# How does potential energy increase, if height increases?

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1. Jan 29, 2015

### sawer

(constants're ommitted)

1-) M=50, m=5, h=5 then Potential Energy = (50*5)/5 = 50

After increasing height

2-) M=50, m=5, h=10 then Potential Energy = (50*5)/10 = 25

Field strength decreases amount of h^2 so according to formulas potential energy decreases as h increases.

But that mustn't be true. But as I showed according to Potential energy and Field strength formulas, potential energy decreases as h increases. What is wrong here?

2. Jan 29, 2015

### PeroK

In that format, the gravitational PE is negative:

$PE = -\frac{GMm}{r}$

Alternatively, for a constant gravitational field, you can have:

$PE = +mgh$

3. Jan 29, 2015

### Quantum Defect

If you write things in the form that you have: V = -G m_1*m_2/r -- there is a very important negative sign that you lost, which you need to keep. What this does is ensure that the potential energy of two bodies at infinite separation is equal to zero, and that as the bodies come together, the potential energy is lowered (more negative). You could tell that you had made a mistake, because as you increased the height, the potential energy went down.

In the other common way of looking aqt things, close to the earth's surface, you are using the simple approximation that V = m * g * h. In using this, you have done a number of things. One, you have redefined the zero of energy -- completely ok to do -- as being at the surface of the earth ==> V = 0 when h = 0. You have also changed the relationship between distance. In the first case, the answer is proportiional to 1/r, in the second case, it is proportional to h.

Can you show how both of these expressions could possibly be true? Hint: Use the first expression, and let r = Re + h, where Re is the radius of the earth. Use an approximation for 1/(1+x), that is valid for small x (x<<1)