In: Physics
A 778 kg satellite is in a circular orbit about the Earth at a height above the Earth equal to the Earth's mean radius.
(a) Find the satellite's orbital speed.
(b) Find the period of its revolution.
(c) Find the gravitational force acting on it.
The mass of the satellite is given as 778kg.
Required constants:
1. Earth mass, M =5.972e24 kg
2. Earth radius, R = 6.371e6 m
3. Gravitational constant, G = 6.674e-11 Nm2kg-2
a) Orbital speed, v is given by the equation
= 7.909e3 ms-1
Satellite's orbital speed = 7.909e3 ms-1
b) To find the period of revolution, we use the equation for speed = distance/ time
Here the speed --> orbital speed,v
distance --> 2R, since the
satellite is travelling in a circular path around the Earth
time --> T, period of revolution, which we have to calculate
==> T = 2R / v = 2 * 3.14 *
6.371e6 / 7.909e3 = 5058.78 s
1.405 hrs (1 hr
= 3600s)
Period of revolution = 5058.78 s or 1.405 hrs
c) The gravitational force equation is
Fgrav = GMm/R2
= 6.674e-11 * 5.972e24 * 778 / (6.371e6)
= 7,639.6 N
Alternatively, the centripetal force equation, mv2/R can also be used since the centripetal force is balanced by the gravitational force acting on the satellite. The final answer will differ by a small unit. However I suggest you use the gravitational force equation.
Gravitational force = 7,639.6 N