Circular Motion and Centripetal Acceleration

In summary, the conversation is about solving a physics problem involving calculating the magnitudes of two accelerations of an object resting on the equator of the Earth due to its rotational and orbital motions. The steps taken by the person asking for help are incorrect, but they eventually figure out the correct formulas and values to use. The conversation also discusses converting units and finding the speed and distance of the Earth's orbital motion around the Sun.
  • #1
aidos
4
0
I’m having trouble with ch3 #73 from Tipler/Mosca Physics for scientists and engineers 5th edition. I’ve tried a few things and I keep getting stuck. Am I taking the right steps?

An object resting on the equator has an acceleration toward the center of the Earth due to the earth’s rotational motion about its axis, and an acceleration toward the sun due to the earth’s orbital motion. Calculate the magnitudes of both of these accelerations and express them a fraction of the magnitude of free-fall acceleration g. Use values from the physical-data table in the textbook.

So we have centripetal acceleration = v^2/r (Ac = v^2/r)
g = 9.81 m/s^2
earth’s radius @ equator = 6370 km


Ac = g^2/6370 = (9.81 m/s^2)^2/6370 km = .015 m^2/Kms^4? (what units/reductions do I need to make? Is this correct?

Period formula
v = 2PIr/T
I have 2 PI and r, but what do I use for T?

Thanks for your help!
 
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  • #2
No, you're not taking the right steps. You need to understand what the factors are in your formula for centripetal acceleration. v is the speed about the axis of rotation and r is the distance from the axis.

The speed is total distance traveled divided by time. Standing at the equator of the rotating Earth you travel a distance equal to the circumference of the planet in a day.
 
  • #3
Ok, so V = (2PI(6370))/86400(secs in a year) = .46 m/s

Ac = (.46)^2/6370 = 3.32 x 10 to the -5 m/s^2? The back of the book has 3.44 x 10 to the -3g.

Help..
 
  • #4
Should you not convert km into m ? And you are asked to find out what fraction of the value of g, this number is.
 
  • #5
The Earth's radius is given in kilometers, not meters. Convert.
 
  • #6
OOOPS, my bad…Thanks so much!

6370 km = 6370000 m
so V = (2PI(6370000m))/86400s = 463.24 m/s

(463.24)^2/6370000 = .0337

.0337/9.81 = .003434 which I can label as 3.43 x (10 to the -3)g which is REAL close to the 3.44 x (10 to the -3)g in the back of the text.

What formula do I use to get the acceleration toward the sun due to the earth’s orbital motion?

I’m assuming it’s At = dv/dt but what is my change in velocity and/or time?

Thanks for your time!
 
  • #7
aidos said:
What formula do I use to get the acceleration toward the sun due to the earth’s orbital motion?

I’m assuming it’s At = dv/dt but what is my change in velocity and/or time?

Thanks for your time!

You should have learned something from doing the first part of the problem. It's exactly the same problem with different input. You know how far the Earth travels around the Sun in a year so you can calculate the speed and use the very same centripetal acceleration formula.
 
  • #8
I didn't understand how to figure out the orbital speed & distance, but I found them in the back of the book so that made it easy to plug it into the Ac formula. Thanks for your time!
 

What is circular motion?

Circular motion is the movement of an object along a circular path at a constant speed. The object's direction is constantly changing, but its speed remains the same.

What is centripetal acceleration?

Centripetal acceleration is the acceleration that keeps an object moving in a circular path. It is always directed towards the center of the circle and is perpendicular to the object's velocity.

How is centripetal acceleration calculated?

Centripetal acceleration can be calculated using the formula a = v^2/r, where v is the velocity of the object and r is the radius of the circular path.

What is the difference between tangential velocity and centripetal acceleration?

Tangential velocity is the speed at which an object is moving tangentially to a circular path. Centripetal acceleration, on the other hand, is the acceleration that keeps the object moving in a circular path.

What is the relationship between centripetal force and centripetal acceleration?

Centripetal force is the force that is required to keep an object moving in a circular path. It is directly proportional to the centripetal acceleration of the object. As the acceleration increases, the force required to maintain the circular motion also increases.

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