#20 Janus

O.K.: potential energy is not an absolute value, but a magnitude always referred to an agreed zero level, and your examples put clear it is positive if we pass from a zero level distance to a higher distance ...

But I´m seeing this concept is more tricky than what it appears.

Gravitational potential energy is linked to a gravitational field, in our case consisting in all "g" values, Earth´s pull on the unit of mass at each point.

Those pulls are forces always radial. in the direction of Earth´s C.G. It is clear that if e.g. we throw a stone vertically upwards in a vacuum (no friction), initial kinetic energy will change into potential energy until maximum hight, due to "g" negative acceleration. And the addition of both energies would keep constant. And similarly when falling back.

But Moon in orbit around Earth has constantly zero "vertical" speed. Earth´s pull does not change any speed vertically, nor does the tangential component of the tidal bulge´s pull we are considering. If this one increases Moon´s tangential speed, we should not consider this increase has to be slowed back because Moon´s distance increases and potential energy increases too.

They are stuff linked to vectors perpendicular to each other, which can operate independently ...

Any acceleration in the direction of orbital velocity will result in an increase in distance from the planet. This is just a fact of orbital mechanics. In fact, acceleration along this direction is the most efficient means of gaining altitude.

You can show this in the following example.

The total energy of a orbiting craft is it is the sum of it kinetic energy and it Gravitational potential energy. In orbital mechanics GPE is taken to be -GMm/r (The zero point is set at infinite distance and become more negative (less) as r becomes smaller)

Thus the orbital energy is mv^2/2-GMm/r

v for a circular orbit can be found by v= sqrt(GM/r) as along as we assume that m is small compared to M

If we substitute this for v we get E= -GMm/2r

Now it also turns out that this equation works for non-circular orbits if use 'a', the semi-major axis of the orbit instead of r (for a circular orbit a=r) a is also the mean orbital radius of the orbit.

Now let's work out an example for a craft of 100 kg with an initial orbit with radius of 7000 km (altutude of ~622 km)

This gives us a orbital velocity of 7547 m/s, and an orbital energy of -2847857143 joules.

Now let's give the craft a instant boost of 1000 m/s. This puts it into a new elliptical orbit with its present altitude the perigee. This also increases its orbital energy to -2043153836 joules ( add 1000 m/s to the the orbital velocity and use the first equation I gave for orbital energy. )

Using this new value, we can solve for the semi-major axis of the new orbit, which turns out to to be ~9735 km. the average radius of the orbit has increased by 2735 km.

Now what happens if we apply this sudden boost in a direction directly away from the Earth? Now we have to use vector addition to get the new orbital velocity. This will be sqrt(1000^2+7547^2)= 7613 m/s

using this new velocity to find the new orbital energy, we get -2797853836 joules.

Solving as we did before for the new semi-major axis, we get 7125 km, or an increase of just 125 km compared to the 2735 km increase we got by adding the boost along the direction of orbital motion.

Also, by thrusting directly away form the Earth, the new orbit will have a perigee closer to the Earth. In fact, in this case, the new perigee would be closer to the center of the Earth than the surface of the earth is. Your craft would not be able to complete one full orbit without crashing into the Earth. So applying a boost away from the Earth could eventually lead to crashing into it.

Orbital mechanics is not always intuitive, and may not get the results you might first expect if you are not familiar with how orbits behave.