Calculating the gravitational field due to a horizontal uniform thin disk

In summary, we are given a task to show the gravitational field of a uniform thin disc at a certain distance above its centre, and we are asked to calculate the number of seconds per year that a pendulum clock in the centre of a room would gain if the floor was covered with a 1cm thick layer of lead. We are given relevant equations for gravitational potential and field, and the steps to take in solving the problem.
  • #1
AaronKnight
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0

Homework Statement


Show that the gravitational field due to a horizontal uniform thin disc (thickness D, radius R and density r) at a distance h vertically above the centre of the disc has magnitude

2πGρd(1-h/(R2+h2)1/2)

A pendulum clock in the centre of a large room is observed to keep correct time. How many
seconds per year will the clock gain if the floor is covered by a 1cm thick layer of lead of density
11350kgm−3?
[Newton’s gravitational constant is G = 6.67×10−11Nm2 kg−2.]


Homework Equations


Gravitational potential, [itex]\phi[/itex]=-Gdm/R
Where dm=2πRDρ.dR
Gravitational field, g= -[itex]\nabla[/itex][itex]\phi[/itex]

The Attempt at a Solution


I have got to [itex]\phi[/itex]=-2πDρGdR and I know I need to integrate with respect to R, then use the g= -[itex]\nabla[/itex][itex]\phi[/itex] but I am unsure what my integration limits should be? I think I need to integrate between 0 and R but then I can't see how I would get the h/(R2+h2)1/2 term?
Any hints would be very useful.
 
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  • #2
In the relevant equations sections ϕ=-Gdm/R should be dϕ=-Gdm/R
 

FAQ: Calculating the gravitational field due to a horizontal uniform thin disk

1. How do you calculate the gravitational field due to a horizontal uniform thin disk?

To calculate the gravitational field due to a horizontal uniform thin disk, you can use the formula:
G * M * sin(theta) / (R^2 + z^2)^3/2
Where G is the gravitational constant, M is the mass of the disk, theta is the angle between the point of interest and the center of the disk, R is the radius of the disk, and z is the distance between the disk and the point of interest.

2. What is a horizontal uniform thin disk?

A horizontal uniform thin disk is a flat, circular object with a uniform distribution of mass throughout its surface. It is assumed to have a negligible thickness and all points on its surface are equidistant from its center.

3. How does the distance from the disk affect the gravitational field?

The gravitational field is inversely proportional to the distance from the disk. As the distance increases, the gravitational field decreases. This means that the gravitational force between the disk and an object decreases as the object moves further away from the disk.

4. What is the significance of the angle theta in the gravitational field formula?

The angle theta represents the direction of the gravitational force acting on an object. It is the angle between the point of interest and the center of the disk. This angle affects the magnitude of the gravitational force, with the maximum force occurring when theta is 0 degrees.

5. Can the gravitational field due to a horizontal uniform thin disk be negative?

Yes, the gravitational field can be negative. This indicates that the direction of the force is in the opposite direction to the positive direction, which is usually taken as pointing towards the center of the disk. This can occur when the point of interest is located below the disk, resulting in a negative value for the gravitational field.

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