A satellite is in a circular orbit around the EArth at a distance equal to twice

Question

A satellite is in a circular orbit around the EArth at a distance equal to twice the radius of the Earth RE, as measured from the center of the Earth, how does its speed v relate to the Earth's radius RE, and the magnitude g of the acceleration due to gravity on the Earth's surface?

I know that I use the equation F = Gm * ME / 2R2.

F = Gm * ME / 2R2, where F = ma and the two m's cancel.

gRE = GME

then I think I use the v = sqrt( gm/r) equation to continue the derivation...this is where I start getting a little confused.

What I got then was: sqrt (g/ME * m/2RE )

sqrt (1/2 gRE) (which is the known right answer) by cancelling out the masses, but I'm not sure how the RE ends up on top...or am I just totally off?

Explanation / Answer

The oribital velocity of the satelite is

centripetal force on the satelite equal for gravitational force between them

mvo^2 /r = GMm/r^2

vo = sqrt GM/r

here r is radius of satelite from the center of the earth

if satelite is placed twice the earth radius distance then orbital speed is

vo = sqrt GM/ 2R_E

-------------------------------------------------

the magnitude of gravity on earth is

g = GM/ R_E^2

the magnitude of gravity on satelite is

g '= GM/ 2R_E^2

compare these two gravities

g'/g = GM/ 2R_E^2/GM/ R_E^2

      g' = g/2

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