Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass m and radius of orbit L. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible. Rank the net force acting on each satellite from their rockets.

Question:

The question is incomplete. The mass and radius of the orbit was not added. Find below the remaining part of the question and the answer.

Rank from largest to smallest.

A) m=200 kg and L = 5000m

B) m=400 kg and L=2500m

C) m=100kg and L=2500m

D) m=100kg and L=10000m

E) m=800kg and L=5000m

F) m=300kg and L=7500m

Answer:

Fe > Ff > Fa = Fb = Fd > Fc

4*10^6 N > 2.25*10^6 > 1 * 10^6 N = 1 * 10^6 N = 1 * 10^6 N > 0.25*10^6 N

Explanation:

The force experienced by the satellite is given by the formula;

Fc = kmL

where;

k = constant = 4π²/T²

m = mass of satellite

L = radius of orbit

Since, all the satellites complete the circular orbit in the same amount of time. The constant k does not affect the force value.

A) The force acting on satellite A from their rocket is given as follows:

Fa = kmL

Fa = 200 * 5000

= 1 * 10^6 N

B) The force acting on satellite B from their rocket is given as follows:

Fb = kmL

=400 * 2500

= 1*10^6 N

C) The force acting on satellite C from their rocket is given as follows:

Fc = kmL

= 100 * 2500

= 0.25*10^6 N

D) The force acting on satellite D from their rocket is given as follows:

Fd = kmL

= 100 * 10000

= 1*10^6 N

E) The force acting on satellite E from their rocket is given as follows:

Fe = kmL

= 800 * 5000

= 4*10^6 N

F) The force acting on satellite F from their rocket is given as follows:

Ff = kmL

= 300 * 7500

= 2.25*10^6 N

The ranking is given as:

Fe > Ff > Fa = Fb = Fd > Fc

4*10^6 N > 2.25*10^6 > 1 * 10^6 N = 1 * 10^6 N = 1 * 10^6 N > 0.25*10^6 N