# Change In Gravitational Energy Percent Error

1. Apr 3, 2013

### Plutonium88

1. The problem statement, all variables and given/known data
There are two equations from which change in gravitaton potential energy of a system can be calculated.

ΔEg = mgh and the other Eg = GmM/R

The first equation is only correct if the gravitational force is constant over a change in height h. The second is always correct. Consider the change in Eg of the mass m when it is moved away from the earth's surface to a height h using both equations, and find the value of h for which the equation ΔEg=mgh is in error by 1%. Express this value of h as a fraction of the earths radius and obtain the numerical value.

m = mass of object M = mass of earth Re = 6.38x10^6meters

2. Relevant equations
ΔEg1= -GmM/(Re+h) + GmM/Re
ΔEg1= GmMh/(Re*(Re+h))
ΔEg2= mgh

3. The attempt at a solution

So
Fg = GmM/(Re+h)^2
mg = GmM/(Re+h)^2
g = GM/(Re+h)^2

ΔEg2 = mGMH/(Re+h)^2

So i have to find the the relationship between these equations

so i say

I)ΔEg1/ΔEg2 = 100/101

II)ΔEg1/ΔEg2 = 100/99

where 100 represents the one which does not have the fraction of error percent and 101 and 99 represent the one with the fraction of error percent

I)
When i solve I my answer is

h = -Re/101
h= -63168.32
which yields a negative number so this is not the answer.

II) When i solve for this
h = Re/99
h= 64444.44 m

When i double check these answers with my teacher she doesn't seem to have the same numbers so i'm just curious what i'm doing wrong?

I know there are 2 other cases i didn't check

III)ΔEg1/ΔEg2 = 99/100

IV)ΔEg1/ΔEg2 = 101/100

where III) yields h = Re/100
h= 63800

IV) h = -Re/100
h = -63800

But i know III) and IV) are not the case because that would mean that Eg = -GmM/Re is the equation off by a fraction of a percent.

Anyways any help is appreciated.

2. Apr 3, 2013

### Staff: Mentor

Your notation is confusing (g depends on h and you have to different formulas for ΔEg2, for example), but the answer looks reasonable. If you fix g to the gravitational acceleration at zero height, the value could be a bit different, that might be the source of the deviation to the teacher's answer.

3. Apr 3, 2013

### Plutonium88

Ah i took g=GM/(Re + H)^2 where its g=GM/Re^2 is this my error?

and sorry about my confusing notation can you tell me what is making it confusing and how to make it better?

4. Apr 3, 2013

### Staff: Mentor

g=GM/Re^2, right.

Introduce things like those energies in words, together with the formulas.

5. Apr 3, 2013

### Plutonium88

Ah I Had a friend show me how to use microsoft word so in the future i will be able to create more visually appealing equations. I would really appreciate so more input because if I consider g=GM/Re^2. Then I get h = +/- 63800 or 1/100Re

My teacher tells me she has decimals in the number for her answer and this number does not

Last edited: Apr 3, 2013
6. Apr 3, 2013

### Staff: Mentor

How are decimals relevant?
63.8km. There are decimals.
63800.000m. Great, more insignificant digits.

A agree with your result of 1/100 Re. Quite interesting, the relative error is just the same as the fraction of the radius of earth.

7. Apr 3, 2013

### Plutonium88

What I am taking from her is that her answer is not an exact whole number that is what I mean by decimals

8. Apr 3, 2013

### haruspex

If you take the radius of the Earth as being a terminating decimal, in whatever units, then 1% of that will also be a terminating decimal. So whether it's a whole number will simply depend on the choice of units. Given 6380 km as the radius, h is 63.8km or 63800 m.

9. Apr 4, 2013

### Plutonium88

Okay I have cleared up my issue with the teacher. Thank you very much for your help. If you are are interested we just had a difference in which g's we had used. She had used 9.8 where this solution replaced g with GM/R^2 which ends up yielding a greater altitude due to the fact GM/R^2<g

Last edited: Apr 4, 2013
10. Apr 4, 2013

### haruspex

That's a strange explanation. The algebra produces the result that an error of x% will arise at a height of x% of Re. That does not depend on what values you use for constants. A different numerical answer should only occur if a different value is taken for Re. Did the teacher compute Re from g, G and Me?

11. Apr 11, 2013

### Plutonium88

I'm sorry for the tardy reply, perhaps the way i explained what i said doesn't make sense

So the reason her number is different than mine is because she didn't replace g with GMe/Re^2

so (mgh)/(GmMh/Re(Re+H)) = 101 / 100

So h = (101*G*M)/(100*g*Re) - Re = 63200.37

So because she chose to keep gravity as g=9.8 in her answer it affected her numbers. She said that subbing in g= GM/R^2 is a better way to do the problem.