[Jinto]

@Aoie_Emesai,

There is just one thing I wasn't very exact about.

To gain a higher orbit you need to speed up to advance from the current orbit. But to hold the new orbit, speed is slower then in the lower orbit.

So higher orbit = lower velocity and lower orbit = higher velocity. (if seen as speed tangential to the orbit)

The potential energy however is higher in higher orbits.
Thats rather hard to explain...



In the example graphic there is given an orbit with the radius r. The object in the orbit has a certain velocity v (that is tangential to the orbit). The attraction/gravitational acceleration is g in this example, and a is the centripetal acceleration. v needs to be high enough to result in an a that is equal to g. This is one way to explain the orbit.

There is another method to explain the orbit. Imagine the object to be constantly in freefall to the center object. Or to phrase it differently:
In the same time the object falls down h (because of g) it moves forward the length of v. Now falling down h and moving forward the length of v is a smooth transition that describes the orbit.

Now I want to explain the thing with the different velocities in different orbits.

First of all the easy part. Since gravity's force decreases by the square of r. One can say, that g decreases when r increases. With decreasing g the object needs less a and therefore less v. So staying in a higher orbit needs less velocity then staying in a lower orbit.

a = v²/r
g=G*m/r² (m is the mass of the centre object)
since g should be equal to a...

v²/r = G*m/r²
v² = G*m/r
v = sqrt(G*m)/sqrt(r) (sqrt(G*m) is const.)
v = const./sqrt(r)

=> higher orbit = less velocity needed to stay in orbit.

To leave an orbit for a higher orbit, an object needs to accelerate however. It needs to accelerate about as much as:

acceleration when falling from higher orbit to lower orbit - acceleration for increasing tangential velocity from higher orbit to lower orbit.



For example if an object is in a very high orbit around earth:

r1 = 384,400,000 m (moon orbit)
m = 5.97*10^24kg (mass of earth)
G = 6.67259 *10^-11 m³/(kg*s²)

v=sqrt(G*m/r1)= 1017.98 m/s approx. 1km/s

in a lower orbit (lets say 7500km above earth's centre):

r2 = 7,500,000 m
v=7287.92 m/s approx. 7.3 km/s

difference r1,r2 = 376900000 m

acceleration from point r1 to r2 will result in a speed of approximately 9 km/s (if the object was not moving in an orbit, but directly falling to earth... this is not easy to calculate because g is increasing while the object falls down to earth (or lets say in the lower orbit))

edit: (there has been still a logical error, now it should be right)

Ekin = 1/2 m1*v² (m1 = mass of object)
WG = G*m1*m2*(1/r2-1/r1) (m2 = mass of earth)
The gravitational energy build up while moving from r1 to r2 is equal to the final kinetic energy of the object (when it reaches r2).

Ekin = WG
1/2*m1*v² = G*m1*m2*(1/r2-1/r1)
v² = 2*G*m2*(1/r2-1/r1)
v = sqrt(2*G*m2*(1/r2-1/r1))
v = 10205 m/s approx. 10.2km/s

WG + Ekin_high - Ekin_low = dEkin

dEkin = 1/2*m1*vG² + 1/2*m1*vhigh² - 1/2*m1*vlow²

vdelta² = vG² + vhigh² - vlow²

vdelta = sqrt(10205² + 1018² - 7287²) m/s
vdelta = 7216 m/s
7.2 km/s + 7.3 km/s = 14.5 km/s

The kinetic energy needed to go from the low to the high orbit is very high. A sudden speed up of 7.2 km/s in the lower orbit would catapult the object into the higher orbit (the object would slow down from 14.5 km/s to 1km/s in the process of reaching the higher orbit).