceleration refers to the size of the acceleration vector.

Now, let's take a look at your calculations. The formula you used to find the velocity (v=(2*(Pi)*(597,000))/5785.2s) is correct, but your unit conversion may have been off. The period should be converted to seconds, so it should be 96.42 min * 60 s/min = 5785.2 s. This may have caused a slight discrepancy in your final answer.

For the acceleration, you are correct in using the formula a=(v^2)/r, but you should use the velocity you calculated earlier (648.39 m/s) instead of squaring it again. This will give you a centripetal acceleration of 0.11 m/s^2, which is a more reasonable answer for the magnitude of the acceleration.

I hope this helps clarify your calculations. Remember to always double check your unit conversions and use the correct formulas for the given situation. Good luck with your studies!

## What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It always acts towards the center of the circle and is responsible for continuously changing the direction of the object's velocity.

## How is centripetal acceleration calculated?

The formula for centripetal acceleration is a = v²/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path. This means that the greater the velocity or the smaller the radius, the greater the centripetal acceleration.

## What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is the actual acceleration experienced by an object moving in a circular path, while centrifugal force is the apparent outward force experienced by the object as a result of its inertia. Centrifugal force is not a real force, but rather a result of the object's tendency to continue moving in a straight line.

## What are some real-world examples of centripetal acceleration?

Some examples of centripetal acceleration in everyday life include the rotation of a Ferris wheel, the movement of a car around a curve, and the orbit of planets around the sun. Any object that moves in a circular path experiences centripetal acceleration.

## How is centripetal acceleration related to centripetal force?

According to Newton's second law, F = ma, the centripetal force is equal to the mass of the object multiplied by its centripetal acceleration. In other words, centripetal force is the force required to keep an object moving in a circular path and is directly proportional to the object's mass and the square of its velocity.