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}
/ 2R^{2}.

F = G_{m} * M_{E} / 2R^{2}, 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

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