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## Main Question or Discussion Point

Using taylor series expansion to prove gravitational potential energy equation, GMm/r=mgh at distances close to the earth.

R= radius of the earth h= height above surface of the earth m= mass of object M= Mass of the earth

U = - GmM/(R + h)

= - GmM/R(1+ h/R)

= - (GmM/R)(1+ h/R)^-1

do a binomial expansion or Taylor's series with (h/R) as the small variable;

= -(GmM/R)[1 - (h/R) + ...].

DOUBT: In this step why do we only use the FIRST order taylor expansion and no more?

The function would not be complete would it?and nothing has been done to bound the error of the polynomial. Why would taking the first non-trivial term (first order term) of the expansion of 1+x (that is 1-x) account for the entire function 1+x or in this case 1+h/r.

PLEASE REPLY ...

Proof (continued): -(GMm/R)(1-(h/R)= -(GmM/R) + GmMh/R^2

In the second term note that GM/R^2 =g

= - (GmM/R) + mgh

So the potential energy at "r" can be written as PE at "R" plus PE at "h"

Now in planetary problems, where the distance h is not small and you use the general formula - GmM/r it is convienient to choose U = 0 at r = infinity

However near the earth we can arbitrarily chose U=0 at U = - (Gmm/r)

Therefore: U = mgh when h<<<R

R= radius of the earth h= height above surface of the earth m= mass of object M= Mass of the earth

U = - GmM/(R + h)

= - GmM/R(1+ h/R)

= - (GmM/R)(1+ h/R)^-1

do a binomial expansion or Taylor's series with (h/R) as the small variable;

= -(GmM/R)[1 - (h/R) + ...].

DOUBT: In this step why do we only use the FIRST order taylor expansion and no more?

The function would not be complete would it?and nothing has been done to bound the error of the polynomial. Why would taking the first non-trivial term (first order term) of the expansion of 1+x (that is 1-x) account for the entire function 1+x or in this case 1+h/r.

PLEASE REPLY ...

Proof (continued): -(GMm/R)(1-(h/R)= -(GmM/R) + GmMh/R^2

In the second term note that GM/R^2 =g

= - (GmM/R) + mgh

So the potential energy at "r" can be written as PE at "R" plus PE at "h"

Now in planetary problems, where the distance h is not small and you use the general formula - GmM/r it is convienient to choose U = 0 at r = infinity

However near the earth we can arbitrarily chose U=0 at U = - (Gmm/r)

Therefore: U = mgh when h<<<R