What would gravity be in these 2 conditions

[HadanIdea]

I was wondering about Earth and it's gravity and I came up with these 2 questions...

1.) What would Earth's gravity be if the Earth stopped spining?

2.) What would Earth's gravity be if the Earth stopped spining and there was no other stars or planets that interfered with there gravity?

As I understand the average standard for gravity is 9.8m/s^square

 

[HallsofIvy]

In Newton's theory, the force of gravity is $-GmM/r^2$. That has nothing to do with the rotation of the earth. And other stars and bodies do NOT "interfere" with gravity.

 

[Bandersnatch]

I'm pretty sure HadanIdea means acceleration experienced by objects on Earth's surface.

If you took away those two factors, you wouldn't notice any difference. Certainly not if you're only measuring it with accuracy of two significant figures. It'd still be 9.8 m/s^2.

 

[adjacent]

I'm pretty sure HadanIdea means acceleration experienced by objects on Earth's surface.

If we were to account for all the forces in the universe,it would be impossible to even think of building structures like Skyscrapers.

 

[HadanIdea]

Would it still be 9.8 m/s^2?

I thought that it would be more, don't we need to take into account the rotating force outwards as the Earth spins?

 

[D H]

Centrifugal force due to the Earth's rotation is a tiny effect, about 0.034 m/s² at the equator (and yes, centrifugal force is a part of g). There is another indirect effect of the Earth's rotation, the equatorial bulge. The Earth would change shape if the Earth stopped spinning, eventually stabilizing to a spherical shape. This reshaping would also result in a slight change in the local gravitational acceleration.

Right now, g at sea level at the North Pole is about 9.832 m/s² while its only 9.780 m/s² at sea level at the equator. If the Earth stopped spinning, gravitation acceleration at sea level would be about 9.807 m/s² everywhere after the Earth has reshaped itself to that spherical shape.

 

[paisiello2]

Does the value of g typically include the centrifugal force?

 

[D H]

Does the value of g typically include the centrifugal force?

Yes. The quantity g is defined from the perspective of an Earth-fixed frame, a rotating with the Earth. This means g is a combination of the acceleration due to gravitation and the acceleration due to the fictitious centrifugal force. Also see [post=1602624]this post[/post].

 

[abitslow]

We should also add that the amount of mass underneath you varies from place to place, even at sea level. Some places have dense rock underneath them, while on the ocean it might be miles of water underneath. Plumes of higher density "lava", mountain ranges, all affect local gravity. Wikipedia has an interesting map of local variations in gravity (at pretty coarse resolution).https://en.wikipedia.org/wiki/Earth_gravity
It is easy for even a grade school child to calculate the force of attraction of any planet (or moon) on us using the simple formula F = G × m × M ÷ d² which can be compared to g, the attraction at sea level by dividing by m giving you g' = G × M ÷ d² where M is the mass of the planet or Moon, d is the distance and G is the Universal Gravitational Constant. The only tricky part is making sure the units of measure are all consistent.
We can compare g' with g by taking their ratio: (g' ÷ g)×100%
The nearest astronomical object to us is the Moon. The closest it gets is 350,000,000 meters, its mass is 7.35E22 kg and G is 6.67E-11 (m²/kg)(m/s²) [where E stands for "× 10^ " like 1.23E4 = 1.23× 10^4 =12,300 or like 5.67E-1 = 0.567] that means g' = 4.00E-5 m/s² or 0.00004. Compare that with g and you get 0.00004÷9.8 is about 0.000004x100% or 0.0004%. The Sun's g on us is about a third of that, and the next significant acceleration is from Venus, but I forget how much that is, its really, really tiny.

 

[nitro21345]

The gravity would be more because centrifugal force that pulls us of the planet, but the gravity keeps us on the plant

 

[phinds]

The gravity would be more because centrifugal force that pulls us of the planet, but the gravity keeps us on the plant

Which part of post #6 did you not understand? As stated in that post, yes, it would be more but only by a negligible amount.

 

[nitro21345]

Which part of post #6 did you not understand? As stated in that post, yes, it would be more but only by a negligible amount.

what part did I not understand

 

[nitro21345]

What about centrifugal force though because the roller coaster ride has a lot of centrifugal force the PULLS you against the sides or when you stop Newtons second law of motion is at play. Gravity pushes on you not pull so if the planet stopped spinning the gravity would be more because to stay on the planet you have to be pushed on harder than pushed with no spinning, therefore if the world stopped gravity would be more.

 

[phinds]

What about centrifugal force though because the roller coaster ride has a lot of centrifugal force the PULLS you against the sides or when you stop Newtons second law of motion is at play. Gravity pushes on you not pull so if the planet stopped spinning the gravity would be more because to stay on the planet you have to be pushed on harder than pushed with no spinning, therefore if the world stopped gravity would be more.

As I said before, yes, it would be more, but by a negligible amount.

 

[nitro21345]

No you need more force to keep you on the planet than if it was not spinning

 

[Nugatory]

No you need more force to keep you on the planet than if it was not spinning

Yes, by about .3% at the equator and tending towards zero at the poles. That's not zero, but it is much less than you would notice or could measure without moderately sophisticated lab equipment.

 

[nitro21345]

No the Earth is spinning faster than us

 

[adjacent]

No the Earth is spinning faster than us

Are we really spinning?