Longest and shortest distance in an elliptical orbit

  • #1

Homework Statement



Two satellites are launched at a distance R from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed v0 and enters a circular orbit. The second satellite, however, is launched at a speed (1/2)v0. What is the minimum distance between the second satellite and the planet over the course of its orbit?

M = mass of planet, m = mass of satellite, R1 = minimum distance, v1 = speed at the minimum distance

Homework Equations



Fg = G(m1)(m2)/(r^2)
Ug = -G(m1)(m2)/r
mvr = const. (conservation of momentum)
PE + KE = const.

The Attempt at a Solution



(Somehow, I got the right answer, but I am confused about the calculations.)

First, I wrote the sum of the potential and kinetic energies for the second satellite.
(1/2)m(v0)^2 - GMm/R ; using the fact that v0 is the speed for a circular orbit, I substituted for M and got U = -(7/8)m(v0)^2

Then, I set that value of U equal to the sum of the potential and kinetic energy at an arbitrary point on the orbit, with distance R1 and speed v1 at that point. My intention was to find the smallest possible value of R1 (the minimum distance.)

-(7/8)m(v0)^2 = (1/2)m(v1)^2 - GmM/(R1)

I substituted for M like I did previously, and for v1 using conservation of momentum
R(v0)/2 = (R1)(v1)

In the end, it was possible to cancel out v0 and m. The simplified equation turned out to be
7(R1)^2 -8R(R1) + R^2 = 0; the two roots of the equation are R1 = R and R1 = R/7; the latter is actually the answer, but I'm confused. It seems like R1 could correspond to any point on the orbit. I expected to get a range of possible values and then find the minimum. Why is it that only the perigee and apogee satisfy the equation? I can't figure out where I made an assumption that only applies to those points. Or, was my expression for mechanical energy incorrect? I noticed that if R1 is in between R/7 and R, my expression for total energy would be less than -(7/8)m(v0)^2. Perhaps there is a form of energy I forgot to consider?

I would really appreciate it if someone can clarify this.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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Homework Statement



Two satellites are launched at a distance R from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed v0 and enters a circular orbit. The second satellite, however, is launched at a speed (1/2)v0. What is the minimum distance between the second satellite and the planet over the course of its orbit?

M = mass of planet, m = mass of satellite, R1 = minimum distance, v1 = speed at the minimum distance

Homework Equations



Fg = G(m1)(m2)/(r^2)
Ug = -G(m1)(m2)/r
mvr = const. (conservation of momentum)
PE + KE = const.

The Attempt at a Solution



(Somehow, I got the right answer, but I am confused about the calculations.)

First, I wrote the sum of the potential and kinetic energies for the second satellite.
(1/2)m(v0)^2 - GMm/R ;
=U? Isn't that for the first satellite? The second satellite has speed [itex]v_0/2[/itex] right?
using the fact that v0 is the speed for a circular orbit, I substituted for M and got U = -(7/8)m(v0)^2

Then, I set that value of U equal to the sum of the potential and kinetic energy at an arbitrary point on the orbit, with distance R1 and speed v1 at that point.
You'd expect the total energy to be the same at all points in the orbit.
My intention was to find the smallest possible value of R1 (the minimum distance.)

-(7/8)m(v0)^2 = (1/2)m(v1)^2 - GmM/(R1)

I substituted for M like I did previously, and for v1 using conservation of momentum
R(v0)/2 = (R1)(v1)
The energy relation fixes the relationship between kinetic energy and potential energy for any distance at all - conservation of angular momentum would make sure the distances are all to points on the orbit.

In the end, it was possible to cancel out v0 and m. The simplified equation turned out to be 7(R1)^2 -8R(R1) + R^2 = 0; the two roots of the equation are R1 = R and R1 = R/7; the latter is actually the answer, but I'm confused.
Well R1=R is reasonable, you'd expect it to return to the start so it's the R/7 that has you foxed.

It seems like R1 could correspond to any point on the orbit. I expected to get a range of possible values and then find the minimum. Why is it that only the perigee and apogee satisfy the equation? I can't figure out where I made an assumption that only applies to those points. Or, was my expression for mechanical energy incorrect? I noticed that if R1 is in between R/7 and R, my expression for total energy would be less than -(7/8)m(v0)^2. Perhaps there is a form of energy I forgot to consider?

I would really appreciate it if someone can clarify this.
Clearly the two roots are the locations of the maxima and minima kinetic or potential energies- looks like you have come up with the derivative of the formula you were looking for.

I think you need to examine your rationale closely at each step - makes sure that step is doing what you think it is in terms of physics: what physical quantity are you describing and for what?

While you are doing that, I'll take a closer look.
 
  • #3
Simon Bridge
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Perhaps there is a form of energy I forgot to consider?
Cannot help notice that the length of the radius changes with time in an elliptical orbit and you have no radial component to velocity (your v's are all tangential). So the only place where your calculations will hold is where the radial component of velocity is zero - which will be right at the start, and at the closest approach. At those positions, the radial speed is changing direction.
 
  • #4
Cannot help notice that the length of the radius changes with time in an elliptical orbit and you have no radial component to velocity (your v's are all tangential). So the only place where your calculations will hold is where the radial component of velocity is zero - which will be right at the start, and at the closest approach. At those positions, the radial speed is changing direction.
Thank you! That would explain the "missing" energy. It seems I forgot that in addition to the tangential speed, there would be a normal component (unless it was an extremum, where the rate of change of the radius is zero.) So, would the total velocity vector be pointing either outward or inward (rather than be tangential) at all the points of the orbit except the perigee and apogee?
 
  • #5
Simon Bridge
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The instantaneous velocity will not be tangential, in general, no.
Only in a circular orbit.

See that second-to-last sentence in post #2 - this is the sort of thing I was talking about - what was being written down was not saying what you thought it was. These things can be hard to spot. As soon as I wrote vT in my treatment it sprang out at me.
 

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