一衛星原本是重mg，現在重mg/2，圍地球轉一圈需時多久？

Let the distance from the earth centre be r when the satellite weights m(g/2), i.e. the acceleration due to gravity g' at distance r from the earth centre is given by g' = g/2 (where g is the acceleration due to gravity on earth surface)

Since the weight m(g/2) provides for the necessary centripetal force
thus, m(g/2) = m.r.w^2
where m is the satellite mass and w is the angular frequency, which is given by 2.pi/T (T is the period and pi = 3.14159.....)

i.e. m(g/2) = m.r.(2.pi/T)^2
thus, T^2 = 2r x (2.pi)^2/g ------------------ (1)

But g' = GM/r^2
where G is the Universal Gravitational Constant
M is the mass of the earth

and g = GM/R^2
where R is the radius of the earth

g'/g = (R/r)^2
or r^2 = R^2(g/g') = 2.R^2
r = square-root(2) x R = 1.414R

Hence, from (1)
T^2 = 2 x 1.414 x (6400 000) x (2 x 3.14159)^2/10 s^2

Taking square root on both sides to find T

MG/2=Mw^2R
G/2=(2丌/T)^2 R
T^2=4(丌^2)*R/5

代入R
這樣便可計出T...T為周期..,便是你要的一圈時間