A comet revolves around the sun

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Homework Statement



A comet revolves around the sun in a closed elliptical trajectory. Ignore any force acting upon it besides gravity. Prove that the angle between the position vector (sun in the origin) and the velocity vector of the comet at its perihelion and aphelion is 90º.

The Attempt at a Solution



I tried to approach this problem by writing down the motion equations for a body on an elliptical orbit under the action of gravity. But it is a second order differential equation, that I later found out has no analytic solution.

So I just thought: when the comet is nearest and furthest from the sun, position r(t) must have a maximum/minimum, hence dr(t)/dt = 0, at those points. That also works for the r(t)^2, so dr(t)^2/dt = 0 <=> r(t)[itex]\cdot[/itex]v(t) = 0 and this happens (since none of them is zero) only when they are both perpendicular.

However, by assuming dr(t)/dt = 0, I'm assuming that v(t) = 0 for some time t. But as we know that is never true, has the comet never stops moving. Is my solution valid? If so, why? If not, could you give me an hint? Does this all come down to the geomtry of the elipse ( because I never studied the equations that describe ellipses)?
 
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I think you're getting mixed up between scalars and vectors. If ##\vec r(t)## is the position vector, what is its magnitude? What do you get when that is at an extreme value?
 
After a few unsuccessful efforts to try to prove it by means of pure geometry, I learned a very useful theorem about ellipses, that the tangent of a point bissects the exterior angle. I just had to prove that. Then I found the answer where I least expected. I learned about Fermat's principle of least time, and with a little bit of thinking I had my proof.

I was really pleased with that, because those are the kind of answers I'm the most attracted to in physics.
 
Glad you got there, but it's not clear to me whether you now understand what you did wrong here:
Calabi_Yau said:
when the comet is nearest and furthest from the sun, position r(t) must have a maximum/minimum, hence dr(t)/dt = 0, at those points. That also works for the r(t)^2, so dr(t)^2/dt = 0 <=> r(t)[itex]\cdot[/itex]v(t) = 0 and this happens (since none of them is zero) only when they are both perpendicular.

However, by assuming dr(t)/dt = 0, I'm assuming that v(t) = 0 for some time t
The first paragraph works if what you mean is:
when the comet is nearest and furthest from the sun, position ##r(t) = |\vec r(t)|## must have a maximum/minimum, hence dr(t)/dt = 0, at those points. That also works for the ##r(t)^2 = \vec r \cdot \vec r##, so dr(t)^2/dt = 0 <=> ##\vec r(t) \cdot \dot {\vec r}(t) = 0## and this happens (since none of them is zero) only when they are perpendicular.​
But what are you saying in the second paragraph? dr(t)/dt = 0 does not imply ##\vec v(t) = 0 ## nor ##|\vec v(t)| = 0 ##. ##\frac {d\vec r(t)}{dt} = 0## would imply that, but that's not what you assumed.
 
Right, I see. I assumed [itex]\frac{dr.r}{dt}[/itex] = 0 does not imply dr(t)/dt = 0.

(Sorry for not writing with the proper symbols, but I am not very used to using those features)
 
Calabi_Yau said:
Right, I see. I assumed [itex]\frac{dr.r}{dt}[/itex] = 0 does not imply dr(t)/dt = 0.

(Sorry for not writing with the proper symbols, but I am not very used to using those features)
If you don't want to be bothered with LaTex for vectors, you can just put the vector variables in bold.
 

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