What will be the trajectory of space objects?

[AbhiFromXtraZ]

Can you please explain why the paths of objects are hyperbolic, parabolic and elliptic for energies positive, zero and negative respectively?

 

[tiny-tim]

Hi AbhiFromXtraZ! Welcome to PF!

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.

 

[AbhiFromXtraZ]

Thank you a lot for responding to my thread.

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).

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??

 

[tiny-tim]

Hi AbhiFromXtraZ!

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

 

[AbhiFromXtraZ]

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...