Find g, measured to be 9.78 at the equator, when Earth is still

[lpettigrew]

Homework Statement

Hello, I have found the following question which appeared to be quite straightforward but the more I think about it the more I complicate matters. If the measured value of gravity, g, at the equator is 9.78 how much g be if the Earth was not rotating?
I have tried to answer this fully and give reasoning to my solution but I have struggled to calculate the correct value of the centripetal force acting. I would be very grateful for any further explanation or advice

Relevant Equations

F=mv^2/r
v=2πr/T

If we think about the forces acting on an object at the surface of the Earth, these forces would be the weight force (mg) pulling downwards and the normal force (N) acting upwards . If the Earth were to stop rotating, then according to Newton's first law the forces must balance, meaning the net force is zero.
However, when the Earth is rotating an object on the surface of the Earth would experience an acceleration, supplied by the net force acting, the centripetal force pointing toward the centre of the Earth.
The centripetal force can be calculated as F=mv^2/r.
Hence, ΣFnet=Fc=Fg-FN=mg-N
To find the value of g if the Earth were still would this be the apparent weight of the object, i.e the normal force?
N=mg-Fc=mg-mv^2/r
g'=mg-mv^2/r
I think mass could cancel here;
g'=g-v^2/r
If g is measured to be 9.78 at the equator when the Earth is spinning then;
g'=9.78-465^2/6.4*10^6
g'=9.74621 ~9.75 ms^-2
This would mean ∆g =9.78-9.75~0.03ms^-2

Or alternatively finding the centripetal force one could subtract this directly from the measured value of g at 9.78ms^-2;
Take v=2πr/T
T=86400s
r=6.4*10^6m
Thus, v=2*π*6.4*10^6/86400
v=465.4211..~465 ms^-1

Then, the force acting on an object of 1kg: F=1*465^2/6.4*10^6=0.0337..~0.034 N
However, this is my point of difficulty, since I have seen elsewhere that the centripetal force acting on an object at the equator is equal to 0.34N, how am I off by a factor of 10? Have I found the centripetal acceleration instead?

 

[haruspex]

g'=9.78-465^2/6.4*10^6

Don't you mean g' (the apparent value at the equator)=g-v²/r, where g is the value with a static Earth?

I have seen elsewhere that the centripetal force acting on an object at the equator is equal to 0.34N

For a 1kg object? No. See e,g, https://pwg.gsfc.nasa.gov/stargaze/Srotfram1.htm

 

[lpettigrew]

Don't you mean g' (the apparent value at the equator)=g-v²/r, where g is the value with a static Earth?

Yes, I did mean that g is the value with a static Earth. I apologise but I am a little confused by your reply, although I appreciate it greatly. Where does my mistake lie?
Is it that this is a very simple question wherein the value of g if the Earth were not spinning would simply be the measured value at the equator (9.78) minus the centripetal force (calculated as ~0.034 N)?
Thus, the value of g when the Earth is not spinning would be 9.746~9.75 to three signifcant figures.

Or after having found ∆g =9.78-9.75~0.03ms^-2 would this mean that the net force acting is;
ΣFnet=mg-N=9.78*1-0.03=9.75 ms^-2 when the Earth is rotating. Since the object would not experience an accleration when it stops rotating this would mean the value of g would increase by the amount of the centripetal force at the equator but not at the poles?
So the value of g when the Earth is not spinning would be 9.78+0.034 N=9.814~9.81 ms^-2 to 3.s.f

 

[PeroK]

Yes, I did mean that g is the value with a static Earth. I apologise but I am a little confused by your reply, although I appreciate it greatly. Where does my mistake lie?
Is it that this is a very simple question wherein the value of g if the Earth were not spinning would simply be the measured value at the equator (9.78) minus the centripetal force (calculated as ~0.034 N)?
Thus, the value of g when the Earth is not spinning would be 9.746~9.75 to three signifcant figures.

Or after having found ∆g =9.78-9.75~0.03ms^-2 would this mean that the net force acting is;
ΣFnet=mg-N=9.78*1-0.03=9.75 ms^-2 when the Earth is rotating. Since the object would not experience an accleration when it stops rotating this would mean the value of g would increase by the amount of the centripetal force at the equator but not at the poles?
So the value of g when the Earth is not spinning would be 9.78+0.034 N=9.814~9.81 ms^-2 to 3.s.f

It's so difficult to see what you are doing with all those numbers. Let's try to get the physics right first:

1) The ##g## given in the problem statement is the measured gravity on a spinning Earth. It would seem more logical to let ##g'## be the measured gravity on a non-spinning Earth. As you shouldn't really change the notation from the problem statement.

2) If the Earth were to spin faster, then the measured gravity would decrease (until we would eventually leave the surface). Therefore, we expect the measured gravity on a non-spinning Earth, let's call that ##g'##, to be greater than ##g##.

3) The difference is, of course, the centripetal acceleration, ##a_c##.

4) Let's calculate ##a_c## and then ##g' = g + a_c##. And we are done.

 

[lpettigrew]

It's so difficult to see what you are doing with all those numbers. Let's try to get the physics right first:

1) The ##g## given in the problem statement is the measured gravity on a spinning Earth. It would seem more logical to let ##g'## be the measured gravity on a non-spinning Earth. As you shouldn't really change the notation from the problem statement.

2) If the Earth were to spin faster, then the measured gravity would decrease (until we would eventually leave the surface). Therefore, we expect the measured gravity on a non-spinning Earth, let's call that ##g'##, to be greater than ##g##.

3) The difference is, of course, the centripetal acceleration, ##a_c##.

4) Let's calculate ##a_c## and then ##g' = g + a_c##. And we are done.

Thank you very much for your concise response.
So, g'=g+v^2/r
g'=9.78+465^2/6.4*10^6
g'=9.8137..~9.81 ms^-2

However, where has the equation ##g' = g + a_c## come from?

 

[PeroK]

Thank you very much for your concise response.
So, g'=g+v^2/r
g'=9.78+465^2/6.4*10^6
g'=9.8137..~9.81 ms^-2

Looks about right.

However, where has the equation ##g' = g + a_c## come from?

From the equation ##g = g' - a_c##.

Which comes from the equation: ##\vec g = \vec g' - \vec a_c##

 

[lpettigrew]

Looks about right.From the equation ##g = g' - a_c##.

Which comes from the equation: ##\vec g = \vec g' - \vec a_c##

Thank you for your reply again. Right but where have these two equations come from in terms of the information given?
Also, is my value for the centripetal force calculated earlier as 0.034 N incorrect? (which I thought may be out by a factor of 10).

 

[PeroK]

Also, is my value for the centripetal force calculated earlier as 0.034 N incorrect? (which I thought may be out by a factor of 10).

That's got to be right. Acceleration is force per unit mass.

Thank you for your reply again. Right but where have these two equations come from in terms of the information given?

Force and acceleration obey the vector laws - every question is not necessarily going to tell you that!

 

[lpettigrew]

That's got to be right. Acceleration is force per unit mass.

But I have calculated the force not acceleration?Force and acceleration obey the vector laws - every question is not necessarily going to tell you that!
[/QUOTE]
Ok, right.

 

[PeroK]

But I have calculated the force not acceleration?

If you tackle a problem using force per unit mass, then that's equivalent to tackling the problem using acceleration. That's true on both a superficial level, and true rather more profoundly.

 

[lpettigrew]

If you tackle a problem using force per unit mass, then that's equivalent to tackling the problem using acceleration. That's true on both a superficial level, and true rather more profoundly.

Thank you for your reply. Right yes I agree with that. However, if I was intending to find the centripetal force specifically would my solution then be incorrect, as it corresponds to the acceleration?

 

[PeroK]

Thank you for your reply. Right yes I agree with that. However, if I was intending to find the centripetal force specifically would my solution then be incorrect, as it corresponds to the acceleration?

Your solution looks right. I'm not sure now what the problem was now.

My comment was only that the physics was getting lost in the numbers and the unclear notation.

 

[lpettigrew]

Your solution looks right. I'm not sure now what the problem was now.

My comment was only that the physics was getting lost in the numbers and the unclear notation.

Oh yes, I can see your point, there are obscured areas where the physics is lost a little.
No, I was just asking if the value I had calculated earlier for the centripetal force as 0.034 N was correct because I have seen it elsewhere given as 0.34 N instead and wondered whether I made a mistake in calculation?

 

[PeroK]

because I have seen it elsewhere given as 0.34 N instead and wondered whether I made a mistake in calculation?

This is wrong. See post #2. It's ##0.34\%## of ##g##.

 

[lpettigrew]

This is wrong. See post #2. It's ##0.34\%## of ##g##.

Oh so the centripetal force is 0.034 N?

 

[PeroK]

Oh so the centripetal force is 0.034 N?

It is indeed!

 

[lpettigrew]

It is indeed!

Oh splendid thank you for your help

 

[haruspex]

Oh so the centripetal force is 0.034 N?

It still bothers me that you seem confused regarding force and acceleration. The centripetal force at the equator depends on the mass. It is only 0.034N if you happen to be considering a 1kg mass. It is more helpful in general to say the centripetal acceleration at the equator is 0.034ms^-2.
And here:

ΣFnet=mg-N=9.78*1-0.03=9.75 ms^-2

you equated a force to an acceleration.