Gravitational Potential Calculation

[21boy]

Homework Statement

A meteorite impacts a flat-tabular Earth (with a density of 6gm/cm^3). During transit, the meteorite is observed to have a density of 8 gm/cm^3 and to be a perfect sphere of radius 1km. The meteorite penetrates the flat Earth to an unknown depth to center mass, z_0; z_0 > 1km. The meteorite is undeformed during emplacement and the Earth that is displaced magically vanishes and the hole above the meteorite is filled again so that the surface is flat and the density uniform everywhere except for the meteorite. Use relevant equations relating Gauss' Law to this geometry to derive teh gravitational potential of the meteorite as a function of position along the Earth's (slab's) surface. Let x=0 be over the center of the mass of the meteorite.

*Positive z is pointing towards the Earth's surface
*z=0 at the surface of the slab
*The thickness of the Earth is t

Homework Equations

Δρ=ρ1-ρ => 8-6=2

g_z = 2.79E-10 (Δρ r_sphere h)/(3 (x^2+h^2)^(3/2) To find gravity in g_z direction

V= ∫g_z d? to find potential of gravity.

Question is I'm not sure of my bounds for the integral and what I should be integrating in respect to. Also, for calculating gravity, should I use x=0?

 

[Simon Bridge]

Question is I'm not sure of my bounds for the integral and what I should be integrating in respect to. Also, for calculating gravity, should I use x=0?

The answer depends on how you are using them - what is your reasoning?

 

[21boy]

Well I'm trying to calculate gravitational potential over the x-surface above a buried object. I would think that I would derive my integral in terms of dx from x=0(area directly above meteorite) to some positive value and do the same for a symmetric negative value. I'd define this area as my equipotential surface that's being affected by the meteorite.

Does that answer the problem statement?

 

[Simon Bridge]

OK - so the x-y plane is the surface, the com of the meteorite is at $(x,y,z)=(0,0,-z_0)$ ... The potential of the flat-Earth is $gz:z\geq 0$ but you want the potential of the meteorite.

You are asked to use Gausses Law - or "the relevant equations relating Gauss' Law to this geometry" - find the potential.

So what is the geometry in question?
What shape will the equipotential surfaces make in the plane?

 

[paranormal]

Thank you for providing the necessary information and equations for this problem. To begin, let's define some variables for clarity:

- ρ1 = density of the meteorite (8 gm/cm^3)
- ρ2 = density of the Earth (6 gm/cm^3)
- Δρ = difference in density (ρ1 - ρ2)
- r_sphere = radius of the meteorite (1 km)
- h = depth of penetration of the meteorite (unknown)
- x = position along the Earth's surface (slab)
- t = thickness of the Earth (unknown)
- g_z = gravitational acceleration in the z direction
- V = gravitational potential

Now, to find the gravitational potential as a function of position along the Earth's surface, we can use Gauss' Law. This law states that the gravitational flux through a closed surface is equal to the enclosed mass divided by the squared distance from the center of mass. In our case, the closed surface will be a cylinder with the meteorite at its center.

First, let's find the gravitational acceleration in the z direction, g_z. Using the equation you provided, we can substitute the known values and solve for g_z:

g_z = 2.79E-10 * (Δρ * r_sphere * h) / (3 * (x^2 + h^2)^(3/2))

Next, we can use this value to find the gravitational potential, V. This can be done by integrating g_z with respect to x, from position x = 0 (over the center of mass of the meteorite) to the edge of the Earth's surface, x = t/2. This will give us the potential at the surface of the slab.

V = ∫g_z dx = ∫2.79E-10 * (Δρ * r_sphere * h) / (3 * (x^2 + h^2)^(3/2)) dx

To find the potential at any point along the surface, we can use the equation:

V(x) = V(0) + ∫g_z dx = V(0) + ∫2.79E-10 * (Δρ * r_sphere * h) / (3 * (x^2 + h^2)^(3/2)) dx

Where V(0) is the potential at the center of mass of the meteorite,