# Investigate how gravity varies over the earth's surface

1. Aug 24, 2007

### Michaelcarson11

I am doing a school project to investigate how gravity varies over the earth's surface. I understand that the force is weaker at the equator due to the earth's rotation but there are a few things that I don't quite understand:
1: Firstly, why is an object on the earth's surface only affected by the earth's mass which is within that radius (sorry for the poor explanation but I hope you understand what I mean).
2: If this is the case, then how can you prove that this effect is lesser than the effect of the distance from the centre (because it must be otherwise the gravitational force would be greater at the equator than at the poles)
3: We have also been asked to investigate the effects of mountains, oceans, varying rock densities etc. I can see why they might affect the gravity in a particular region but again, I can't fully understand it

Any help will be greatly appreciated!!

2. Aug 24, 2007

### DaveC426913

Not true.
The force may, in theory, be partially cancelled by an opposing force due to inertia from rotation, but that does not make the force of gravity weaker, and it's really a very, very small amount.

In fact, every point on the Earth's surface at sea level is at gravitational equipotential.

I think what you mean is: why do object that are below Earth's surface only affected by the mass within that radius. (i.e. if you are halfway to Earth's core, you only feel gravity from a mass that's half the radius).

The reason for this is that there's a lot of Earth above you pulling up. It can be demonstrated mathematically that the gravitational force within a hollow sphere cancels to zero. So, If you are halfway to the core of Earth, you can divide the Earth into 2 pieces: 1, a ball below you, half Earth's radius, and 2, a hollow shell which you are inside. Now work out the gravitational force you experience.

Uh, you lost me.

Yep, you're on the right track here. Satellites can map the densities of the Earth under the surface (such as, say, the massive Canadian Shield or mountain ranges) by taking careful readings of their altitude changes as they pass over denser areas of the Earth.

3. Aug 24, 2007

### arivero

4. Aug 24, 2007

### AlephZero

It may be small in absolute terms, but historically it was large enough to be important.

The first attempts at sea navigation using pendulum clocks as a time reference to find longtitude were seriously affected by the variations of the resultant "gravitation + rotation" force with latitude - all at sea level, of course.

One of Newton's achievements in "Principia" was making a mathematical model of this that was consistent with the experimental facts, and which therefore eliminated other hypotheses that were being proposed - for example that the force of gravity at the equator was weaker because the temperature at the equator was higher than at the poles.

As for the diffierence being "small", Newton's models gave a change in the resultant force, at sea level, from equator to poles as 0.43%. That meant a pendulum clock on a ship might accumulate an error of about 1.5 minutes a day depending what latitude it was at, so the effect wasn't so small it could be ignored!

Ref: "Principia" Book III Proposition XX Problem IV: "To find and compare with each other the weights of bodies in the different regions of our earth".

5. Aug 24, 2007

### mgb_phys

Not quite, if you are next to a million square km of granite in Canda or the Himalyas in India the difference between astronomical 'UP' as a line from the Earth's centre and gravitational 'UP' from a spirit level can vary by several arc-sec.

The spinning of the Earth would give you less 'weight' on the equator when compared to the poles. This is magnified because this effect has changed the shape of the earth. At the equator you weigh less because of the spin but this spin has pushed the ground out 20km further from the centre than at the poles. Similairly at the poles you weigh more because you are 20km closer to the centre of the Earth.

6. Aug 24, 2007

### rbj

[strike]i don't believe that is true at all. the Earth is far more spherical than that.[/strike]
(does anyone know the PF markup for strike out?)

what source do you use to come up with the 20 km figure?

okay, i looked it up myself and stand corrected. the figures i find are polar diameter: 12,714 km and equatorial diameter: 12,756 km, a difference of 42 km.

Last edited: Aug 24, 2007
7. Aug 24, 2007

### mgb_phys

8. Aug 24, 2007

### rbj

i hope you jest! (but we all have our conspiracy theories. i, personally, cannot believe the "single bullet theory", from a physics and common sense POV.)

i was remembering a deviation from spherical of hundreds of meters for the ocean surface (clearly a faulty memory).

9. Aug 24, 2007

### D H

Staff Emeritus
You are mistaking equipotential to mean "equal force". Equipotential means "equal potential energy". Gravitational acceleration is the gradient of the scalar potential field, and this varies over the surface geoid, even though the potential is by definition constant on the surface of the geoid.