An astroid moving due to Earth's gravity

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The discussion centers on determining the minimal distance between an asteroid in a hyperbolic orbit and Earth, with the vertex of the hyperbola believed to represent this closest point. To prove this, one can derive the distance from the focus (Earth's center of mass) and show it is minimized at the vertex using the hyperbola's equation. The conversation also touches on conservation laws, calculating velocities at specific points, and confirming that the asteroid's total energy indicates it will not return to its initial position. The participants share insights on the mathematical approach needed to solve the problem, emphasizing the importance of understanding the hyperbolic orbit's properties. The discussion concludes with a focus on the significance of radial velocity in identifying the minimal distance.
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Hello!
I think it's a bit unnecessary to describe the whole problem, so I'll just focus on my question:
There's an astroid moving in an hyperbolic orbit due to Earth's gravity (I found that)...
I have been asked to find the minimal distance between the astroid and the Earth.
It looks pretty obvious that this minimal distance is at the vertex (I hope I'm right with the word) of the hyperbula. If I'm right - how do I prove this? The velocity there isn't 0 or anything and I've wondered for some time what's special about that point and came to no cunclusion.
If I'm wrong - I'm in even bigger trouble :)

I'd really appreciate a nudge :)

Thanks,
Tomer.
 
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Well, I wonder how you got the curve to be a hyperbola. Just out of curiosity, if you could post it, I would appreciate it. If you want to prove that the vertex is the closest point, you could find the distance from the focus ( I think the Earth's COM will be the focus), and the distance is would come out to be a(1-e) (x^2/a^2-y^2/b^2=1 being the equation of the hyperbola and e the eccentricity) which is as small as it could get.

Or you could find the distance of a general point on the hyperbola and use calculus to prove that its smallest at the vertex.
 
I'll post it, ok.
I actually Have a date on this task, so I'm solving it faster than understanding it - which is horrible but necessary sometimes. So I got to it by using the equations I read in the book, flipping fastly on the material. I have only two questions left, and it's already after last date for delivering it, so... :)
There's the Earth. And there's an astroid.

at point A, the astroid has velocity V1, and is situated to the "left" of earth, it's velocity V1 perpendicular to the distance vector from the Earth (the COM of the earth), and the distance between it and the Earth at point A is 4Re (Re - radius of Earth)
At point B, the astroid has velocity V2 which forms an angle of 30 degrees with the "north" direction (like the positive Y direction). It is situated right above earth, and the distance from it to the Earth is 12Re.
(the astroid travels from A to B... and of course continues)
The mass "m" of the astroid is a given.

I hope the drawing is understandable from my description.

first thing they want is for me to write the conservation laws that apply here (which are, and this is the time to say if I'm wrong, mechanical Energy and ang. momentum)
Then, they ask to calculate V1 and V2, which I did using the conservation laws.
then they ask what's the total energy of the astroid. I caulculated the energy in point A -potenital + kinetic... This should be the energy at every point.
Then - they ask if the astroid will ever return to point A (like an eliptical orbit).
This is where I have no actual understanding of the case so I just opened the book and saw that if the total energy is greater than 0 (and it is, this case... if I'm not wrong) then for sure e>1 (the eccentric thingy), therefore the orbit is an "open" one, meaning the astroid will not return to point A.

Last question is the minimal distance...

Thanks...
 
Imagine a hyperbola with asymptotes at the x and y axis.
It's formula is y = k/x
If you draw a square or rectangle from the (0,0) and the opposite corner touches a part of the hyperbola, then the area of the hyperbola is k. This will work for ANY situation. So, if you can find the equation for the hyperbola, then the minimum distance is:

\sqrt{2k}
 
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Thank you ! :)
I actually already did that by saying that "The minimal distance must happen when there's no radial velocity". Since there's only one point where that happens...

But I'll read the link anyway.
Thanks again.
 
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