Example
An Earth satellite of mass 200 kg is in
stable circular motion orbit, calculate the launch speed at the
distance of 2 REarth from
centre of Earth.
Answer: 5600 m/s
Calculate the potential energy, kinetic
and total energy in this orbit.
Answer: PE = -6.27x107 J, KE =
3.14x107 J, TE =
-3.13x107 J.
calculate the value of the gravitational
potential at this point.
Answer: φ = 3.14x10 5 J/kg
calculate the value of the gravitational
acceleration at this point and state its direction.
Answer: a = 2.46 m/s2 towards
centre of Earth
calculate the value of the gravitational
force on the satellite at this point and state its direction.
Answer: F = 492 N towards centre of
Earth
At this point of the motion, the
satellite fires (click the fire- button) backward thrusters to
reduce its speed.
The orbits path decays to a point where
it shoots some circular motion thrusters and goes on a lower
circular motion.
Calculate the new potential energy,
kinetic and total energy in this lowered orbit, given the new
radius of orbit is approximately 8.68x106 m
Answer: PE = -9.22x107 J, KE =
4.61x107 J, TE =
-4.61x107 J.
Hence or otherwise, calculate the
changes in the potential, kinetic and total energies of the
satellite as a result of this transition. (Mass of Earth = 6.0 x
1024 kg)
Answer: ΔPE = -2.95x107 J, ΔKE =
1.47x107 J, ΔTE =
-1.48x107 J.
With the diagrams below or otherwise,
describe
the significance of the different parts
of the graph of PE for the complete motion above.
suggested answers
1. fire reverse thrusters to begin decay
path that results in decrease of KE
2. increase in speed as PE decrease and
KE increase
3. fire circular motion thrusters to
begin lower circular motion orbit to meet the model
the significance of the different parts
of the graph of PE for the complete motion above.
suggested answers
the PE graph move according to the
equation = mφ from r = 2Rearth to the
lower orbit from right to left
the significance of the different parts
of the graph of TE for the complete motion above.
suggested answers
1. fire reverse thrusters to begin decay
path result in decrease of KE thus a corresponding decrease in TE
2. TE is constant
3. fire circular motion thrusters to
begin lower circular motion orbit, results in decrease of KE and a
corresponding decrease in TE to meet the model
Java
Model
http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_NewtonsMountainwee06.jar
Model
https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_gravity08_2/gravity08_2_Simulation.xhtml