During High school, we learned (or you will learn) that
Ep = mgh
where
"Ep" is the Potential Energy (or Work that has to be done in order to reach that height),
"m" is the mass of the object,
"g" is the gravitational acceleration and
"h" is the distance (or the height position).
But this is only true when the distance is small enough such that "g" will be treated as a constant.
During high school, we also know that
g = GM/(r ²)
where
"G" is the Gravitational Constant,
"M" is the mass of the other object and
"r" is the distance from it.
The further you get, the lower the value it has. As you can see the value of "g" changes as "r" changes.
If the distance is extremely long, is Ep really = mgh ?
"g" will change as "h" is changing during the process.
In this case, we will have to use integral to solve this problem.
Since we know that if "r" (or the distance) is small enough, "g" can be treated as a constant.
And we know that Ep = mgh, so
dEp = m g dr = (m)(GM/(r ²))(dr)
where
"dEp" is a small "piece" of the whole Work that has to be done or the Potential Energy it has,
"dr" is a small "piece" or a small change of the whole distance.
now let's start integrating:
dEp = (m)(GM/(r²))(dr)
∫dEp = ∫(m)(GM/(r²))(dr)
∫dEp = GMm∫(1/(r²))(dr)
say the starting position is "a" away from the massive object and the ending point is "b" away from the massive object, then we can get:
∫dEp = GMm[-1/r]
Ep = GMm[(-1/b)-(-1/a)]
Ep = GMm[(1/a)-(1/b)]
Ep = GMm[(b-a)/ab]
Conclusion:
If the distance is short enough, we can use Ep = mgh, but if the distance is long enough, we will have to use Ep = GMm[(b-a)/(ab)] instead.
EXTRA:
Just looking at
g = GM/(r ²)
we know that the closer we get, the larger the value of "g". If we place an Apple so near to the core of the Earth, it should has a "g" approaching to an infinite amount.
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