What will be the trajectory of space objects?

In summary, the paths of objects are hyperbolic, parabolic and elliptic for energies positive, zero and negative respectively. The parabolic path is the limiting case between an ellipse and a hyperbola, depending on the eccentricity of the ellipse.
  • #1
AbhiFromXtraZ
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Can you please explain why the paths of objects are hyperbolic, parabolic and elliptic for energies positive, zero and negative respectively?
 
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  • #2
Hi AbhiFromXtraZ! Welcome to PF! :smile:
AbhiFromXtraZ said:
Can you please explain why the paths of objects are hyperbolic, parabolic and elliptic for energies positive, zero and negative respectively?

By "energy", you mean total mechanical energy, = kinetic energy plus gravitational potential energy.

Potential energy has to be measured relative to an arbitrary zero-level.

In this case, we choose "at infinity" to be the zero-level of potential energy.

Consider a parabolic orbit.

As it "approaches infinity", its speed becomes smaller and smaller, and tends to zero … ie, its speed (and KE) is zero "at infinity".

So, anywhere along its trajectory, its KE is minus its PE, ie its total energy is 0 (and its speed is always equal to the local escape velocity).

If its speed "at infinity" is positive, then it's hyperbolic.

And if its speed "at infinity" is negative, then obviously it can't reach infinity! So it's ellipitc. :wink:
 
  • #3
Thank you a lot for responding to my thread.

tiny-tim said:
By "energy", you mean total mechanical energy, = kinetic energy plus gravitational potential energy.

Potential energy has to be measured relative to an arbitrary zero-level.

In this case, we choose "at infinity" to be the zero-level of potential energy.

Consider a parabolic orbit.

As it "approaches infinity", its speed becomes smaller and smaller, and tends to zero ¡­ ie, its speed (and KE) is zero "at infinity".

So, anywhere along its trajectory, its KE is minus its PE, ie its total energy is 0 (and its speed is always equal to the local escape velocity). :wink:

According to your answer, if the total energy is zero, then the path will be parabolic...but why? It could be hyperbolic...as it also an unbounded path and ends at infinity just like a parabola.

The total energy in an attractive field is given by,
E = 1/2mv^2 + 1/2Iw^2 - k/r
now if the total energy is positive such that 1/2mv^2 term is greater than the sum of other two terms (negative)...Then??...if the centre of force is earth, it will attack earth...then where will be the turning point??
 
  • #4
Hi AbhiFromXtraZ! :smile:
AbhiFromXtraZ said:
The total energy in an attractive field is given by,
E = 1/2mv^2 + 1/2Iw^2 - k/r

We can ignore the 1/2Iω2, it makes no difference since it is constant over the whole trajectory
According to your answer, if the total energy is zero, then the path will be parabolic...but why? It could be hyperbolic...as it also an unbounded path and ends at infinity just like a parabola.

negative energy is obviously an ellipse

positive energy is obviously a hyperbola

zero energy is the limiting case, and therefore has to be the limit between an ellipse and a hyperbola, which is a parabola
now if the total energy is positive such that 1/2mv^2 term is greater than the sum of other two terms (negative)...Then??...if the centre of force is earth, it will attack earth...then where will be the turning point??

negative total energy tells you that the trajectory is an ellipse, but it does not tell you the eccentricity of the ellipse

if two ellipses have the same energy but different eccentricities, one may have a "turning point" outside the earth, and the other inside the earth: so the first is an orbit while the second is a crash
 
  • #5
tiny-tim said:
negative total energy tells you that the trajectory is an ellipse, but it does not tell you the eccentricity of the ellipse

Oh...you fell in a misunderstanding...sorry...it was my fault..
Actually I asked for positive mechanical energy...and ''(negative)'' means the sum of 1/2Iw2 and k/r is negative...and the magnitude is less than 1/2mv2 such that E becomes positive...

And 1/2Iw2 is actually 1/2m(wr)^2...my book says this the centrifugal potential energy (sorry for my fault)...I think this term prevents the object from moving along straight line...
 
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1. What factors determine the trajectory of space objects?

The trajectory of space objects is primarily determined by the object's mass, velocity, and the gravitational forces acting upon it. Other factors such as atmospheric drag and the influence of nearby objects can also affect the trajectory.

2. How do scientists calculate the trajectory of space objects?

Scientists use mathematical models and equations, such as Newton's laws of motion and the law of universal gravitation, to calculate the trajectory of space objects. They also use data from observations and simulations to refine their predictions.

3. Can the trajectory of space objects be changed?

Yes, the trajectory of space objects can be changed through various methods such as thrust from rockets, gravitational slingshot maneuvers, and aerobraking. However, the feasibility and effectiveness of these methods depend on the specific characteristics of the object and its surroundings.

4. Are there any potential dangers associated with the trajectory of space objects?

Yes, there are potential dangers associated with the trajectory of space objects. If an object's trajectory intersects with Earth, it could potentially cause damage or harm. Additionally, the increasing amount of space debris in orbit poses a risk for collisions that could alter the trajectory of objects.

5. How do scientists track the trajectory of space objects?

Scientists use various methods to track the trajectory of space objects, including radar, telescopes, and satellites. They also collaborate with international organizations, such as NASA and ESA, to share data and track objects in orbit.

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