Potential energy of earth and gravity

AI Thread Summary
The discussion revolves around calculating the potential energy of a mass m at a height R above Earth's surface. The initial calculation yielded mgR/4, while the teacher's answer was mgR/2, leading to confusion. The key issue identified is the assumption of where potential energy is zero; the teacher uses infinity as the reference point, while the initial calculation assumes zero potential energy at Earth's surface. The correct formula for gravitational potential energy at that height should account for the variable gravity, leading to the realization of the mistake in applying mgh directly. Ultimately, both answers can be correct depending on the reference point chosen for potential energy.
Bandarigoda
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Assume Earth's radius R and there is a mass m. We put that mass in height of R from the Earth's surface. I want to calculate it's potential energy.

I calculated it and got mgR/4 but my teacher said the answer is mgR/2 . Why my answer is wrong?

I just calculated the gravity in the height of R and applied it to e = mgh
 
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Hi Bandarigoda! :smile:
Bandarigoda said:
I calculated it and got mgR/4 but my teacher said the answer is mgR/2 . Why my answer is wrong?

I just calculated the gravity in the height of R and applied it to e = mgh

?? :confused:

Show us your full calculations. :smile:

(and at that distance you can't treat gravity as constant)
 
Either answer can be right or wrong, depending on where the potential energy is supposed to be zero. Unless you fix that, the discussion is meaningless.
 
Here
 

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You are assuming zero potential energy at the surface of Earth while your teacher's doing it at infinity. Both answers are correct but at these distances infinity is usually used.
Your teacher is using the formula:
$$U= \frac {Gm_1m_2}{R}$$ [edited]
Assuming U=0 at R=##\infty##
 
Last edited:
ahhh now i understand …

you've correctly found the different values of g(r) (as a function of radial distance r),

but then you've used mgh (= mg(r - R)) for potential energy,

instead of -MmG/r :wink:
 
Oh thank you very much guys. I got it now.
 
Enigman said:
Your teacher is using the formula:
$$U= \frac {Gm_1m_2}{R^2}$$
Assuming U=0 at R=##\infty##

for the record: that should be
##U= -\frac {Gm_1m_2}{R}## :wink:
 
Uncaffeinated brain fart.
:redface:
 
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