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.

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

igneous rocks also form where plates converge. the subducting plate melts as it sinks into the crust of the earth, and the melt rises into the overriding plate forming volcanoes. metamorphic rocks are formed mainly in the lithosphere, wherever there is high pressure and high temperature. -woods