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why gravitational potential energy is negative?

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- #1

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why gravitational potential energy is negative?

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Integral

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That depends on the coordinate system.

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George Jones

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The gravitational potential energy is taken to be zero as a matter of convenience - so that that zero potential energy is at an infinite distance from the centre of a spherical object.why gravitational potential energy is negative?

This is a choice, and other choices can be made. For example, the zero for potential energy can be chosen to be at the centre of a spherical object. Then, gravitational energy is positive at all other locations. This is a valid, but unusual, choice.

In both cases, the derivative of gravitational potential with respect to distance from the centre of the object is negative, as it must be for the force of gravity to be attractive.

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T+U=C

Kinetic energy is always positive, and will increase as the body falls faster and faster towards your source of gravity. To compensate, U is going to have to be zero. If it was positive, the total energy C would not be constant.

Sort of a fudge, but neccessary if you want the conservation of energy to work. You get around it by making the gravitational force equal to

A negative U makes sense in some way, because your gravitational energy, though always negative,

The choice of negative potential energy is really the best of a bad bunch. Try graphing the equation, then graphing it's negative. Move both graphs up and down by constants to get a feel for why the canonical option really is the lesser of many evils.

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nrqed

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Actually, as George correctly pointed out, there is no direct connection between the sign of the potential energy and the direction of the force! One

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jtbell

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[tex]U(r) = G m_E m \left( \frac{1}{r_E} - \frac{1}{r} \right)[/tex]

which would give U = 0 at the earth's surface ([itex]r = r_E[/itex]) and [itex]U = +G m_E m / r_E[/itex] at [itex]r = \infty[/itex], and the same [itex]\Delta U[/itex] between two points as the standard definition.

But this definition is more complicated than the standard one, algebraically speaking.

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HallsofIvy

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It think it should be pointed out that "gravitational potential energy is negative" is only true for problems with distances large enough to require the inverse square law. Consider the problem, "A 1 kg mass is dropped from the top of a 100 m cliff. What is its speed just before it hits the ground?"

A fairly standard way to do that is to say that the potential energy is mgh= 1(9.8)(100)= 980 Joules (positive!) while the kinetic energy is 0 so the total energy is 980 Joules at the top of the building. At the bottom of the building the potential energy is 0 while the kinetic energy is (1/2)v

But we could just as easily take the "reference" point to be at the top of the building. Then, initially, we have both kinetic and potential energy to be 0: the total energy is 0 Joules. At the base of the building, the potential energy is -mgh= -980 Joules. Conservation of energy now give (1/2)mv

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reilly

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Regards,

Reilly Atkinson

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