# Why is potential energy negative when orbiting in a gravitational field?

I had to do a problem, and part of it was to find the mechanical energy of satellite orbiting around mars, and I had all of the information I needed. I thought the total mechanical energy would be the kinetic energy+ the potential energy, or $KE+PE$. However, I had the answer sheet and it said that I had to do $KE-PE$, because when you integrate $Gm_1m_2/r^2$ you get a negative sign. I can see why it works out mathematically, but I don't understand why you are actually losing energy when orbiting in a gravitational field.

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First things first: the total mechanical energy is always kinetic energy plus potential energy. So if your answer sheet actually said $KE - PE$, it's wrong. But what I suspect it really said is that the potential energy is negative, so the formula you wind up with is

$$\underbrace{\frac{1}{2}mv^2}_{KE} \underbrace{- \frac{Gm_1 m_2}{r}}_{PE}$$

Now, the negative sign doesn't mean that you're losing energy. It just means that the amount of energy happens to be less than zero.

Consider this: the formula that works on the Earth's surface, $PE = mgh$, makes sense, right? It seems intuitive that potential energy should get larger as you go higher above Earth, because you have to put energy into something to raise it up, and when it's higher it has more potential to do work by falling. That same principle should hold for the general $1/r$-type formula: the potential energy should get larger the higher you go. But at larger values of $r$, the reciprocal of $r$ gets smaller, which is the wrong trend. The easy fix is to make it negative. And the math works out to support that.

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Well ok let's say the satellite is not orbiting; it is just falling down towards Mars, and it has an initial kinetic energy if A and an initial potential energy of B. Wouldn't you agree that by the time the satellite reaches the (lets says there's a hole) center of Mars, its kinetic energy would be equal to A+B, where A and B are positive? If you say that B is negative, then you have a final KE of A-B, which doesn't make sense since it is less than when you started. By conservation of energy, the initial energy also had to be A+B, where A and B are positive. Why does this change here then? –  Ovi May 11 '13 at 20:52
If the satellite has an initial potential energy of B, then B is negative, because gravitational potential energy is negative. So no, I do not agree. Your condition that both A and B are positive does not hold. On another note, the total energy is always numerically equal to A+B (not A-B). If the satellite falls inward, the potential energy becomes more negative, i.e. less than B, so the kinetic energy must become larger than A to compensate. (I'm not sure I really understand your confusion yet.) –  David Z May 11 '13 at 21:08
Ok let's go by what you said and say the KE is 50J, and the PE is -50J. Mechanical energy=KE+PE=50J+(-50J)=0J, which obviously wrong. I think the object would have to have 100J of total mechanical energy. –  Ovi May 12 '13 at 4:21
Why is that obviously wrong? Because there's nothing wrong with it to me. It's perfectly reasonable for an object to have 0J of total mechanical energy. Objects in hyperbolic orbits, or those which are launched at escape velocity, do have zero total mechanical energy. –  David Z May 12 '13 at 4:30
That situation isn't directly comparable, though, because when you're using $mgh$, the zero point of potential energy is not the same as when you're using $-Gm_1m_2/r$. In the former case, the zero point is the floor, or wherever you measure height from, but in the latter case, the zero point is infinitely far away. Your argument does validly show why an object can't have zero mechanical energy when you set the reference (zero) point for PE on the floor, which the object can never go below, but it doesn't apply when the zero point is way out in space. –  David Z May 12 '13 at 4:43