How fast does the Earth need to move to provide centripital acceleration

[Makonia]

Tltr: The quesion d makes no sense to me and I don't know what my teacher mean.
1. Homework Statement
The Earth turns once around its axis in 24.0 hours. We will assume that it is perfectly spherical, with radius 6400 km. The mass of the Earth is taken to be 6.00 × 1024 kg. The gravitational acceleration on the surface is taken to be g = 9.80 m/s2 .

a) What is the centripetal acceleration of a person at the equator? How big a fraction of the gravitational acceleration does this correspond to? Remember to draw a sketch of the situation.

b) How fast would the Earth have to turn for the centripetal acceleration to be exactly equal to gravity? Give the answer in revolutions per day. Remember to draw a sketch of the situation.

c) Define θ to be the latitude, so that θ = 0 corresponds to the equator, and θ = 90◦ is the North Pole. What is the centripetal acceleration of a person standing on the surface at a given value of θ? Remember to draw a sketch of the situation.

d) Taking into account that the centripetal acceleration is towards the rotation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)? Remember to draw a sketch of the situation.

e) Using the force of gravity |Fg| = G (Mm)/r^2 , where M is the mass of the Earth, m the mass of the orbiting object. Find the required speed vorbit, as a function of r, so that the object performs uniform circular motion around the Earth, under the influence of gravity. How long does it take to go round once in an orbit 900 km above ground? Remember to draw a sketch of the situation.

Homework Equations

a=v^2/r
v = (2*pi*r)/T , where T is the time it takes for the Earth to pass one round around it's own axis (24*60*60)s

The Attempt at a Solution

A)
I used a=v^2/r and found it to be 0.0338 m/s^2
9.8/0.0338 = 290 thus the acceleration is 1/290 of g
B)
By setting 9.8 = (V^2)/r and solving it for v i got 7920 m/s
Bearing in mind that T=(2*pi*r)/v I got 5078 sec. or 84.6 min.
(24*60)/84.6=17 rounds per 24 hours
C)
This one is difficult to explain here but I'll try my best.
Thinking that the radius for the person gets smaller as one gets further north (because the Earth is smaller there) I found that the "new" radius is r*cos(tetta) where r is the original radius
Putting this into the a=(v^2)/r gives this witch I hope is the correct one :
(4*pi^2*r*cos(tetta))/T^2
I've allso solved e but the problem is for me d

"how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)?"

I thought that i could decompose g in x- and y-direction because as the task sais, a is in horizontal direction and g is in both x and y since it towards the center of earth. I tried to find then when the x-component of g (gcos(tetta)) is equal to my answer from task c. This resulted in the angle dissaparing completely from my equation and so it must be wrong.

Quite francly I don't understand the question. How can g produce the necessary centripetal acceleration when that should come from the movent of the Earth around its own axis. Those two accelerations has nothing to do with each other as far as I can understand. Also what is the necessary centripital acceleration?

 

[haruspex]

I see no reason to assume your answer to (d) is wrong. It looks right to me. If you think of the centripetal acceleration in the form $\omega^2r$ you can see that the higher the latitude the less acceleration is needed.
The centripetal acceleration does not "come from" the rotation of the Earth. It is that component of acceleration required to keep a body rotating about the Earth (at the given rate and distance).

 

[Makonia]

I see no reason to assume your answer to (d) is wrong. It looks right to me. If you think of the centripetal acceleration in the form $\omega^2r$ you can see that the higher the latitude the less acceleration is needed.
The centripetal acceleration does not "come from" the rotation of the Earth. It is that component of acceleration required to keep a body rotating about the Earth (at the given rate and distance).

I see, but the reason I thought it was wrong is because the task ask for the answer as a function of θ and when I solve the equation (4*pi*r*cosθ)/((2*pi*r)/v)^2=gcosθ for v the θ dissapairs and thus I haven't answered the task. But mabye I did it wrong. I'll do it again to check

 

[haruspex]

I see, but the reason I thought it was wrong is because the task ask for the answer as a function of θ and when I solve the equation (4*pi*r*cosθ)/((2*pi*r)/v)^2=gcosθ for v the θ dissapairs and thus I haven't answered the task. But mabye I did it wrong. I'll do it again to check

If I write that y is a function of x, y=y(x), and later arrive at a solution that y is a constant, that is not a contradiction. It is just a trivial function of x.

 

[Makonia]

If I write that y is a function of x, y=y(x), and later arrive at a solution that y is a constant, that is not a contradiction. It is just a trivial function of x.

Okey, well if I do that i get that v = sqrt((r^2*g)/pi) Does that mean that the angle is irrelevant? Like if I'm at the equator or further north it dosn't matter?

 

[haruspex]

Okey, well if I do that i get that v = sqrt((r^2*g)/pi) Does that mean that the angle is irrelevant? Like if I'm at the equator or further north it dosn't matter?

Yes.

 

[Makonia]

Yes.

Awesome! Thank you soooo much! :)

 

[insightful]

v = sqrt((r^2*g)/pi)

Do the units work out correctly here?

 

[haruspex]

Do the units work out correctly here?

Sorry, I didn't notice you had r² there. Go back to your equation in post #3 and work it through again.

 

[Makonia]

Do the units work out correctly here?

I forgot to check that. Also see that I forgot to write pi^2 in the equation. But anyways I did it again an got v = sqrt(g*r) , an answer that seems really simple for me considering the complexity of the task. Does this answer make sense to you guys? At least now the units are correct.

 

[haruspex]

I forgot to check that. Also see that I forgot to write pi^2 in the equation. But anyways I did it again an got v = sqrt(g*r) , an answer that seems really simple for me considering the complexity of the task. Does this answer make sense to you guys? At least now the units are correct.

That's what I got yesterday, if I recall correctly.

 

[vetsi]

In your answer c. shouldn't also rcos(tetta) be (rcos(tetta))^2?

I get the same as you in d, only that then it's: v=sqrt(g*r^2)

 

[Makonia]

In your answer c. shouldn't also rcos(tetta) be (rcos(tetta))^2?

I get the same as you in d, only that then it's: v=sqrt(g*r^2)

In c this is my calculation:

When I put this equal to gcosθ i get v = sqrt(g*r) and I've tried several times with same result. Still don't think its the correct answer though.

 

[Makonia]

If you have v=sqrt(g*r^2) your answer isn't in m/s so it makes no sence

 

[insightful]

i get v = sqrt(g*r) and I've tried several times with same result.

I agree with you for c). For d), doesn't your answer give v at the equator? I would think you're looking for an the answer for any location on the Earth, i.e., solve for revolutions per day as in b).

 

[vetsi]

In c this is my calculation:

When I put this equal to gcosθ i get v = sqrt(g*r) and I've tried several times with same result. Still don't think its the correct answer though.

Seems it went a bit to fast when I did it. Then I get the same as you two

 

[Makonia]

For d), doesn't your answer give v at the equator? I would think you're looking for an the answer for any location on the Earth, i.e., solve for revolutions per day as in b).

Thats what I'm concerned about, acording to haruspex earlier in this thread it doesn't matter if your on the equator or not, but to me that seems weird. if it was like sqrt(g*r*cosθ) i would actually have a function of θ which is what the task asks for. Honestly I don't know anymore. The task makes little sense to me :P

 

[insightful]

Thats what I'm concerned about, acording to haruspex earlier in this thread it doesn't matter if your on the equator or not, but to me that seems weird. if it was like sqrt(g*r*cosθ) i would actually have a function of θ which is what the task asks for. Honestly I don't know anymore. The task makes little sense to me :P

Well, if it's independent of theta, then isn't the answer the same as b) or 17 revolutions/day?

 

[Makonia]

Well, if it's independent of theta, then isn't the answer the same as b) or 17 revolutions/day?

Maby, but the bigger θ is the smaller the radius for which the centripital acceleration is conserned gets. in other words gcosθ gets smaller and thus the centripital acc. also gets smaller. Then the necessary acc. produced by g should get smaller.
Also from the way my teacher has formulated the question I find it hard to believe that its the same as b. Plus his asking for speed not revolutions/day

 

[insightful]

d) Taking into account that the centripetal acceleration is towards the rotation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)?

It appears he is asking for revolutions/day, just like in b).

 

[Makonia]

It appears he is asking for revolutions/day, just like in b).

Awesome! You managed to see what i couldn't. But still, does it make sense to you that its the same anywhere on Earth because to me that's weird. If the centripital acceleration changes with the angle I would think that the amount of rotations would also change

 

[insightful]

Awesome! You managed to see what i couldn't. But still, does it make sense to you that its the same anywhere on Earth because to me that's weird. If the centripital acceleration changes with the angle I would think that the amount of rotations would also change

Try this visual:
Anywhere on the Earth, set up a post that points directly away the axis of rotation of the Earth. The post will be vertical at the equator and horizontal at the north pole. Put a frictionless donut on the post. At 17 revolutions per day, everywhere on the Earth, the donut will begin to slide on the post away from the Earth.

 

[Makonia]

Try this visual:
Anywhere on the Earth, set up a post that points directly away the axis of rotation of the Earth. The post will be vertical at the equator and horizontal at the north pole. Put a frictionless donut on the post. At 17 revolutions per day, everywhere on the Earth, the donut will begin to slide on the post away from the Earth.

I guess that makes sense.
So you feel that by answering the same as in b I've done everythin I need to? I mean, then I still haven't explained it as a function of θ. Why would he write that if it wasn't depending on the angle?

 

[insightful]

I guess that makes sense.
So you feel that by answering the same as in b I've done everythin I need to? I mean, then I still haven't explained it as a function of θ. Why would he write that if it wasn't depending on the angle?

Yes. Remember, a constant is a valid (and quite common) function.

Why ask for a function of theta? Maybe to introduce you to the "trick question."

 

[Makonia]

Yes. Remember, a constant is a valid (and quite common) function.

Why ask for a function of theta? Maybe to introduce you to the "trick question."

Okey then, thank you very much for the help. Have been stuck on this for a long time

 

[SanDiss]

I know this is off topic, but how did you solve e) ? And could you give me a detailed step by step on d) ? I can't understand it...

 

[Makonia]

I know this is off topic, but how did you solve e) ? And could you give me a detailed step by step on d) ? I can't understand it...

For e you know that (G*M*m)/r^2 is the force and since F=m*a where a=v^2/r you can solve the problem by solving it for v and then setting in the formula for T instead. Then you will get the time for one lap. If you are a UiS student and that's why you're wondering about d I would suggest you ask the student helpers this week. They can probably explain it way better than me since writing step by step here would take a long time for me

 

[coffeemanja]

In c) the answer is 0, since a=v^2/r and r=0

 

[haruspex]

In c) the answer is 0, since a=v^2/r and r=0

You seem to have left out a step there. v depends on r, so you need to write it as $a=\omega^2r$ first. (Otherwise, the conclusion would have been a goes to infinity as r goes to zero!)

 

[coffeemanja]

You seem to have left out a step there. v depends on r, so you need to write it as $a=\omega^2r$ first. (Otherwise, the conclusion would have been a goes to infinity as r goes to zero!)

Alright, alright! No need to shout;)

 

[Sabalaba]

How did u do c,d and e??

 

[hefalomp]

For e you know that (G*M*m)/r^2 is the force and since F=m*a where a=v^2/r you can solve the problem by solving it for v and then setting in the formula for T instead. Then you will get the time for one lap. If you are a UiS student and that's why you're wondering about d I would suggest you ask the student helpers this week. They can probably explain it way better than me since writing step by step here would take a long time for me

Did you get the periode for the object to do one lap equal to 268,7s ?? Felt that this answear was very small, but i got the formula T = 2*pi*r / (sqrt(G*M/r)

 

[Makonia]

Did you get the periode for the object to do one lap equal to 268,7s ?? Felt that this answear was very small, but i got the formula T = 2*pi*r / (sqrt(G*M/r)

No I got 6000 and something s, your formula looks wrong so you should try to do it again

 

[Makonia]

How did u do c,d and e??

How to do both c and e is written in this thread so just read it. For d it's difficult for me to explain because I'm not sure myself. Thats why I asked here but don't know if I'm all that wiser on it anyways

 

[Sabalaba]

How to do both c and e is written in this thread so just read it. For d it's difficult for me to explain because I'm not sure myself. Thats why I asked here but don't know if I'm all that wiser on it anyways

ok, thank u!
The problems are so hard