# Homework Help: Can someone guide me?

1. May 20, 2010

### kevin heck

1. The problem statement, all variables and given/known data

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
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

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?
2. Relevant equations

3. The attempt at a solution

2. May 20, 2010

### 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 )

3. May 20, 2010

### 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.

4. May 20, 2010

### Gokul43201

Staff Emeritus
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.

Last edited: May 20, 2010
5. May 20, 2010

### kevin heck

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

6. May 20, 2010

### kevin heck

or a= 4(pi)squared r/ Tsquared

7. May 20, 2010

### kevin heck

is that the step in the right direction?

8. May 20, 2010

### 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.

9. May 20, 2010

### kevin heck

i am doing my physics 20 by corrospondance right now.

10. May 20, 2010

### Gokul43201

Staff Emeritus
Correct so far - haven't checked the numbers.

11. May 20, 2010

### Gokul43201

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

12. May 20, 2010

### kevin heck

thank you very much :)

13. May 20, 2010

### 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?

14. May 20, 2010

### 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?

15. May 20, 2010

### kevin heck

hello?

16. May 20, 2010

### Gokul43201

Staff Emeritus
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.

17. May 20, 2010

### kevin heck

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

18. May 20, 2010

### Gokul43201

Staff Emeritus
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.

19. May 20, 2010

### 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.

20. May 20, 2010

### Gokul43201

Staff Emeritus
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.