Question

derivation of gravitational potential energy with diagram​

Answer 1

Explanation:

Expression for Gravitational Potential Energy at Height (h) – Derive ΔU = mgh. When, h<<R, then, R + h = R and g = GM/R2.

Answer 2

Short bit of calculus.

Ug = − ⌠

⌡ F · ds

r

Ug = − ⌠

⌡ − Gm1m2 dr

r2

∞

Ug = − Gm1m2 ⎛

⎝ 1 − 1 ⎞

⎠

r ∞

and here it is…

Ug = − Gm1m2

r

where…

Ug = gravitational potential energy

m1m2 = masses of any two objects

r = separation between their centers

G = universal gravitational constant (6.67 × 10−11 N m2/kg2)

Note that there is no Δ in this expression. Discuss here.

Discuss potential vs. potential energy somewhere.

What about the old equation?

ΔUg = mgΔh

It's hidden in the new equation.

Ug = − Gm1m2

r

Let me show you.

ΔUg = Uf − Ui

ΔUg = Ug(r + Δh) − Ug(r)

ΔUg = − Gm1m2 + Gm1m2

r + Δh r

Combine terms over a common denominator.

ΔUg =

Gm1m2((r + Δh) − r)

r(r + Δh)

ΔUg =

Gm1m2Δh

r(r + Δh)

Multiply by "one".

ΔUg = Gm1m2Δh r

r(r + Δh) r

Swap terms in the denominators.

ΔUg = Gm1m2Δh r

r2 r + Δh

Factor some stuff out of the numerator.

ΔUg = m2h Gm1 r

r2 r + Δh

Do you see it? If r is the radius of the Earth, m1 is the mass of the Earth, and m2 is the mass of something being lifted, then…

g = Gm1

r2

is the acceleration due to gravity on the Earth's surface. Making this substitution (and dropping the subscript, since we only have one mass left), we get…

ΔUg = mgΔh r

r + Δh

The first part of this expression is our old friend, the original equation for gravitational potential energy. The second term is a correction factor. For ordinary heights, this term is essentially one. Let's confirm this using a really high height — the top of the spire on the Burj Khalifa in the United Arab Emirates (818 m).

r = 6,371,000 m

r + Δh 6,371,000 m + 818 m

r = 0.999872

r + Δh

The engineers who designed the Burj would have an error in the fourth decimal place of their calculations. This deviation is probably smaller than the uncertainty in the mass of the girders used to construct this building, which is why ΔUg = mgΔh is totally acceptable for most down-to-earth applications

Now let's try something astronomical. Can ΔUg = mgΔh be used to measure the gravitational potential energy of the moon? The Earth-moon distance (384,400,000 m) is measured from the center of the Earth, not it's surface. In this case, r + Δh will actually be a difference in two numbers.

r = 6,371000 m

r + Δh 384,400,000 m − 6,371,000 m

r = 0.016853

r + Δh

This number is obviously closer to zero than to one, which is why…

Ug = − Gm1m2

r

…is used for astronomical applications.

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