# 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.

• Commented Oct 28, 2019 at 7:06

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.

• 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
Commented May 11, 2013 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.) Commented May 11, 2013 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
Commented May 12, 2013 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. Commented May 12, 2013 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. Commented May 12, 2013 at 4:43

Here is a simple model as explanation:

Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing.

When you want to get rid of gravitation, returning where you came from, you have to restitute this energy, for leaving the gravitational field. Then your potential energy account is at zero again.

• It sort of makes sense, but the idea of "negative" potential energy still makes me queasy. You should either have stored potential energy (in which case it would be positive), or you have nothing at all (zero). To be converting 0 potential energy into kinetic energy means you would have to be borrowing energy from something that is already zero. Commented Jun 28, 2015 at 22:01

It is simply a matter of definition. It is defined in a way such that in infinite distance the potential energy is 0, therefore as you get closer, the potential energy is expressed in a form of kinetic energy and the amount of potential energy "available" decreases. Just definition.

In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa.

Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth doesn't exist at all. Here, your $KE + PE =0$. Now, what so erratic with $0$ ? He is representing quantity(ok! Nothing;) . Suppose, you and earth constitute an isolated system. Now, when you come closer, gravitational force starts to rise and you start to accelerate and your kinetic energy starts increasing. But since you are moving downwards, your velocity must be negative. When you are $R$ units away from the earth, your total energy( since the earth is massive, most of the potential energy of the system is yours.) can be given by $$KE + PE = \dfrac{1}{2} m(-v)^2 + U$$ . But since the system is isolated and the force is conservative, mechanical energy must be conserved at any moment which is equal to $0$ ( don't get bothered by zero, it's just a number!) . So, $$U = - \dfrac{1}{2}mv^2$$ . Thus it is negative. Now remember while sign means direction for the vectors, for scalar quantities, it means the quantity is decreasing . So, potential energy is decreasing as you're approaching the earth towards the gravitational force. Now, if one decreases from $0$, what does the number become?? Negative! So, potential energy is decreasing which is indicated as above. As you approach closer and closer, the Potential Energy becomes more & more negative as it's decreasing. As you go farther & farther, the PE becomes less & less negative ultimately to zero. And just dump that book which is telling $KE - PE$;it's a blunder! As said by other answerer, it is always $$KE + PE \implies \dfrac{mv^2}{2} + (-\dfrac{GMm}{R})$$ . It's just playing with numbers!

• It sort of makes sense, but the idea of "negative" potential energy still makes me queasy. You should either have stored potential energy (in which case it would be positive), or you have nothing at all (zero). To be converting 0 potential energy into kinetic energy means you would have to be borrowing energy from something that is already zero. Commented Jul 14, 2015 at 9:08
• @1110101001: Try to acclimatize the fact; I tried to write it intuitively but I am really sorry if it didn't help you:\
– user36790
Commented Jul 14, 2015 at 9:13

As @David Z explained, total energy E=kinetic energy T+potential energy V. The expressions for KE and PE are then substituted with their signs in this equation.

Theoretically

All of mechanics has lagrangian as its starting point which deal with potentials. Consider any potential energy V of the form (potential is just V per unit charge) $$V=\frac{k}{r}+C$$ where k and C are constants in r
The associated force field generated by this potential is \begin{align} \mathbf{E}&=-\nabla{V}\\ &=-\nabla{(\frac{k}{r}+C)}\\ &=\frac{k}{r^2}\hat{r} \end{align}

Now for sources which attract, the force must point towards the source($$\hat{r}$$ points away) so k must be negative.This is why all attractive potential expressions have -ve signs.

But wait a second..does this mean that attractive PEs are always negative? No,because you can arbitrarily choose a C i.e the boundary condition for the potential--it need not always be $$0$$ at infinity. For eg. let

$$V=\frac{-|k|}{r}+\frac{|k|}{r_{0}}$$

An object in this attractive potential has positive energy beyond $$r_{0}$$.

So the negative sign of the potential is to supplement the fact that the force is attractive. Its negative everywhere because of the "$$0$$ at $$\infty$$" condition. (If |k|=$$G M_{earth}m_{test}$$ and $$r_{0}=R_{earth}$$, then this V is approximately $$m_{test}g(r-R_{earth})$$ which as you know is positive above earth's surface.

Qualitatively

As an object falls from rest at infinity to the point r in the field it gains the energy stored in the field in the form of its increased kinetic energy. The field on the other hand has lost this much energy to the mass by working to accelerate it from infinity to r. Since the energy at $$\infty$$ was $$0$$,the losses appears as -ve field potential energy at r.

as we know the energy of an elctron increases as it moves away from the nucleus..that means potential energy is directly depending on distance between electron and nucleus..but when we derive an mathematical expression we get energy inversly proportrional to the radius...to compensate diz we add MINUS sign to show that lesser negative enrgy means more energy

• Is the electron bound to the atom gravitationally? If not, how does this address the question at hand? Commented Apr 20, 2015 at 2:32