Geostationary Satelitte (1 Viewer)

[wrxsti]

Given that the mass of the earth is 6 x 10^24 and raduis of the earth is 6370000m determine the height above the earths surface for a geostationary satelitte

i know period is 24 hours for a geostationary orbit and i know we need to use keplers law somewhere but i dont how to apply it

 

[jonni]

Force(centripetal)=Force(gravity)

(mv^2)/r=(GMm)/r^2

because v=(2pi r)/T
substituting v and re-arranging:

r^3=(GMT^2)/(4pi^2)

r^3=(6.67*10^-11*6*10^24*(24*60*60)^2) / (4pi^2)

r=4.23*10^7m

from earth surface= 4.23*10^7 - 6.37*10^6
=3.59*10^7m above earth's surface

 

[wrxsti]

thanks but isnt the force of gravity... g = GM/r^2 not g= GMm/r^2
(First line)

 

[jonni]

g= GMm/r^2 is the force of attraction due to two bodies. ie. (Satelitte and earth)
the mass of the Satelitte(m) cancels out in the equations, leaving the mass of earth (M).

 

[xiao1985]

@jonni: correct, though I refer F = GMm/r^2

since g= gravitational acceleration usually.