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Gravitational Potential Energy - OZO

There are two equations from which a change in the gravitational potential energy U of the system of mass m and the earth can be calculated.  One is u = mgy the other is U = -(Gm_e * m)/r_e (where m_e and r_e are the mass & radius of the earth respectivly). As shown the first equation is correct only if the gravitational force is a constant over the change in height Δy. THe second is always correct. Consider the difference in U between a mass at the earth's surface and a distance h above it using both equations, and find the value of h for which u = mgy is in error by less than 1%. Express this value of h as a fraction of the earth's radius, and also obtain a numerical value for it.


I thought something like this:


| GM_e * mh            |
| -------------- - mgh  | < .01
| r_e(r_e + h)           |



I'm not sure if thats correct however.
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BrianGEFF719
Asked:
BrianGEFF719
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1 Solution
 
Harisha M GCommented:
Hi,

Let F be the first and S be the second equation. Then, the formula needed is..

Error/Actual

which is nothing but

(F-S)/S < 0.01


---
Harish
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Harisha M GCommented:
and of course, you need to have the modulus..

| (F-S)/S | < 0.01
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BrianGEFF719Author Commented:
i see what your saying :0, I will give it a shot  and write back if I have a problem
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BrianGEFF719Author Commented:
Maybe I'm not doing this right, but I end up with something like:


 | (r_e + h)gh + Gm_e |
 | ----------------------- |    < 0.1
 |         -Gm_e            |



Is this right?
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ozoCommented:
((-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e)) - mgh)
/
(-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e))
=
(((Gm_e * m) * (1/r_e - 1/r_e+h))  - m(Gm_e/(r_e^2))h)
/
((Gm_e * m) * (1/r_e - 1/r_e+h))
=
(
(Gm_e * m) * ((r_e+h)/(r_e*(r_e+h)) - (r_e+h)/(r_e*(r_e+h)))
-
m(Gm_e/(r_e^2))h
)
/
((Gm_e * m) * (1/r_e - 1/r_e+h))
=
((1/(r_e*(r_e+h))) - 1/(r_e^2))
/
(1/(r_e*(r_e+h)))
=
(1/(r_e*(r_e+h)) - 1/(r_e^2))
/
(1/(r_e*(r_e+h))
=
1 - 1/(r_e^2))/(1/(r_e*(r_e+h))
=
1 - (r_e*(r_e+h))/(r_e^2)
=
1 - (r_e+h)/r_e
=
- h/r_e
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d-glitchCommented:
If you take the derivative of the potential energy equations, you get the simpler force eqations:

    F  =  mg  =  (G Me m) / Re²      Where Me and Re are the mass and radius of the earth

          g(R)  =  (G Me) / R²          Where g is now a function of R  (G and Me are constant)



For small perturbations around Re

                                     g(Re)
           g(Re + h)  =  ---------------------
                                Re² + 2Re*h + h²


For 1% accuracy          |2Re*h|  <  Re²/100     ==>     |h|  <  Re/200  =  4000 mi/200  = 20 mi


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BrianGEFF719Author Commented:
Ozo, thanks for the response.

Please explain your first line:

((-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e)) - mgh)
/
(-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e))


How you set this up like that. So I can understand how this works.
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BrianGEFF719Author Commented:
d-glitch, thats a good idea.
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ozoCommented:
-(Gm_e * m)/(r_e)  is the potential at r_e
-(Gm_e * m)/(r_e+h)  is the potenial at r_e+h
(-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e)) is the change in potential when you raise m from r_e to r_e+h
mgh assumes the force is constant
((-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e)) - mgh)
is the absolute error
((-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e)) - mgh)
/
(-(Gm_e * m)/(r_e+h) - -(Gm_e * m)/(r_e))
is the relative error
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ozoCommented:
You must mean
                                     g(Re) * Re²
           g(Re + h)  =  ---------------------
                                Re² + 2Re*h + h²
But that only gives you F, the question asked for U
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