What is the Centripetal Acceleration at Earth's Equator?

[kevin heck]

Homework Statement

Earth is commonly thought of as a sphere, but this is not true. Because of Earth's spin, it closely resembles an oblate spheroid, which is just a fancy name for "shaped like a squashed orange". The effect of this spin is a tidal bulge that forms at the equator so that the equatorial radius of Earth is about 21km greater than the polar radius of 6370km. The mass of Earth is 5.979*10^24kg.

a) calculate the accel. due to gravity at the North Pole using the definition of gravitational field.
mass Earth=5.979*10^24kg
radius Earth=6.370*10^6m
G=6.67*10^-11Nm2/kg2

accel. due to gravity=G(mass Earth)/radius Earth2
=9.83N/kg


b) same but for values at the equator. Accel. due to gravity=9.76N/kg.


c) Calculate the centripetal acceleration at the equator?
So i know my mass of Earth and radius of Earth. Also have the acceleration due to gravity for the equator. They formula I went with was

Fc=(mass Earth)(4)(pie)squared(radius Earth)/(T)squared
after substituting in all my numbers i got a solution that came out as N. Can you help steer me straight? Where did I go wrong?

Homework Equations

The Attempt at a Solution

 

[mgb_phys]

The centrepedal acceleration doesn't depend on the mass of the earth,
It's simply F = m r omega^2, or F = m v^2/r

The overall force at the equator is the gravity force you just found minus this.

Where omega is the angular velocity in rad/sec = 2pi / (24*60*60)
(or more precisely 23 hours, 56 minutes, 4sec )

 

[kevin heck]

I'm not really clear on what the omega is all about. I haven't come across this term yet. what i understand is i should use the equation F = m v^2/r, but i thought the mass had nothing to do with the centripetal acceleration. I'm confused.

 

[Gokul43201]

The centrepedal acceleration doesn't depend on the mass of the earth,
It's simply F = m r omega^2, or F = m v^2/r

mgb's had a long day () - that's the formula for the centripetal force.

Kevin, what's the general formula for centripetal acceleration? You are right that the mass doesn't figure in it.

 

[kevin heck]

I believe it is a= v^2/r.

 

[kevin heck]

or a= 4(pi)squared r/ Tsquared

 

[kevin heck]

is that the step in the right direction?

 

[kevin heck]

what i did was substitute in my values for the second equation previous. so i got
a= 4 (pi^2) (6.391*10^6m)/ (3.1536*10^7m/s^2)^2 and came up with an answer of
2.537*10^-7 m/s^2. i was wondering if i am going in the right direction? i am a 30 year old man who is going back to school for my power engineering and really need help. due to where i live access to the college i got my books from is not easy and quite expensive for continuous travel. i appreciate any help along the way. i am willing to do the work myself but sometimes i just need a little push in the right direction.

 

[kevin heck]

i am doing my physics 20 by corrospondance right now.

 

[Gokul43201]

or a= 4(pi)squared r/ Tsquared

Correct so far - haven't checked the numbers.

 

[Gokul43201]

Your value for T^2 seems to be off (and its units too).

 

[kevin heck]

thank you very much :)

 

[kevin heck]

then it goes on to tell me to calculate the net acceleration at the equator. i believe i would do this by using formula

net acceleration= mass* acceleration?

 

[kevin heck]

my units for T^2 are off. is this not the period of Earth? so i took 365 days and broke it down into seconds and came up with 3.5136*10^7 seconds. then i squared that number. where did i go wrong?

 

[kevin heck]

my units for T^2 are off. is this not the period of Earth? so i took 365 days and broke it down into seconds and came up with 3.5136*10^7 seconds. then i squared that number. where did i go wrong?

hello?

 

[Gokul43201]

Patience! We're just volunteers who pass by and help folks out when we find some time.

You want the period for rotation of the Earth about its own axis, not the period for revolution around the sun. And by "units", I meant that the unit for time is seconds, and not m/s^2.

 

[kevin heck]

so to get the rotation of Earth i would take its radius and divide by 9.80 m/s^2?

 

[Gokul43201]

then it goes on to tell me to calculate the net acceleration at the equator. i believe i would do this by using formula

net acceleration= mass* acceleration?

That's the equation for "net force", not "net acceleration". In fact, there really isn't such a thing as "net acceleration" - that's a somewhat ambiguous terminology.

Is this part d of the question? Please write it down EXACTLY as it appears in the book/notes you are using.

 

[kevin heck]

Sorry for the impatience. I appreciate the help you have given me especially since you are just volunteering. Thank you. :)

d) Calculate the net acceleration at the equator.

That is exactly how it appears.

 

[Gokul43201]

Sorry for the impatience. I appreciate the help you have given me especially since you are just volunteering. Thank you. :)

d) Calculate the net acceleration at the equator.

I do not feel confident in divining what is actually implied by that question. I'll reserve my best guess for the moment and instead ask you what you think it should be.

 

[Gokul43201]

so to get the rotation of Earth i would take its radius and divide by 9.80 m/s^2?

Not sure what you are referring to here, or where you get that equation from. Are you asking about what value to use for T?

How long does it take for the Earth to spin once about it's axis? Hint: it's not a year.