# Energy of planetary orbits

1. Apr 9, 2009

### marmot

Suppose you have a star of mass M and a planet of mass m orbiting around it. The orbit is circular. m<<<M. U of course is -GmM/r^2.

E of course is going to be -U/2 and kinetic energy U/2 due to the virial theorem.

What would happen if the mass of the star suddenly doubles?

This is a practice test question. Now, if it doubles, the kinetic energy is going to stay the same because the angular momentum won't suffer a torque (because force is radial). However U will double.

So we are going to have a new E of -3/2U where U=-GmM/r^2.

Now this seems clear to me however. The test then asks the for the kind of orbit the planet is going to experience. The question does not make sense because if it was circular before, and now E is more negative, then the eccentricity will be an imaginary number so it would not be physically possible.

Now, I did a similar homework problem to this but the star instead halves. If the star halves and the orbit was originally circular, E is gonna be 0 so the orbit will be a parabola and unbound. However, here I am experiencing instead that the star doubles, so E is going to be more negative than before. What am I supposed to answer here?

2. Apr 9, 2009

### Cantab Morgan

Hi, marmot. Your dimensions don't match. The formula you provide for energy doesn't have units of energy. So you need to do some rethinking.

3. Apr 9, 2009

### D H

Staff Emeritus
That's not quite right. You have a non-physical situation here (masses don't just suddenly double, after all). You have to make some physically real assumptions, and conservation of linear (and angular) momentum being key. In other words, you are implicitly assuming momentum is conserved. Make that assumption explicit and be done with it.

Another way to look at this: To get an instantaneous change in velocity you need an impulsive force. Are there any impulsive forces involved here? (The answer is no.)

That is not correct. Look at the units. What is the equation for potential energy?

What is the new Vmin, the minimum possible energy? The reason that orbital energy cannot be less than Vmin is because that would require an imaginary speed. The speed hasn't changed: It is still a real number.

4. Apr 9, 2009

### marmot

oh geez. sorry! i know that U=- GmM/r, i poisted force not energy! sorry i am sleepy right now.

5. Apr 9, 2009

### marmot

First the practice test is:

http://www.pa.msu.edu/courses/2009spring/PHY321/handouts/PHY321-PracticeMT3-09.pdf

I am referring to the last part of the first page.

Isnt that what I am doing though? t=rxF, so if the force is parallel to the lever there is no torque and what changes angular momentum is torque. So if that force field changes, i.e. mass increases, it would not affect the momentum because it exerts no torque.

Another way to look at this: To get an instantaneous change in velocity you need an impulsive force. Are there any impulsive forces involved here? (The answer is no.)

yeah i know

What is the new Vmin, the minimum possible energy? The reason that orbital energy cannot be less than Vmin is because that would require an imaginary speed. The speed hasn't changed: It is still a real number.[/QUOTE]

So is the orbit still circular? I was referring to this when I said imaginary:

http://img15.imageshack.us/img15/7063/eccentricity.png [Broken]

If the eccentricity is already zero for the circle, and the only thing that changes in this equation is E (which becomes more negative) then the square root will give imaginary.

EDIT: Actually, nevermind. myu also changes. But still, I had the impression that for a certain angular momentum, the more negative the energy the more circular the orbit, and in this case wouldnt the orbit be more negative than circular itself?

Last edited by a moderator: May 4, 2017
6. Apr 9, 2009

### D H

Staff Emeritus
No, it isn't. Torque is not needed and does not answer the question. For example, invoking torque as an explanation does not explain why the radial component of velocity doesn't instantaneously change.

What is the new circular orbital velocity at the given radial distance? Is the planet's velocity vector equal to this? If not, the orbit cannot be circular.

What is the value for Vmin after the star's mass doubles?

7. Apr 9, 2009

### marmot

Well

Vmin=-1/2(.k^2*m)/(L^2*r^2)

We know that the new k=2*GMm

But we know that k^2/r^2=4U^2

So Vmin=-4/2U^2m/(m^2v^2)

Vmin=-4/2U^2/(U)

which is -2U

That is less than the new energy, which is -3/2U

So it is eliptical because they are different but E is more but is still negative.

8. Apr 9, 2009

### marmot

I dont get this though. If there is a sudden change of potential, there is a sudden change of force - doesnt that constitute an impulse?

9. Apr 9, 2009

### D H

Staff Emeritus
Correct.

No. The force undergoes a step change, not a delta function. Jerk (time derivative of force) is a delta function.

10. Apr 9, 2009

### marmot

Oh so basically, you are saying that if there is an abrupt and discontinous change of force - like some mass magically doubling, that is not impulse? If the mass just kept growing continuously then it would exert an impulse right?

Sorry for being a dummy!

11. Apr 9, 2009

### D H

Staff Emeritus
Then two perfectly rigid billiard balls collide the balls undergo an instantaneous change in momentum. Since force is the time derivative of momentum, that means that the contact force is instantaneous infinite at the instant the balls collide and zero at all other times. This is an example of an impulsive force. Impulsive forces can be useful fictions. (Fictions because force is always finite in reality, useful because they simplify away a lot of unnecessary detail.)

In this case the momentum is continuous. While the time derivative (i.e., the force) undergoes a step change at the instant the mass doubled, it does not become unbounded. There is no impulsive force here. Looking at it the other way around, since the force is bounded, there are no instantaneous changes in velocity.