Question

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

I know that I use the equation F = G_m * M_E / 2R².

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

gR_E = GM_E

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/M_E * m/2R_E )

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

Answer 1

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