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Q & A: Gravitational field variation

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Most recent answer: 10/22/2007
Is not the gravitational force acting at the two arms of a balance when using a balance and a standard 1kg mass to measure out a mass of 1kg of anything the same? Because in my text-book it is given that the gravitational force difference between the two ends of a balance is negligible in a particular area(thereby implying that there is a differnce)since the radius of the earth is longer than the arms of the balance.
- Debbie (age 21)
Your book is just being careful, and the authors are perfectly correct in worrying about a small detail like that.

The gravitational force between two point particles, or more generally, two spherically symmetric objects, has a strength of G*M1*M2/r^2 where r is the distance between their centers. The farther the two objects are from each other, the weaker their gravitaitonal force gets.

If your balance was so big that the end of one arm was noticeably farther away from the center of the earth than the end of the other arm, you would expect a difference in the gravitational forces on identical masses. Most balances are not this big, but a similar argument explains why there are tides on the earth. The earth is big enough that the gravitational pull of the moon (and the sun) is different enough from one side to the other that water sloshes around in the tides (we should have a more complete explanation elsewhere on this site).

The book probably said that you can neglect the spatial variation of the gravitational field strength if the radius of the earth is *much* longer than the arms of the balance.


(published on 10/22/2007)

Follow-Up #1: gravity inside sphere

why the gravitational field strength at any point inside a uniform sphere is proportional to the distance from the centre of the sphere
- pepe (age 18)
If you assume that the density of the matter, rho,  is a constant independent of radius then the total mass enclosed at a radius, R,  is  4/3 * pi * rho * R^3.   The gravitational force, F,  is proportional to the total enclosed mass divided by R^2.   Net result:  F is proportional to R.
The fact that one can ignore the mass outside the radius R is true but is a bit tricky to prove.


(published on 10/22/2007)

Follow-up on this answer.