Asatellite of mass 5500 kg orbits the earth (m circular motion and has a period of 6200 s. determiner the magnitude of the earth's gravitational force on the satellite. show your working.

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.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y