Question

mercury rotates about its own axis once every 58.6 Earth days. a satellite orbits mercury once every murcurian day. how high is this satellite above mercury's surface? mass of mercury = 3.2x10^23 kg. the radius of mercury is 2400km.

Answer 1

change period from Earth days to seconds.

let k be the distance from center of Mercury to the satellite

GM/k^2=v^2/k=(2pi*k/period)^2 / k

now if my mental math is correct..
k^3=GM*period^2 check that...as did Kepler

then solve for k, and subtract r (radius of mercury, and you have altitude

h=k-r