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

• lpettigrew
In summary, if we consider the forces acting on an object at the surface of a spinning Earth, the weight force and the normal force must balance in accordance with Newton's first law. However, due to the rotation of the Earth, there is a centripetal acceleration acting on the object. This can be calculated using the formula F=mv^2/r, and the value of g' (the measured gravity on a non-spinning Earth) can be found by adding the centripetal acceleration to the measured value of g on a spinning Earth. Therefore, g' = g + a_c.
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?

lpettigrew said:
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?
lpettigrew said:
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

PeroK and lpettigrew
haruspex said:
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

Last edited:
lpettigrew said:
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
PeroK said:
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?

lpettigrew said:
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

lpettigrew said:
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##

PeroK said:
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).

lpettigrew said:
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.

lpettigrew said:
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!

PeroK said:
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.

lpettigrew said:
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
PeroK said:
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?

lpettigrew said:
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.

PeroK said:
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?

lpettigrew said:
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##.

PeroK said:
This is wrong. See post #2. It's ##0.34\%## of ##g##.
Oh so the centripetal force is 0.034 N?

lpettigrew said:
Oh so the centripetal force is 0.034 N?
It is indeed!

lpettigrew
PeroK said:
It is indeed!
Oh splendid thank you for your help

lpettigrew said:
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:
lpettigrew said:
ΣFnet=mg-N=9.78*1-0.03=9.75 ms^-2
you equated a force to an acceleration.

## 1. How is the value of g measured at the equator?

The value of g at the equator is measured using a device called a gravimeter. This device measures the gravitational acceleration, or the rate at which objects fall towards the Earth's surface, at a specific location. The value of g is affected by factors such as altitude, latitude, and the density of the Earth's surface.

## 2. What is the significance of measuring g at the equator?

Measuring g at the equator is important because it helps us understand the Earth's gravitational field. The Earth is not a perfect sphere and has variations in its density, which can affect the value of g. By measuring g at different locations, we can gain a better understanding of the Earth's internal structure and composition.

## 3. How does the Earth's rotation affect the value of g at the equator?

The Earth's rotation causes a centrifugal force that counteracts the force of gravity at the equator. This results in a slightly lower value of g at the equator compared to other latitudes. This effect is also influenced by the Earth's shape and rotation speed.

## 4. Does the value of g change over time at the equator?

The value of g at the equator is relatively constant over time. However, it may vary slightly due to changes in the Earth's rotation, tectonic movements, and other factors. Scientists continue to monitor and study these variations to gain a better understanding of the Earth's dynamics.

## 5. How does the value of g at the equator compare to other locations on Earth?

The value of g at the equator is slightly lower than at other latitudes due to the Earth's rotation and shape. As we move towards the poles, the value of g increases because the centrifugal force decreases. The highest value of g is at the Earth's poles, where the centrifugal force is negligible.

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