Question about mechanical energy and orbits

[nhmllr]

I was watching something on elliptical orbits. Early on, I saw something that really bugged me and I can't continue until I figure out what it's about:
Mechanical Energy = Kinetic Energy + Potential Energy
I know that KE = 1/2*mv^2, which makes sense and I know why
I know that PE = mgh = mMG/r, which makes sense and I know why
So why did the lecturer write
ME = 1/2*mv^2 - mMG/r
?
Shouldn't it be a +?

 

[Vagn]

I was watching something on elliptical orbits. Early on, I saw something that really bugged me and I can't continue until I figure out what it's about:
Mechanical Energy = Kinetic Energy + Potential Energy
I know that KE = 1/2*mv^2, which makes sense and I know why
I know that PE = mgh = mMG/r, which makes sense and I know why
So why did the lecturer write
ME = 1/2*mv^2 - mMG/r
?
Shouldn't it be a +?

Potential energy is usually taken to be negative by convention, this is because you need to do work on an object in a gravitational field to move it against the field.

 

[nhmllr]

Potential energy is usually taken to be negative by convention, this is because you need to do work on an object in a gravitational field to move it against the field.

But then ME would not stay the same if it was a minus sign. m, M, G, and r are all ≥ 0, so it does stay negative.

 

[BruceW]

PE = -mMG/r
This is the actual equation, you seemed to have missed the minus sign. The way I remember it is that in a gravitational potential, the object is in a potential well, and the top of the well is at r=infinity, where the potential is zero.

 

[Mathusalem]

Hi !
You can actually derive the form of the potential energy going out from its definition.
Place a corpse of mass M in a polar coordinate system, for simplicity's sake, and look at the gravitationnal force F(r) acting on a mass m.

Then, W(r) = $$- \int \limits_{\infty}^{r} \vec{F(r)}\cdot\vec{dr}$$
That's the definition, and if you're careful with the scalar product, you'll see that what comes out is $$W(r) = -\frac{GMm}{r}$$

The sign difference with the usual W(h) = mgh comes from the fact that the zero of the potential is set at h = 0 in this case. In the case of the above problem, W(r) = 0 iff r tends to infinity. Inversely, at r = 0, the potential is - infinity, while it is infinity at h = infinity.

Setting the 0 of the potential at 0 or infinity implies a change of signs. (be careful, this is not a very accurate statement ;) )

 

[AlephZero]

The potential energy increases when you move further away from the central mass. That is consistent with the approximation that potential energy = mgh close to the surface of the earth.

When you increase r, the value of mMG/r decreases, so you need a minus sign.

It doesn't matter at what position you take the potential energy to be 0. For "mgh", you take it as 0 at the Earth's surface (or whatever position you choose to measure h from). For -mMG/r, you take it as 0 when r is infinite, so the numerical value is always negative.

 

[nhmllr]

Thanks guys, I understand the minus now!
What I do not understand is that the kinetic energy is always positive. (The v^2 takes care of that) If the object's velocity is going up OR down, it is positive. So if I throw an object in the air, as it goes up the total mechanical energy would be the vector of KE + vector of PE
But when the object comes down, the KE vector should be pointing down, but the PE vector is still pointing down, so I do not see how the mechanical energy is conserved.

 

[BruceW]

Kinetic and potential energy are not vectors. They are simply values. Just add them together (remembering the minus sign) and you'll get the right answer.
Think of it this way: when you throw the object up, its KE is positive and PE is negative.
Then when the object comes back down, its KE is positive and PE is negative. So when it gets back to the same height, it has the same KE and PE, but it is moving in the opposite direction to when you threw it.

 

[nhmllr]

Kinetic and potential energy are not vectors. They are simply values. Just add them together (remembering the minus sign) and you'll get the right answer.
Think of it this way: when you throw the object up, its KE is positive and PE is negative.
Then when the object comes back down, its KE is positive and PE is negative. So when it gets back to the same height, it has the same KE and PE, but it is moving in the opposite direction to when you threw it.

They're not vector!? But aren't they exerting energy in a direction? Something speeding to right can make other things speeding to the right go faster, and things going to the left go slower, and when it speeds to the left, it slows down things going to right, and speeds things going to left. It seems as though the direction matters!

 

[Redbelly98]

No, energy does not have a direction associated with it and is not a vector. Force is exerted in a direction, but not energy.

If something has a kinetic energy of 100 Joules, that is it's kinetic energy whether it is moving up, down, sideways, or in any other direction.

 

[BruceW]

No, neither kinetic nor potential energy are vectors.
For example, if the velocity of an object was given as:
$$\underline{v} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$$
(where the i,j,k represent the right, forwards and up directions).
Then the Kinetic energy would be:
$$\frac{1}{2} m ( 3^2 + 6^2 + 2^2 )$$
which is not a vector

Did my latex work?

 

[nhmllr]

No, neither kinetic nor potential energy are vectors.
For example, if the velocity of an object was given as:
$$\underline{v} = 3 \hat{i} + 6 \hat{j} + 2 \hat{k}$$
(where the i,j,k represent the right, forwards and up directions).
Then the Kinetic energy would be:
$$\frac{1}{2} m ( 3^2 + 6^2 + 2^2 )$$
which is not a vector

Did my latex work?

So you're saying that the units come off when you square, making it a scalar?
But why do the units come off anyway? I don't really see why it's a scalar, wouldn't it change if you rotate the system? How does v^2 change?

 

[BruceW]

The i,j,k are unit vectors (symbols to represents right, forwards, up). They're not units as in m/s (just to make sure that's understood).
$$KE = \frac{1}{2} m \underline{v} \cdot \underline{v}$$
So, v^2 is defined as the dot product of the velocity with itself (which is why the unit vectors i,j,k come off). When the system is rotated, v^2 doesn't change. This is a very important property of vectors.

 

[fluidistic]

So you're saying that the units come off when you square, making it a scalar?
But why do the units come off anyway? I don't really see why it's a scalar, wouldn't it change if you rotate the system? How does v^2 change?

No, the units get indeed squared.
In the SI, the unit of energy is the joule (J). It's worth $\frac{ 1 kg m^2} { s^2 }$. And energy is not a vector as they explained to you.
$\underbrace {\vec v }_{\text {vector} } \underbrace{ \cdot }_{\text {dot product} } \underbrace{ \vec v }_{\text { vector }} = \underbrace {v^2 }_{\text {scalar }}$.

Hmm if latex is a little weird, $\vec v \cdot \vec v = v^2$.

 

[BruceW]

Yeah, fluidistic, your latex looks weird on my computer...

 

[fluidistic]

To take Bruce's example more rigorously, $\vec v =\frac { 3m}{s} \hat i +\frac {6m}{s} \hat j +\frac{2m}{s} \hat k$.
If you do $\vec v \cdot \vec v$ you'll get $\frac{9m^2}{s^2} + \frac{36 m^2}{s^2} + \frac{4 m^2}{s^2} =\frac {49m^2}{s^2}=v^2$.
So that when you calculate the kinetic energy ($\frac{m v^2}{2}$), you get units in joule as it should.

 

[BruceW]

Yes, true. I always got told off in school for forgetting to write the units.

 

[nhmllr]

To take Bruce's example more rigorously, $\vec v =\frac { 3m}{s} \hat i +\frac {6m}{s} \hat j +\frac{2m}{s} \hat k$.
If you do $\vec v \cdot \vec v$ you'll get $\frac{9m^2}{s^2} + \frac{36 m^2}{s^2} + \frac{4 m^2}{s^2} =\frac {49m^2}{s^2}=v^2$.
So that when you calculate the kinetic energy ($\frac{m v^2}{2}$), you get units in joule as it should.

Ohhhh I see. That makes a ton of sense.
Thanks everybody for your help!