Answer:
a) v = 7533.3 m/s
b) T = 54677.0 seconds
Explanation:
let Fc be the centripetal force and Fg be the gravitational force.
m be the mass of the satelite and M be the mass of the earth.
a) A satelite in orbit experiences two forces, centripetal force and gravitation force due to the center of mass.
this forces are balanced, meaning at each point in the orbit:
Fc = Fg
m×v^2/r = G×m×M/(r^2)
because r is the same and m is the same:
v^2 = G×M/r
= [(6.67408×10^-11)×(5.976×10^24)]/(6378×10^6 + 650×10^3)
= 56750572.11
then v = \sqrt{56750572.11} = 7533.3 m/s.
therefore, the speed of the satelite in it's orbit will be 7533.3 m/s.
b) for a given circular orbit, the satelite will cover a displacement of 2πr in speed of 2πr/T where T is the period of the orbit:
then.
2πr/T = \sqrt{G×M/r} = v
T = 2πr/(v)
= 2π(6378×10^6 + 650×10^3)/(7533.3)
= 54677.0 seconds
therefore, it will take the satelite 54677 secords to orbit earth.
