# Finding the apparent acceleration due to gravity versus latitude

• pandafish
In summary, the conversation involves solving a force balance equation for apparent gravity, which is defined as the net force per unit mass. There is some discussion about the equations used and the direction of the force balance, as well as finding an expression for g. Ultimately, it is clarified that the apparent force is made up of the normal force and the component of the centrifugal force tangential to Earth's surface.f

#### pandafish

Homework Statement
A mass at rest on the surface of the earth at latitude λ , experiences a reaction from the earth that consists of a normal component S per unit mass and a tangential frictional component F per unit mass, directed towards a point vertically above the north pole. The earth is assumed to be a uniform sphere of radius R and mass M rotating with an angular velocity ω about its axis. How is the magnitude and direction of the apparent ’acceleration due to gravity’ g , acting on the mass at latitude , related to these forces?

Show that:

$$g^2 = [\frac{GM}{R^2}-R\omega^2cos^2\lambda]^2 + [\frac{1}{2}\omega^2Rsin2\lambda]^2$$
Relevant Equations
$$g=-\frac{GMm}{R^2}$$
I began by drawing a diagram and resolving the forces. Since the question asked for 'apparent gravity' I tried to find the normal force.

I started with the equations:

$$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$
$$\\(\frac{GM}{R^2}-N)sin\lambda-Fcos\lambda=0$$

Solving simultaneously, I ended up with:

$$\\N=\frac{GM}{R^2}-R\omega^2cos^2\lambda$$

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$$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$
The terms on the left cannot both be sine. F is tangential while the other force sum is radial. I don’t think having trig terms both sides is right either.
Which direction do you intend this force balance to be in?

My bad, its suppose to be:

$$\\(\frac{GM}{R^2}-N)cos\lambda+Fsin\lambda=m\omega^2Rcos\lambda$$

This force balance is suppose to be perpendicular to the axis of Earth's rotation.

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My bad, its suppose to be:

$$\\(\frac{GM}{R^2}-N)cos\lambda+Fsin\lambda=m\omega^2Rcos\lambda$$

This force balance is suppose to be perpendicular to the axis of Earth's rotation.
Then why the cos on the right?

Then why the cos on the right?
The right side is the centripetal acceleration,

$$a = \frac{v^2}{r} = \omega^2r$$

I believe r is distance from the mass to the earth's axis of rotation, so

$$a = \omega^2(Rcos\lambda)$$

I made a mistake by including mass on the right.

The right side is the centripetal acceleration,

$$a = \frac{v^2}{r} = \omega^2r$$

I believe r is distance from the mass to the earth's axis of rotation, so

$$a = \omega^2(Rcos\lambda)$$

I made a mistake by including mass on the right.
Sorry, I read R as though it was r.

You need an expression for g. Note how the problem statement defines g. It is not as used in your "relevant equation".

• scottdave
I see what you mean. Working backwards from the solution, I realised that the 'apparent gravitational acceleration' was made up from the normal force , and the component of the centrifugal force tangential to Earth's surface. However, I'm don't quite understand why this is the case.

I see what you mean. Working backwards from the solution, I realised that the 'apparent gravitational acceleration' was made up from the normal force , and the component of the centrifugal force tangential to Earth's surface. However, I'm don't quite understand why this is the case.
Consider a mass placed on the surface. Apparent gravity is the net force per unit mass. This corresponds to the weight that would be recorded by a conventional weighing machine.

• scottdave and pandafish
Thank you, that explanation makes it a lot clearer what the apparent force is.