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I'm pretty sure it's something to do with rocket science.... it's on a plaque at the Kennedy Space Center. It may be a dumb question here, but can anyone tell me what the formula
Vc=Ro√(g/Ro-h)
means?
Vc=Ro√(g/Ro-h)
means?
Looks like some form of the equation for escape velocity. I have seen something similar, but not what you have here. The form I am used to seeing is
[tex]u_e = R_o\sqrt{\frac{2g_o}{R_o+h}}[/tex]
Ro is the radius of the Earth
h is altitude
go is the acceleration due to gravity at the surface
Quite a few formulae are found by just guessing from dimensional analysis.Relations like the one you mentioned are not found by just randomly guessing or by experimenting but through mathematical derivation
Formula's do not 'evolve' and they don't have 'pre-cursors'. Relations like the one you mentioned are not found by just randomly guessing or by experimenting but through mathematical derivation. (I'm sorry if that sounds a little naggy but it bugs me when people think that something like E=mc^2 comes from like randomly putting together variables and seeing what works). Now as for the meaning of the one you've posted your representation is a little ambigious. Was the formula:
a) [tex]R_o \sqrt{\frac{g}{R} - h}[/tex]
or was it
b) [tex]R_o \sqrt{\frac{g}{R-h}[/tex]
Now when something is in a stable circular orbit the centripetal acceleration it feels will be exactly equal to the force of gravity pulling it inwards (i.e. If the amount it wants to fly away from the planet is exactly equal to the amount it wants to fall down towards it then the net result will be that it neither flys away or fall towards).
Now the force of gravity is given by [tex]\frac{GMm}{(R+h)^2}[/tex] where M is the mass of the earth, m is the mass of the satellite, R is the radius of the earth and h is the height above the earth that the orbit is (therefore, R+h is the distance from the satellite to the center of the earths mass). This is Newton's Law of Gravitation and is basically just a mathematical way of saying that if you double the distance between two masses the gravitational force between them will be a quarter of what it was (the gravitationaly force, like the coulomb force decreases proportional to 1/r^2 where r is the difference between the masses). G is a 'proportionality constant' and this is determined from experiment. So we know the force of gravity and now the force of centripetal acceleration is [tex]\frac{mv^2}{R+h}[/tex] this is just a result from calculus. A derivation of it can be found here http://en.wikipedia.org/wiki/Centripetal_acceleration#Geometric_derivation . This is a purely mathematical result of the 'perceived' force that one will experienced if one assumes that a rotating frame of reference is not rotating. So finally we have our relations and we know that they must be equal for stable orbit so we have:
[tex]\frac{mv^2}{(R+h)}=\frac{GMm}{(R+h)^2}[/tex]
We multiply both sides by (R+h) to get:
[tex]mv^2=\frac{GMm}{R+h}[/tex]
Divide both sides by m:
[tex]v^2=\frac{GM}{R+h}[/tex]
and take the square root of both sides to finally get:
[tex]v=\sqrt{\frac{GM}{R+h}}[/tex]
Therefore, we see that a satellite which has a velocity (in the direction which is 'parallel' to the earths surface) [tex]\sqrt{\frac{GM}{R+h}}[/tex] will stay in geosynchronous orbit. Of course we didn't consider things like air resistance (most satellites aren't in 'outer space' but the exosphere so there is still some atmosphere) and radiation pressure from cosmic rays and such but all those are very very minor. Therefore we have a 'formula' for what the orbital velocity of a satellite needs to be without ever doing an experiment or fiddling or the likes. And also we can tell things like the mass of the satellite (which was m) doesn't matter at all for what the speed needs to be and such. This is how physics is done, experiment is used for two things, firstly to determine constants of proportionality (although this is very minor) and far more importantly to verify a theoretical assumption (i.e. if a particle which emits two photons of equal energy in opposite direction must have its momentum conserved then one must conclude that it must have had its mass decreased by x amount and then you check and you find that it has and then you rejoice).