Linking gravity, angular velocity and radius

[jonnybmac]

Homework Statement

I am having trouble linking gravity to the radius of the Earth and angular velocity. I was using this as a solid method to confirm the equation for values of a different sort based on centripetal acceleration. When inputting the values though it does not add up and I cannot see where and why

Homework Equations

w = θ/time
α = w²r = centripetal acceleration
α = g
g = w²r = acceleration due to gravity
r = g/w²

The Attempt at a Solution

Since α has the values m s^-2 and g = m s^-2
α = gif
w = 2π/86400 (rotation of Earth per day)
≈ 7.272 * 10^-5 rad s^-1

and
g = 9.81 m s^-2

then aranging
g = w²r

to

r = g/w²
= 9.81/(7.272 * 10^-5)² m
≈1.86 * 10⁹ m

Now if I google radius of Earth I get 6.37 * 10⁶ m

This is nowhere near the radius that I should be getting with the above equation, so where am I going wrong?

Many thanks for reading

 

[Orodruin]

First of all: Always write out your units. A statement such as g=9.81 is utterly meaningless unless you write out the m/s^2.

To answer your question, what you computed is not the radius of the Earth. It is the radius at which the gravitational acceleration at the Earth’s surface would be sufficient to keep an orbit with a one day period.

At the Earth’s actual surface, the gravitational acceleration is far too large on its own to be in a circular orbit with that period. That is why you feel a normal force from the ground that generally is sufficient to just balance out the gravitational acceleration. The net acceleration is a result of both the gravitational and the normal force.

 

[jonnybmac]

First of all: Always write out your units. A statement such as g=9.81 is utterly meaningless unless you write out the m/s^2.

To answer your question, what you computed is not the radius of the Earth. It is the radius at which the gravitational acceleration at the Earth’s surface would be sufficient to keep an orbit with a one day period.

At the Earth’s actual surface, the gravitational acceleration is far too large on its own to be in a circular orbit with that period. That is why you feel a normal force from the ground that generally is sufficient to just balance out the gravitational acceleration. The net acceleration is a result of both the gravitational and the normal force.

Thank you for your response, and apologies for the lack of units - I have amended the original post.

So centripetal acceleration is not the same as gravitational acceleration? how would you link the radius of the Earth and its gravitational acceleration at its surface then?

I was of the understanding that at whatever radius and speed the particle is travelling, it would produce a different centripetal acceleration.
In the case of earth, I presumed the 9.81 m s^-2 was relative to the distance from the centre and the angular speed at which it is travelling. Since angular speed is a constant, the only thing that can change gravitational acceleration is the radius?

 

[PeroK]

Thank you for your response, and apologies for the lack of units - I have amended the original post.

So centripetal acceleration is not the same as gravitational acceleration? how would you link the radius of the Earth and its gravitational acceleration at its surface then?

I was of the understanding that at whatever radius and speed the particle is travelling, it would produce a different centripetal acceleration.
In the case of earth, I presumed the 9.81 m s^-2 was relative to the distance from the centre and the angular speed at which it is travelling. Since angular speed is a constant, the only thing that can change gravitational acceleration is the radius?

https://en.wikipedia.org/wiki/Gravity_of_Earth#Variation_in_gravity_and_apparent_gravity

The gravity at the Earth's surface depends only on the mass and radius of the Earth. The Earth's rotation has the effect of reducing the "apparent" gravity slightly (and changing its direction slightly) because of the centripetal force needed to keep things on the surface.

If the Earth were spinning fast enough, then the apparent gravity at the Equator could be zero. Although it would be unaffected at the Poles.

 

[jonnybmac]

Thank you for clearing that up for me

 

[CWatters]

Centripetal force mv^2/r is the _net_ force that must be applied to an object to make it move in a circle of radius r and Velocity v.

If the applied force is greater than mv^2/r the object would normally move with a smaller radius than r. You can prove this by dropping an object. Gravity at the Earth's surface provides more force than mv^2/r so dropped objects fall, reducing r.

Objects in contact with the Earth's surface experience a normal force which the reduces the net force to exactly mv^2/r so the radius r remains constant.