# Why gravitational potential energy is negative?

why gravitational potential energy is negative?

Integral
Staff Emeritus
Gold Member
That depends on the coordinate system.

In simple terms, if you solve for the force it must be negative for an attractive potential to reflect this. In a repulsive potential the sign must be changed to get the correct direction of the force.

George Jones
Staff Emeritus
Gold Member
why gravitational potential energy is negative?

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.

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.

For the conservation of energy to work, the sum of the Kinetic energy "T" of a body pulled by gravity, and it's gravitational potential energy "U", must be a constant.

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 minus the gradient of the potential energy. $$F=-\nabla U$$.

A negative U makes sense in some way, because your gravitational energy, though always negative, increases as you move away from the body, i.e. upwards. You expect gravitational potential to do this, increase, as you move up, so that you'll gain energy as you fall. If U was always positive, but decreasing, it would mean that potential energy would decrease as you moved away from the body. Also, if you tried to make it always positive but increasing, though the potential energy would approach a maximium, your effort would be confounded as you moved close to the body, i.e. as x->0.

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.

nrqed
Homework Helper
Gold Member
In simple terms, if you solve for the force it must be negative for an attractive potential to reflect this. In a repulsive potential the sign must be changed to get the correct direction of the force.

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 could choose the ground such that potential energy would be positive everywhere. It's the change of potential energy that dictates the direction of the force. It's all about the choice of zero of potential (one could even make a choice of zero such that the potential energy is positive for some points and negative for some other points (and zero on some surface). Only a change of potential energy matters, physically. I think it's important to make that clear to the OP!!!

• donaldparida and Janiceleong26
potential energy is sorta a relative quantity. i think that, for objects that attract each other, it's just convenient that they defined the potential energy when the objects are spaced apart by an infinite distance, they assigned that potential energy as 0. then for closer distances, the potential energy is less than it would be at infinity.

jtbell
Mentor
Right, one could in principle define the potential energy of an object in the Earth's gravitational field as

$$U(r) = G m_E m \left( \frac{1}{r_E} - \frac{1}{r} \right)$$

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

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

HallsofIvy
Homework Helper
The important point to understand is that potential energy is always relative to some point. What the potential energy is at a point really doesn't matter- it is the change between to points that is important. We can always choose a reference point at which the potential energy is 0. It happens that to be simplest, for gravitational problems, to take potential energy to be 0 "at infinity". Since potential energy increases as you move away from the center of a gravitating system, the potential energy is negative at any finite distance.

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)v2 and so the total energy is (1/2)v2. Neglecting air resistance, by "conservation of energy" we have (1/2)mv2= 980 so v2= 1960 and v= 44.2 m/s. We have, implicitely, taken the "reference" point, for 0 potential energy, to be at the bottom of the building.

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)mv2- 980= 0 giving exactly the same answer.

• 