Finding the apparent acceleration due to gravity versus latitude

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.
  • #1
pandafish
11
2
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:

$$\\N=\frac{GM}{R^2}-R\omega^2cos^2\lambda$$
 
Last edited:
Physics news on Phys.org
  • #2
pandafish said:
$$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$
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?
 
  • #3
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.
 
Last edited:
  • #4
pandafish said:
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?
 
  • #5
haruspex said:
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.
 
  • #6
pandafish said:
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".
 
  • Like
Likes scottdave
  • #7
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.
 
  • #8
pandafish said:
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.
 
  • Like
Likes scottdave and pandafish
  • #9
Thank you, that explanation makes it a lot clearer what the apparent force is.
 

Related to Finding the apparent acceleration due to gravity versus latitude

What is the apparent acceleration due to gravity?

The apparent acceleration due to gravity is the effective acceleration experienced by an object due to the combination of gravitational force and the centrifugal force resulting from the Earth's rotation. This value varies with latitude and altitude.

How does latitude affect the apparent acceleration due to gravity?

Latitude affects the apparent acceleration due to gravity because the centrifugal force, which opposes gravity, is maximum at the equator and zero at the poles. As a result, the apparent acceleration due to gravity decreases as one moves from the poles to the equator.

Why is the apparent acceleration due to gravity different at the poles and the equator?

At the equator, the centrifugal force due to Earth's rotation is greatest, reducing the apparent acceleration due to gravity. At the poles, there is no centrifugal force acting outward because the rotational velocity is zero, so the apparent acceleration due to gravity is highest.

What formula is used to calculate the apparent acceleration due to gravity at different latitudes?

The apparent acceleration due to gravity \( g' \) at a latitude \( \phi \) can be approximated by the formula: \( g' = g - \omega^2 R \cos^2(\phi) \), where \( g \) is the standard gravitational acceleration (9.81 m/s²), \( \omega \) is the angular velocity of Earth's rotation, and \( R \) is the Earth's radius.

How does altitude influence the apparent acceleration due to gravity?

Altitude influences the apparent acceleration due to gravity because gravity decreases with the square of the distance from the center of the Earth. As altitude increases, the distance from the Earth's center increases, leading to a reduction in gravitational acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
974
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
217
  • Introductory Physics Homework Help
Replies
2
Views
781
  • Introductory Physics Homework Help
Replies
3
Views
240
  • Introductory Physics Homework Help
3
Replies
99
Views
9K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
5
Views
648
  • Introductory Physics Homework Help
Replies
10
Views
944
Back
Top