I was thinking about gravity, and it seems strange that point masses really work, one question, since I am not that great with relativity is: does GR take density into account? For instance in Newtonian gravity, the distance from the planet's surface doesn't make a difference, it seems in the swartzchild solution it doesn't either. Is this the case?(adsbygoogle = window.adsbygoogle || []).push({});

Why doesn't the arcsecond size of the planet make a difference in gravity?

If all the mass of the planet were within a 10° range less would be cancelled than if it were spread out over a 160° range.

I was trying to figure out how much of a circle could be said to be "under" a point(P) on it's perimeter, "under" meaning what component of it was parallel to a line going through the center and the point.

The area of the sector is A = ∏r^{2}a/360

Since we are going across the circle the radius we want is actually r = 2rcos(a)

So A = ∏r^{2}cos(a)^{2}a/90

This cuts off a bit, but with calculus that is ok since as da get smaller the amount it is off approaches 0

To make it all calculus-y what I wanted first is to prove I get half a circle if i do:

Lim da→0 Ʃ (∏r^{2}cos(a_{i})^{2}/90)da

90

∫(∏r^{2}cos(a_{i})^{2}/90)da

0

you can factor out the constants

∏r^{2}/90 ∫ cos(a_{i})^{2}da

90

∏r^{2}/90 | (sin(2a) + 2a)/4

0

(∏r^{2}/90)[(sin(2*90) + 2*90)/4 - (sin(0) + 0)/4]

(∏r^{2}/90)[(0 + 180)/4 - (0+0)/4]

(∏r^{2}/90)(180/4 - 0)

(∏r^{2}/90)(45)

∏r^{2}/2 which is half a circle.

So If I then took the parallel component of a vector whose magnitude was the area of that sector, and whose direction was the angle a(a was the angle from center, da was the angle used for the sector)

The parallel component is Acos(a)

So I take my A:

A = ∏r^{2}cos(a)^{2}a/90 * cos(a)

That gives me:

So A = ∏r^{2}cos(a)^{3}a/90 right?

So with basically the same integral, constants out I have:

∏r^{2}/90 ∫ cos(a_{i})^{3}da

90

∏r^{2}/90 | (sin(a)^{3}+ 3sin(a))/3

0

∏r^{2}/90(.6558 - 0)

That gives me .0073∏r^{2}

I believe this could be expanded to 3D using sort of a solids of revolution concept but I don't have that done yet. So far as 2D goes does that mean that only .73% of a circle is "under" any given point?

Edit: hmm, .6558 is what i got from the integral calculator, but sin(90) is 1, so (1+3(1))/3 should be 1.333 which would be .0148∏r^{2}, like twice as much, but still REALLY small... As a side question is (sin(a))^{3}the same as sin^{3}(a)?

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# Gravity and vector component

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