Hi,

I am currently in an analytical mechanics class and came across an
interesting question well studying for my exam. I was wondering, if my
solution to this question makes any sense to you guys and if you think
it is correct. Here is the question,

Question:

"A spacecraft payload is to be placed in low circular orbit about a
planet of a radius Rp. Let DeltaE1 Represent the additional
(mechanical) energy the payload will have in orbit compared to what it
has on the planet's surface. Later, after the payload is in orbit, its
rocket engine is fired again, increasing the (mechanical) energy of the
payload by DeltaE2 over its energy in orbit. The additional DeltaE2 is
-just- sufficient so that the trajectory of the payload would
eventually take it to an infinite distance from the planet. What is the
numerical value of the ratio DeltaE2/DeltaE1."

-We neglect gravitational effects of any other masses
-We assume the planet is not rotating, so just PE matters when it is on
the planet.

My Solution:

Ok, from classical physics we know the initial energy (E1_initial) on
the planet is going to be equal to,

E1_initial=-k/Rp , where k is equal to GmM and M is mass of planet and
m is mass of payload.

Next, we have for the energy (E1_final) of the circle orbit,

E1_final=-k/(2*(Rp+R1), where R1 is distance above the planet.

NOTE: if you do not believe the above is true read over
http://hypertextbook.com/physics/mec...l-mechanics-2/

The above gives us,

DeltaE1=E1_final-E1_initial=-k/(2*(Rp+R1)+k/Rp

Now, next, we find DeltaE2,

E2_initial=-k/(2*(Rp+R1)

and

E2_final=0 since it will be inifintly far away.

Thus,

DeltaE2=E2_final-E2_initial=k/(2*(Rp+R1)

Thus, using what we found for DeltaE2 and DeltaE1 we come to the
following relationship after a little algebra,

DeltaE2/DeltaE1=Rp/(Rp+2R1)

Now, since R1Rp (low orbit) we have that, (I am not sure if this is
the correct logic here)

DeltaE2/DeltaE1=(roughly)=1

Wow! Is that correct?

Thanks.

Just a small correction, R1 is the distance above the planet to the
circle orbit (ie Rp+R1=the radius of the circle orbit)

Mechanical energy altitude variation is a kick. Is'nt.

Just pretent it is a valid variable. The scientist needs it to make
actual experimental orbits exist. It is a term unique to your kind of
science. Get though the class without alot of theory thinking, because
the effect of General Reltivity is my understanding of its actual
reason to exist.

Douglas Eagleson
Gaithersurg, DMUSA

"robison" wrote in message
oups.com...
Hi,

I am currently in an analytical mechanics class and came across an
interesting question well studying for my exam. I was wondering, if my
solution to this question makes any sense to you guys and if you think
it is correct. Here is the question,

Question:

"A spacecraft payload is to be placed in low circular orbit about a
planet of a radius Rp. Let DeltaE1 Represent the additional
(mechanical) energy the payload will have in orbit compared to what it
has on the planet's surface. Later, after the payload is in orbit, its
rocket engine is fired again, increasing the (mechanical) energy of the
payload by DeltaE2 over its energy in orbit. The additional DeltaE2 is
-just- sufficient so that the trajectory of the payload would
eventually take it to an infinite distance from the planet. What is the
numerical value of the ratio DeltaE2/DeltaE1."

-We neglect gravitational effects of any other masses
-We assume the planet is not rotating, so just PE matters when it is on
the planet.

My Solution:


[snip derivation]


DeltaE1=E1_final-E1_initial=-k/(2*(Rp+R1)+k/Rp

Now, next, we find DeltaE2,


[more snipping]


DeltaE2=E2_final-E2_initial=k/(2*(Rp+R1)

Thus, using what we found for DeltaE2 and DeltaE1 we come to the
following relationship after a little algebra,

DeltaE2/DeltaE1=Rp/(Rp+2R1)

Now, since R1Rp (low orbit) we have that, (I am not sure if this is
the correct logic here)

DeltaE2/DeltaE1=(roughly)=1

Wow! Is that correct?

I suppose it depends upon your definition of "low Circular orbit".

For us here on Earth we take LEO to be about 100 to 300 nautical miles
up from the surface (limited by atmospheric drag on the one hand and
the Van Allen Radiation Belts on the other). If you plug, say, 150 nmi
in for R1 (about 278km) and take Rp to be 6378km, then the value for
your derived expression is 0.92 .

yeah, I was just looking at very low earth orbit case. That is crazy
that the value still comes out so close to 1 when you look at the
distance you input. Do you think the my logic is correct through the
steps of my solution?

"robison" wrote in message
oups.com...
yeah, I was just looking at very low earth orbit case. That is crazy
that the value still comes out so close to 1 when you look at the
distance you input. Do you think the my logic is correct through the
steps of my solution?


It looked okay to me.