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Change In Gravitational Energy Percent Error

  • #1
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Homework Statement


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

Homework Equations


ΔEg1= -GmM/(Re+h) + GmM/Re
ΔEg1= GmMh/(Re*(Re+h))
ΔEg2= mgh


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.
 

Answers and Replies

  • #2
33,797
9,509
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
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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.
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
33,797
9,509
g=GM/Re^2, right.

and sorry about my confusing notation can you tell me what is making it confusing and how to make it better?
Introduce things like those energies in words, together with the formulas.
 
  • #5
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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:
  • #6
33,797
9,509
My teacher tells me she has decimals in the number for her answer and this number does not
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
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What I am taking from her is that her answer is not an exact whole number that is what I mean by decimals
 
  • #8
haruspex
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What I am taking from her is that her answer is not an exact whole number that is what I mean by decimals
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
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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
 
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  • #10
haruspex
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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
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
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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?
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.
 

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