Question

A satellite orbits the Earth at a height of 810 km above the Earth Surface. The mass of the satellite is 2900 kg.

a. Determine the force of gravity acting on the satellite.

b. Determine the speed of the satellite.

c. Determine its orbital period.

Please show all steps in the calculation.

Answer 1

Given: • h=810km

• Mass of satellite: m= 2900kg.

Let : M be the mass of earth.

m be the mass of satellite.

Fg = gravitational force.

v = speed of satellite.

a) : The force of gravity acting on satellite will be equal to the centrifugal force acting on the satellite i.e.

Fg = ( m*v^2)/(R+h)

= (2900*v^2) /(6400+810) ×10^3 /*(10^3 is multiplied to

to convert km in meters)*/

= 4.02*10^-4*v^2. eq. 1

•To calculate speed of satellite (v) we have formula

v = √(GM/R+h)

= √{(6.67*10^-11 *6*10^24)/(7.21*10^6) }

= √(5.55*10^7)

= 7.45×10^3 m/sec

Putting value of v in eq.1

Fg =(4.02×10^-4) × (7.45×10^3) ^2

The force of gravity acting on satellite = 2.23×10^4 N.

•b) the speed is calculated in part a) is

v = 7.45×10^3m/s

•c) For orbital period T we have formula

T^2/(R+h)^3 = 4×π^2 / GM.

T = √[(4 × π^2 × (R+h)^3) / (G*M)]

=√{(39.43×(7.21×10^6) ^3) /(6.67×10^-11×6×10^24) }

3.69×10^7 s.