Gravity

Agnishom · 2011-11-18 01:31:25

Find the depth or height at which the value of g is less than the original g by 1%
Radius of earth = 6400km

Last edited by Agnishom (2011-11-18 02:09:27)

Bob · 2011-11-18 02:23:55

hi Agnishom

Above the Earth's surface, g drops off following an inverse square law, so using R for the radius of the Earth, h for the height above this, and g' for the gravity at a point h above the surface

That should enable you to do that case.

I seem to remember that for points below the Earth's surface the matter above a given point reduces the gravitational pull from the centre of the Earth by a linear law. So that would give

which should enable you to do that case.

This linear law can be found by integration, using Newton's law of graviation. As it's about 43 years since I derived this formula I'll have to do a bit of work before I can come up with a proof. Back later with this, I hope.

Bob

Last edited by Bob (2011-11-18 02:25:27)

Bob · 2011-11-19 06:14:31

hi Agnishom,

Got it ... Complete I think, but I'll check it over tomorrow. CHECK NOW COMPLETE. 20/11/11.

Background.

Newton's law of universal gravitation.

F is the force of gravitational attraction between two masses, m and M.

d is the distance between them.

G is a constant.

The reason this theory made such an impact was that this one formula gave the mathematical basis for the motion of objects just above the Earth, travelling under the influence of the Earth's gravity, and the motion of the Planets around the Sun. Thus, he was able to tie together and explain the experimental work of both Kepler and Galileo.

In what follows I will use m for the mass of an object and M for the mass of the Earth.

I am also assuming the the Earth is a sphere with uniform density. (Neither is strictly true but the error is small for the purposes of our calculations.)

The proof is in three parts.

(i) To show that an object inside an infinitesimally thin hollow sphere has no gravitational force on it at all. I think this bit is the hardest to follow. Rather than give up with the whole thing, there is no reason why you shouldn't take it 'on trust' for now and aim to follow the other parts first. The diagram below relates to this part.

(ii) Hence show that an object inside a hollow sphere of thickness h also has no gravitational force acting upon it from the sphere. This part is fairly obvious.

(iii) To derive a formula for the gravitational force on an object below the surface of a solid sphere of uniform density. This part is of medium difficulty.

............................................

(i) Hollow spheres have no gravitational effect on objects inside the sphere.

Refer to diagram.

The circle represents the hollow sphere, radius R

m is a mass inside the sphere at a distance r from the centre.

The yellow strip is a thin circular portion of the surface. (imagine it comes out of the circle in a 3rd dimension)

The angle between the 'horizontal' and the strip at m is phi and at the centre is theta.

The distance from m to the strip is x. (remember the strip goes round in a circle and the distance is x, at every point on the strip.)

The angle the strip makes at the centre is 'a lttle bit of theta' and hence is shown as dθ.

Using the law (dM is a little bit of M and dF is the little bit of force dM makes on m):

If the force is resolved into components in the direction of m and perpendicular to this, these second components will all cancel out during this calculation so, from now on, I will only show the component towards m.

So we can get the total force on m, by integrating these bits of force. I'll sort out the limits later.

Now to change the dM into dx so we can integrate the expression.

For the strip

And the connection between the width of the strip and dθ is

so

Now these two areas give us the proportion of M for dM so

Cosine rule:

Cosine rule again.

At this point I can integrate this, so it's time to consider the integration limits.

When the strip is at it's rightmost position x = R + r. And at the left most x = R - r.

so

but

So there is zero gravitational force from the thin sphere.

(ii) If the hollow sphere has a thickness, just imagine it broken up into infinitesimally thin shells. Each shell has no gravitational effect, so they will have no gravitational effect in summation.

(iii) Finally, suppose the sphere is solid and we want to know the gravitational effect at a depth h below the surface.

Consider the sphere as made up of two parts: a hollow shell starting at h below the surface and extending to the surface; and a solid smaller sphere with radius R - h.

Using (i) the first makes zero contribution to the gravitational effect. The second can be considered as a mass equal to the mass of the smaller sphere and acting from it's centre (see ps.)

Thus the force is proportional to (R - h).

Bob

ps. I also stumbled upon the proof that it is ok to treat the force as all acting from the centre when outside the sphere.

It's fairly short, after the above, since most of the work has already been done. I'll do it if you ask, but I need a few days off from this problem to let my fingers recover!!

Last edited by Bob (2011-11-19 21:54:52)

Math Is Fun Forum

#1 2011-11-18 01:31:25

Gravity

#2 2011-11-18 02:23:55

Re: Gravity

#3 2011-11-19 06:14:31

Re: Gravity

Board footer