Question

A 590-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius.

(a) Find the satellite's orbital speed.
m/s

(b) Find the period of its revolution.
h

(c) Find the gravitational force acting on it.
N

Answer 1

We have the following data:

height = 6370 km (Earth´s mean radius) = 6.37 x 10 ^6 m

Earth´s mass = M E= 5.98X   10 ^{^}   24 kg

Earth´s radius= RE = 6370 km = 6.37 x 10 ^6 m

Case b)

Let´s obtain first the period of its revolution:

As we know the orbital RADIUS (RE+ height) and the mass of the Earth (ME) and want to get the period of the satellite (T), we use the formula that relates the three dimensions.

Relationship between period and the orbital RADIUS:

T = Period = 2((RE + height)^3 / (G ME)) where G = constant of universal gravitation= 6.67 x 10 ^-11 N m^2/ kg^2

Substituting values and solving:

T = Period = 2(( 6.37 x 10 ^6 m + 6.37 x 10 ^6 m)^3 / ((6.67 x 10 ^-11 N m^2/ kg^2)(5.98X   10 ^{^}   24 kg)

T = 14, 306.07513 seconds = 3.973909758 hours

case a)

The speed of the satellite is given by:

v = G (ME / RE) = (6.67 x 10 ^-11 N m^2/ kg^2) ((5.98X 10 ^{^} 24 kg)/ (6.37 x 10 ^6 m))

v = 7913.04 m/s

Case c) Let´s occupy the following formula to solve this question:

w = G (m (object) x m (Earth)) / ((Earth´s radius) (height)) ^ 2

w = (6.67 X 10 ^-11 N m^2 / Kg^2) ( ( 5.98 X 10 ^24 Kg) ( 590 Kg) / ((6.37 x 10 ^6 m + 6.37 x 10 ^6 m ))^2

Solving the last expression:

w = 1499. 90709