Answer:
The earth's gravitation force on the satelite is 41193.51 N.
Explanation:
if v is the orbital speed of the satelite and r is the radius of orbit of the satelite then, the period of a circular motion is given by:
T = 2πr/v
T^2 = [4×(π^2)×(r^2)]/(v^2)
v^2 = [4×(π^2)×(r^2)]/(T^2)
but if G is gravitational constant and M is the mass of the earth then, the orbital speed of the satelite is given by:
v = √[(G×M)/r]
v^2 = G×M/r
then :
G×M/r = [4×(π^2)×(r^2)]/(T^2)
r^3 = [G×M×(T^2)]/(4×(π^2))
= [(6.67×10^-11)×(5.972×10^24)×((6200)^2)]/(4×(π^2))
= 3.8785×10^20
r = 7292723.678 m
the Earth's gravitation force on the satelite is given by:
Fg = G×M×m/(r^2)
= (6.67×10^-11)×(5.972×10^24)×(5500)/((7292723.678)^2)
= 41193.51 N
Therefore, the earth's gravitation force on the satelite is 41193.51 N.
