Expert: Jayendra Upadhye - 6/4/2008

Question
QUESTION: Jayendra:

If a person moved from the center of the earth outward, by what factor will that person's weight change?

ANSWER: Hi,
Gloria!
Never ask imperfect questions!
Because when you ask result will change by "what factor" you are not saying "input condition" will change by what factor!

It is a rather loose question, requiring a loose answer!

F = (G*M1*M2)/(R*R)
where
G = 6.67 * 10 raised to -11 or Universasl gravitation constant.

If to quantify properly on unit basis, we convert M2 = 1 (unit mass),
& M2 to just M, a gravitating mass much larger than unity,
we get
F/M2 = F/1 = F = G*M/(R*R)

As F/M1 = g gravitational acceleration on M1 due to M2.

g = G*M/(R*R).

Let earth mass be M.

Let g, be the gravity by earth at R units form its center. (not hieght above surface mind you).

using differential calculus

dg/dR = rate of change of g with respect to R
     = - (2*G*M)/(R)
[BECAUSE d(x squared)/dx = 2x & d(x^n)/dx = n*d(x^n-1)]

Likewise weight is nothing but product of g * M1 !!

so for unit mass F = 1*g = g = G*M/(R*R) [same value as g numerically]
therefore dF/dR = - (2*G*M)/(R*R*R) but in force units Newton or Kg-met.

So in inexact terms, weight will reduce inversely with THE CUBE of increasing radius (hyperbolically) with the constant of proportionality being -2*G*M = 2*6.67*10^-11 * 5.9736*10^24*(1/R*R*R) Newton/met change at R units distance from its mass center.
[as long as you are "outside the earth". Inside, by another set of equations & thanks to Newton, the value of g falls off linearly with radius to 0]!!

dF/dR = -(79.687*10^13)*(1/R*R*R)  .... eqn (1)

This is very interesting.
As we reside near the surface which is 6371000 met away from center,
this rate at earth's surface is
dF/dR = -(79.687*10000000000000)*[1/(6371000*6371000*6371000)]=
     = -(79.687*10000)*[1/(6371*6371*6371)]=
     = -3.08151*10^-6 Kg-met/met or Kg for meter travelled
     = -3.08151 milli-gms per Kg meter travelled
So 1 kg will weigh 3 milli-grams less with every meter travelled, in the vicinity of the earth's surface.

This drop will approximate the hyperbolic drop by a linear drop in a
narrow range of a few 100s of meters before appreciable diviations from hyperbolic true valus begin to appear.

ref:- http://en.wikipedia.org/wiki/Earth

Hope that sufices
Please do arte the answer if you find it interesting.

Jayen

---------- FOLLOW-UP ----------

QUESTION: Jayen:

Suppose that you move three times as far from the center of the Earth as you are now. By what factor will your weight change?

Answer
Hi,
Come on Gloria!
Unless you want poor old me to slog, you wouldnt be asking such a simple question! :)

It is all about ratios.
weight at surface = W1 = K (1/Re*Re)
weight 3 earth radii away = W2 = K/[(3*Re)*(3*Re)]
Where Re = earth's mean radius
and K = [G*Me*(My Mass)]
As you take ratio K will disappear
WI/W2 = (Re*Re) / [9*(Re*Re)]
Again Re will disappear and straightforward answer that is independent of your weight, earth's radius, and G appears!!
W1/W2 = 1/9
or W2 = (1/9)*W1
or change = W1 *[1 - (1/9)] = W1*(8/9)

whichever way you want to interpret it.
if you use calculus,
W2 = Definite integeral of [-2*K/(r cubed) dr] over range Re to 3 * Re
  = -[K/[9*(Re squaed)] - K/(Re squared)]
  =  K/(Re squared) - [K/[9*(Re squaed)]
  = W1(8/9)
Same answer if you integerate the "point rate of change" from earlier answer over given range!

Hope that helped.

Please do rate the asnwer if you found it helpful.

Regards
Jayen