Expert: Paul Soderman - 12/3/2011

Question
QUESTION: Sir,

Consider an mass m( say a rocket) being launched from the surface of earth. Initially the KE=0 while PE = -GmM/Re, Where Re( radius of earth). As the rocket moves upwards, external force acts on it ( in the form of propellant momentum) and keeps increasing the KE and PE of the body.  This means that the total energy of the object is continuously increasing.If this mass is on escape trajectory, then such a trajectory should be parabolic and parabolic trajectories have total energy =0. How then can we say that escape trajectories are parabolic?

ANSWER: Sangeeth

In the gravitational two-body problem, the specific orbital energy of the two bodies is the sum of their mutual potential energy and their total kinetic energy, divided by the reduced mass.  That total energy is constant and can be negative, zero, or positive.  

Consider the International Space Station. Before launch it has zero kinetic energy and a potential energy of approximately -63 mJ/kg (coordinate system is positive upward).  The launch rocket adds kinetic energy and puts it into elliptical orbit.  At some point in orbit it has a kinetic energy of 30 mJ/kg and a potential energy of -59 mJ/kg for a total specific orbital energy of -29 mJ/kg. Note that both potential and kinetic energy have increased, but the total is negative in the conventional coordinate system.

To escape the Earth's gravitational field, more energy must be added to achieve a parabolic trajectory.  In that case the two-body equations result in potential energy and kinetic energy that are equal in magnitude but of opposite sign.  The total is zero.

For a hyperbolic trajectory, the total specific orbital energy is positive.

Paul

---------- FOLLOW-UP ----------

QUESTION: Sir,
I analyzed your answer and the associated data and came to this conclusion. Please correct me if I am wrong.
I am using the same values as you have used.

A rocket at rest on the surface of earth has 0 KE and -65MJ/kg PE; TE= -65MJ/kg.

As the rockets is fired, new energy is added in form of propellant which burns and produces KE and as the rocket climbs up, this KE is converted to PE. Thus KE first increases then as it is converted; decreases, but PE increases continuously, TE increases.

At some point, as it enters the orbit ( say elliptical) PE has increased to -59MJ/kg, KE has increased ( effectively) to 30MJ/kg and TE = -29MJ/kg.

Thus in effect TE has increased.

This TE value of  -29MJ/kg is what will remain constant in the orbit ( neglecting orbit debris!)

This points to one understanding: That the path taken by the rocket from the surface of earth to the perigee is not included in the orbital trajectory. Is this right?

As for parabolic, We have to give KE=63MJ/kg if we want the body to escape completely.

For hyperbolic trajectory, we have to give in excess of 63MJ/kg.

Answer
Yes Sangeeth, you are basically correct, but a few things could be stated more clearly.  As the rocket accelerates upward both kinetic and potential energy increase. There is no reason that the kinetic energy must decrease unless the rocket slows or the mass decreases faster than V^2 increases. In the Kepler orbit (two-body problem), the rocket is no longer firing and the total energy of the body remains constant. Technically, anything including a rocket moving along a path is following a trajectory. It is when the body is freely following a Kepler orbit that we say it is in orbital trajectory. So, yes, the rocket acceleration from Earth is not considered part of the orbital trajectory, but could be said to be following a launch trajectory.  Your KE numbers for parabolic and hyperbolic trajectories seem reasonable though the minimum value of KE required to achieve escape velocity depends on the elliptic orbit altitude and location in the orbit as the PE and KE depend on altitude and location in the orbit. In your example above, I should think that increasing KE from 30 MJ/kg to 59 MJ/kg would achieve a parabolic orbit.  

Paul