Newtonian Gravity Question

[BigMoosie]

[math]F = G\frac{M_1 M_2}{d^2}[/math]

 

Does this equation make the presumption that the Earth is a point gravity source? It seems so becuase taking d down to low values (i.e. close to the center of the Earth) there would be practically no gravity due to it being cancelled out in every dirction, though the formula suggests that it should approach infinite force. Is there a modification of this formula that takes into account the radius of the masses? (presuming they are spherical)

[jordan]

As to the first question, yes, it assumes point masses. As to the second, I don't know if there is a way to take a planet's radius into account. Good question.

[Xyph]

You could assume a uniform density, in which case I think the formula would be [math]F = \frac{4\pi Gdrm_{2}}{3}[/math]. In this case, d = the density of the object and r = the radius of the object (d in your version of the formula).

 

Of course, this will only work below the surface of an object, and assuming a uniform density probably won't be that useful, but I don't see any other way to do it in a simple manner.

[krisse]

[math]F = G\frac{M_1 M_2}{d^2}[/math]

 

Does this equation make the presumption that the Earth is a point gravity source? It seems so becuase taking d down to low values (i.e. close to the center of the Earth) there would be practically no gravity due to it being cancelled out in every dirction' date=' though the formula suggests that it should approach infinite force. Is there a modification of this formula that takes into account the radius of the masses? (presuming they are spherical)[/quote']

 

No, you are missing something here. When you move in through the earth the mass responsible for the attractive force ([math]M_2[/math]) towards the center ([math]d->0[/math]) approaches 0 - such that the whole fraction approaches 0 <=> net gravitational force at center -> 0.

 

The density is another matter. Fluctautions in the mass density would cause the gravitational field to differ from place to place and thereby move the particle in irregular paths through the body.

 

(Sorry - I really suck at the [ math ] tags)

[swansont]

Gauss's law, which applies to any 1/r² force, shows that distributed mass and point mass give the same answer, as long as you have angular symmetry in the mass distribution. The mass inside of r behaves as if it were at r=0. The mass outside doesn't have any net effect.

[BigMoosie]

Cheers!

[Klatu]

Bigmoosie,

 

The given newtonian formula determines the force of attraction that both masses experience proportional to the square of the distance between them.

 

To determine the force of attraction using this formula in a subterrian exploration you will need to integrate the net forces acting on m2, if m1 in the planet mass.

The simplest way to achieve the results you maybe looking for is to diminish the value of m1 for every meter you travel below the earth surface(you will need to know how much mass to remove using simple geometry).

M2 would be your mass and would remain constant, m1 would become less and less as you approach the center of m1.

 

It maybe interesting to note your acceleration and velocity during the trip.

 

I like the Einstienian approach, it would go something like... imagine a giant hole drilled through the plant from end to end. You approach the hole's opening at one end in your space suit and jump in. You would fall at a predictable rate of 32 feet per second squared, since there is no atmosphere there will be no areodynamics to hinder your fall. As you desend past the crust (1000 miles or so)and approach the mantel your acceleration and velocity continue to increase rapidly.

The mantel is the thickest part of the planet many thousands of miles so this might take a few hours so sit back and enjoy the ride.

As you approach the center of the planet your accelerometer is declining but because of your accumulated momentum your velocity remains at it's maximum.

Being unable to slow yourself you would pass right by the center point and begin your journey out the other side. At this point you would notice your velocity decreasing rapidly. Before you got one third the way past the center point your velocity would slowly decrease to zero and would you would start falling back toward the center again.

This might happen a number of times until eventually you would find yourself at the center point with the net gravitational effects are the same in all directions leaving you weightless at the dead center of the planet.

 

Of coarse this is an imaginary ride but it is based on the use of Newton's formula.

 

If you program in basic or C it should'nt be more than 2 dozen lines of code.

 

happy landings!

 

K