Orbit of almost 1 eccentricity

In summary, the orbiting particle changes course almost 180º when it is at a distance of 1 m of the focus of the ellipse, but changes almost 0º course when it is at a distance of the focus of 1.m in the way up.
  • #1
luxtpm7
7
0
You make a 100 m diameter tunnel that goes from the north pole to the south one.
You put into orbit a tennisball by dropping it in the tunnel from the north hemisphere
Supposing it makes an ellipse 99 m the minor axe and the Earth radius the bigger axe wouldn't the ball get into an orbit of eccentricity almost one which focus would be the center of the earth
Wouldnt this mean that a ball dropped from the north hemisphere would get such an orbit it would come out the northern hemisphere?
 
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  • #2
The ball would come out at both poles. You wouldn't get a standard eliptical orbit where the center of the main mass (in this case, The Earth) is located at one of the foci of the elipse, instead, it would be located at the center of the elipse. This is because, in this case, you cannot treat the entire Earth's mass as a point mass at its center.
 
  • #3
Janus said:
in this case, you cannot treat the entire Earth's mass as a point mass at its center.

More specifically, when the ball is a distance r from the center of the earth, you can treat the mass of that part of the Earth that is inside radius r as a point mass at the center, and ignore the rest of the Earth. (assuming spherical symmetry)

Therefore, as the ball falls towards the center of the Earth, the "effective mass" of the Earth acting on it decreases.
 
  • #4
Imagine a body with the Earth's mass and 1 cm radius you drop a ball10kmfromthenorth zenith of the body and make it an ellipse with a minor axe of 1 m and a bigger axe of 10km,an ellipse with an eccentricity of almost one

According to kepler the ball will orbit in an ellipse shape and the masive body will be at one of the focus of the ellipse

This means the ball will drop almost straight 10 km from the zenith of the north of the body will rise 10 cm in the south zenith and will return to the north zenith 10 km up and so on

My question is isn't gravity so odd it reverses the direction of the ball almost 180º when its a full speed? just like an ufo

How do Newtonian physics deal with this revers of sense at full speed?
 
  • #5
luxtpm7 said:
How do Newtonian physics deal with this revers of sense at full speed?

Calculate the acceleration of the ball, produced by the gravitational force exerted on it by an earth-mass located at a distance of 1 m. That will give you a clue. :smile:
 
  • #6
Imagine a body with the Earth's mass and 1 cm radius you drop a ball10km from the north zenith of the body and make it an ellipse with a minor axe of 1 m and a bigger axe of 10km,an ellipse with an eccentricity of almost one

This does not model the situation you have set up. Note that as mentioned in previous posts that the mass effecting the falling object changes with the radius, and is exactly zero when the falling body passes the center of the earth. This would not be a Keplerian orbit which assumes a constant mass as the attracting body.
 
  • #7
My point is to understand orbits with an eccentricity close to one.

Lets take two particles under a common gravity field and with an orbit of eccentricity 0.999periodic.

If we consider one of the particles fixed, the fixed particle will be at the focus of an ellipse that will be almost a segment

The orbiting particle will almost perfectly fall and rebound up(trully it will be an allipse of eccentricity almost one)

my question is how is it possible the orbiting particle changes course almost 180º when its at a distance of 1 m of the other particle in the way down but changes almost 0º its course when its at a distance of the focus of 1.m in the way up(ellipses are symmetrical)?
 

FAQ: Orbit of almost 1 eccentricity

1. What is meant by an "orbit of almost 1 eccentricity"?

An orbit with an eccentricity of almost 1 means that the orbit is highly elliptical, with a shape that is very close to a perfect circle. In other words, the path of the orbiting object is almost a perfect circle, but with a slight elongation or "flattening" on one side.

2. How is eccentricity calculated for an orbit?

Eccentricity is calculated by taking the ratio of the distance between the foci (the two points where the orbiting object is closest and farthest from the object it is orbiting) to the length of the major axis (the longest diameter of the ellipse). This value is always between 0 and 1, with 0 representing a perfect circle and 1 representing a parabola.

3. What causes an orbit to have an eccentricity of almost 1?

An orbit with an eccentricity of almost 1 is typically the result of a strong gravitational pull from another object, which causes the orbiting object to deviate from a perfectly circular path. The closer the orbiting object is to the center of the gravitational pull, the more elongated its orbit will be.

4. What are some examples of objects with orbits of almost 1 eccentricity?

Some examples of objects with orbits of almost 1 eccentricity include comets, which are pulled in by the strong gravitational force of the sun, and some dwarf planets, such as Pluto, which have highly elliptical orbits due to their distance from the sun.

5. How does an orbit's eccentricity affect its speed?

An orbit's eccentricity does not necessarily affect its speed, as the speed of an orbiting object is determined by its distance from the object it is orbiting and the strength of the gravitational force between them. However, an object with a high eccentricity may experience variations in its speed as it moves closer to and farther away from the object it is orbiting.

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