Orbit and Energy

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Homework Statement


A satellite in low-Earth orbit is not truly traveling through a vacuum. Rather, it moves through very thin air. Does the resulting air friction cause the satellite to slow down?


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The Attempt at a Solution


Evidently, the answer is, "Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameter of the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit, however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!"

However, when I consider the relationship [itex]E = \frac{-Gm_1m_2}{2r}[/itex], which becomes [itex]r = \frac{-Gm_1m_2}{2E}[/itex], I am not convinced that the radius becomes smaller, thereby causing the diameter to become smaller. Am I considering the wrong relationship in order to solve this problem?
 

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  • #2
TSny
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Keep in mind that E is negative. Does losing energy due to friction cause E to become more negative or less negative?
 
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Hmm, it would seem as though E would become less negative, that is, approaching zero; as to why, I am not certain.
 
  • #4
TSny
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If a number becomes less negative would you say the number increases or decreases?
 
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By becoming less negative it increases. My guess as to what E becomes less negative was, well, a guess. Could you explain to me why E becomes less negative?
 
  • #6
TSny
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Yes, if a number becomes less negative it increases in value. But, what should friction do to the energy E? Increase it or decrease it? Therefore, should E become less negative or more negative?
 
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So, a more negative number corresponds to a higher energy, simply by definition?
 
  • #8
TSny
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So, a more negative number corresponds to a higher energy, simply by definition?
No. Does friction take away energy from the satellite or add energy to the satellite?
 
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I'm sorry, I wasn't taking about E in terms of friction. I was just trying to get a feel for the conventions used when considering E in general. To answer your question, friction removes energy.
 
  • #10
TSny
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OK, friction takes away energy from the satellite. Since the energy E was already negative before friction acted, does E become more negative or less negative as friction acts?
 
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  • #11
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E becomes less negative.
 
  • #12
TSny
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Suppose E = -10. Then I take away 2. Does x become more negative or less negative?
 
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By take away 2, do you mean subtract 2? If so, then E would be -12.
 
  • #14
TSny
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That' right. If friction takes away some energy from the satellite, then E will become more negative. Looking at your equation for r in terms of E, what happens to r as E becomes more negative?
 
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  • #15
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r in fact becomes smaller. Thank you for your time and effort!
 
  • #16
TSny
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You are welcome. Enjoy your studies.
 
  • #17
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Hello TSny ,

From an energy perspective it is clear that a decrease in energy , causes the radius of satellite to decrease and therefore an increase in speed . But how should I look at this qualitatively from a force perspective .

The tangential speed decreases due to air friction .The radial speed (towards earth) increases due to gravitational force .The resultant net speed increases .

Does that seem correct ?
 
  • #18
TSny
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Yes, basically I think that's correct. At least "on the average" the friction will cause the radius to decrease and the speed to increase.

I plotted some orbits assuming a friction force that is opposite to the velocity direction and proportional to the speed. Hope I don't have a bug in the program. The first image shows plots for a relatively small amount of friction. The second shows plots for a larger amount of friction. The speed increases on the average and the radial distance, r, decreases on the average. But you can see that the speed can actually decrease at times and r can increase at times. But the energy is always decreasing.
 

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  • #19
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Thank you TSny .

There is another point I would like to clarify . Suppose the satellite is moving in a circular orbit around the earth . Now it undergoes a sudden small increase in speed so that it starts moving in a circular orbit ( although elliptical) of larger radius . From the non inertial reference frame of the satellite ,the increased centrifugal force outweighs the gravitational force such that the satellite moves radially outwards . But how to explain this from the perspective of an inertial frame where there is no centrifugal force ?

I think ,initially the gravitational force provides just enough centripetal force for the satellite to move in an orbit of radius R1 . But when speed is increased (and no change in gravitational force) , the radius also increases to some value R2 , such that mv12/R1 = mv22/R2 .

Does that seem correct ?
 
  • #20
TSny
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I think ,initially the gravitational force provides just enough centripetal force for the satellite to move in an orbit of radius R1 .
Yes.
But when speed is increased (and no change in gravitational force) , the radius also increases to some value R2 , such that mv12/R1 = mv22/R2 .
Assume the sudden increase in speed is due to an impulsive force tangent to the initial circular motion. So, just after the increase in speed the satellite is still moving along a tangent to the initial circular motion. The increase in speed means the gravitational force is not enough to keep the satellite in its circular orbit and the satellite begins to move in an elliptical orbit. Initially the radial distance r increases for half an orbit until the satellite reaches apogee. Then r decreases for the second half of the orbit until the satellite returns to its starting point (perigee). The radial distance at the end of the elliptical orbit will be the same as the radius of the initial circular orbit.

The satellite will now keep orbiting on this elliptical path. It does not move to a new circular orbit of some radius R2. So, I don't understand the equation mv12/R1 = mv22/R2.
 
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  • #21
jbriggs444
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The satellite will now keep orbiting on this elliptical path. It does not move to a new circular orbit of some radius R2.
As I am sure TSny knows, you can apply a brief forward thrust to produce an elliptical orbit. Then 1/2 orbit later you can apply another brief forward thrust to circularize the resulting elliptical orbit. Despite having applied a forward thrust both times, the resulting orbital speed will be lower than the initial orbital speed.
 
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