how does U = mgh when h is smaller then r?

Why does the potential energy; U, equal mgh when h is smaller then r?

use equation

F = GmM/r^2 which then leads to,

F = -GmM/r

.... Why do we say that U = mgh when h<<r?

please explain with math!!

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  • 8 years ago
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    g = GM/r^2 is the gravity field, as in PE = mgh, where R - r >= h is the height above r, the radius of the source mass M (e.g., Earth). R is the radius struck from the center of the source mass to the location of the target mass having the potential energy.

    Work done on the mass, m, in lifting, is found from dPE = m d(gh) = m (dg h + g dh) = m (2GM/r*h + GM/r^2 * dh) = m [2GM(R - r)/r + GMdr/r^2)] = m [2GM(R/r - 1)] as dr = 0 since r is the radius of the source mass and it's assumed fixed.

    And there you are. When R < r, so that R/r < 1.00 and (R/r - 1) < 0, the change in potential energy goes negative. That is, negative work is done; work done by the target mass as it falls below the surface of the source mass.

    So as h goes inside r, the PE decreases in value (i.e., work done by the mass). Note, the starting point, mgh, has not changed, but the direction of the PE change has as h < r gets farther away from r, the source mass radius.

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  • 8 years ago

    wher you define the the potential energy to be zero is largely arbitrary. we use U = mgh when we are talking about objects near the surface of the earth. All of the mass of the earth is below the object, In fact you can show that g acounts for the radius of teh earth as follows:

    F = mg = GMm/(R+h)^2 where M = mass of earth, R = radius of earth and h << R

    g = GM/(R+h)^2 = GM/R^2 1/(1+(h/R))^2 ~ GM/R^2 *(1 - 2(h/R)^2 + O((h/R)^4) ~ GM/R^2

    g = GM/R^2

    So the potential energy of amass at a height h above the earth's surface is:

    U = mgh = GMmh/R^2 which assumes the zero of potential energy is at h = 0, i.e. the earth's surface.

    When we do porblems with satellites, we usually assume the zero of potential energy t be at r = infinity so that:

    U = GMm/r increases as you get close to the earth's center. THis really is true only as long as you are above the earth's surface. So let r = R + h then

    U = GMm/r = GMm/(R+h) and assume h << R

    U = GMm/R 1/(1+h/R) ~ GMm/R (1 - h/R) = GMm/R - GMmh/R^2 = GMm/R - mgh

    So what this did is add the potential required to move the object from the surface of the earth to r = infinity (first term) and subtract off the potential relative tot he surface of the earth assuming the surface has zero potential energy. You'll get a different value for U but when you do conservation of mechanical energy problems, you'll end up with the same speeds, etc since that cannot change.

    So it all has to do with where you pick U(r) = 0 to be.

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  • 8 years ago

    2 + 2 = 5!

    Source(s): RADIOHEAD! :P FEAR THE Derivative...MY SOURCE IS MY BRAIN....
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  • 3 years ago

    This undertaking is consistent with regulation of conservation of skill Loss in ability skill (PE) = income in Kinetic skill (KE) Loss in PE = mgh - a million/4*mgh = 3/4mgh = KE (preliminary KE =0 for the rationalization that V=0 intially)

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