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.
[Physics] Why is potential energy negative when orbiting in a gravitational field
conventionsenergynewtonian-gravitynewtonian-mechanicspotential energy
Related Solutions
About negative energies: they set no problem:
On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.
However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:
let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:
$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$
as expected: we lose $PE$ and win $KE$.
Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.
Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $ r = r_{earth}$.
The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).
By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:
$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.
As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$
Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:
$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.
Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.
So, from what I understood, your logic is totally correct, apart from two key points:
energy is defined apart of a constant value.
in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.
"the gravitational potential energy should DECREASE as it is converted into kinetic energy, but I guess I'm not really sure why it's negative at position 3."
To try and explain this I'll use an analogy. Imagine for a second you have a cat, and you throw the cat up into the air. If you choose to use the ground as a reference point, then the cat starts off with some vertical displacement, and will arc its way downwards until it hits zero, when it hits the ground. Instead, lets try taking the moment the cat leaves our hands as a reference point. Instead, the cat's distance away from you will increase for a while, come to a standstill, drop PAST your hands, and become negative. In a sense, that's the theory behind the statement you've made. Gravitational potential energy really doesn't have any meaning unless you state from which reference point your'e taking the measurement from, just as the cat's height has no meaning unless you say whether you are measuring from your hands or from your feet.
"but why is −mgxsinθ negative and why isn't kx2 negative?"
Because xsinθ is a distance, it will be positive. But that's not what we really want here. Since the block is BELOW the reference point we've taken (point 2), its distance is actually negative. Take the cat example I used, if we take our hands as the reference point, and the cat is below them, it follows that the distance in the upward direction is negative. kx^2 isn't negative, because the distance in this case is assumed to be inwards toward the spring, as it represents the elastic potential energy. Let's take an analogy. Say the cat that we have thrown arcs, and instead of landing on the floor lands on a trampoline. Logically, the more that the trampoline bends inward, the greater force the cat will feel, and the greater potential the cat will have to fly off into space. This means that the cat's Elastic potential energy will have some correlation to the distance travelled into the trampoline. It's the same principle for the spring, the greater value of x, the more you travel into the spring, the more potential energy you have.
"And finally, how come one side is equal to the other if the right side involves a completely different aspect to it (the spring) that isn't found in the initial conditions. "
Oh, but it IS found in the initial conditions. Here's what the full equation would look like.
Gravitational Potential energy + Kinetic energy + Elastic potential energy (reference point 1)
= Gravitational Potential energy + Kinetic energy + Elastic potential energy(Reference point 3)
But then where did it go on the left hand side? Simple, the block was just not touching the spring yet, so the value of the Elastic potential energy at point one was ZERO, (similar to how the value of the kinetic energy at point 3 is zero).
Hope I answered your question, if something needs clarifying reply, it's my first answer here so it might not be very clear.
Best Answer
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.