- #1

petreza

- 8

- 0

Orbital mechanics is a bit of mystery to me.

Let us have a particle with mass

**m**orbiting the Sun (with mass

**M**)in a circular orbit with radius

**R**. The velocity vector

**V**is always tangential to the orbit. Let us also have a weak, constant ("drag") force

**F**that is always pointing in the direction opposite of

_{d}**V**.

At a first glance it would seem that

**F**would reduce the magnitude of

_{d}**V**(less

**K**- kinetic energy) and as a result push the particle to a higher orbit (more

**U**- potential energy). What actually happens is the opposite: the particle speeds up and moves to a lower orbit. And since the force is constant, it keeps loosing

**R**, spiraling into the Sun.

Is there a simple geometric explanation why this happens? After all, there are only two forces here

**F**and

_{d}**F**- Sun's force of gravity.

_{G}I know that what happens is that the particle looses angular momentum -

**L = mVR**(in this case). Why the increase of

**V**is proportionally less than the decrease of

**R**?

I know that for circular orbits

**K = -(1/2)U**and that total mechanical energy

**E**is:

**E = K + U = (1/2)(GMm)/R - (GMm)/R**

Still, conceptually I don't see why the particle would react in that particular way to those two forces.

Thanks for your help!