Taylor series expansion for gravitational potential energy. GMm/r=mgh near the earth

  • #1

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
 

Answers and Replies

  • #2
Doc Al
Mentor
44,865
1,112


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.
Succeeding terms would be higher order powers of h/R and thus increasing insignificant when h/R << 1.
 
  • #3


Wouldn't that mean that h/r would be off by all the succeeding terms all the way to infinity?
 
  • #4
1,006
104


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.
If you're worried, you can write down the second order term, and then figure out what condition needs to hold in order for the second order term to be at most, say, 1% of the first order term.
 
  • #5


Okay... So your trying to imply that when h/r is way less than 1, higher order powers of h/r increase it insignificantly?
 
  • #6
Doc Al
Mentor
44,865
1,112


Wouldn't that mean that h/r would be off by all the succeeding terms all the way to infinity?
The terms get smaller and smaller, so they can be safely ignored.

Imagine if h/R = 1/1000

Then (h/R)2 = 10-6, (h/R)3 = 10-9, and so on.

Write out the next few terms of the expansion and see how small they are compared to the first.
 
  • #7


Thanks a lot! That explains it.
 
  • #8


Also that implies that U = mgh is just an incredibly good approximation right? Not exactly the potential energy.
 
  • #9
Nugatory
Mentor
12,496
4,992


Also that implies that U = mgh is just an incredibly good approximation right? Not exactly the potential energy.
Yes. It's a good approximation anytime that h (which is really the difference between two values of the distance from the center) is small compared with the distance from the center. This will be the case for most reasonable heights from the surface of the earth.
 

Related Threads for: Taylor series expansion for gravitational potential energy. GMm/r=mgh near the earth

Replies
17
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
6
Views
283
  • Last Post
Replies
3
Views
691
Replies
3
Views
872
Replies
6
Views
1K
Replies
2
Views
4K
Replies
10
Views
515
Top