Question 1090222
The assumption here is that the Earth is a point mass.
All of the mass of the Earth is concentrated at the point where the radius is equal to zero (or a small 
value compared to the radius of the Earth and the values where the measurements are taken).
Weight is inversely proportional to the distance from the center of the earth.
{{{W=k/D^2}}}
{{{k=WD^2}}}
{{{k=200*(4000)^2}}}
So,
{{{W=(200*(4000)^2)/D^2}}}
{{{W/200=(4000/D)^2}}}
Then for example, when {{{D=500}}}
{{{W/200=(4000/500)^2}}}
{{{W/200=(8)^2}}}
{{{W=64*200}}}
{{{W=12800}}}{{{lbs}}}

.
.
.
Uniform density solution:
So the gravitational force is a function of the two masses involved and the distance between them.
{{{F=GMm/R^2}}}
If you allow the mass of the Earth to change as a function of its radius (R) (assuming uniform density)
{{{M=rho*(4/3)pi*R^3=kR^3}}}
Substituting,
{{{W=a*R^3*m/R^2}}}
{{{W=a*mR}}}
Where "a" is a constant (from all of the constants put together).
So then your mass doesn't change so it's also a constant.
{{{W=bR}}}
You can then use the value at the earth to calculate the value of b.
{{{200=b(4000)}}}
{{{b=1/20}}}
So then,
{{{W=R/20}}} valid for {{{0<R<4000}}}
Again that's a simple solution using uniform density model.