Download Fall 2013 Physics 172 – Recitation 9 Using both

Fall 2013
Physics 172 – Recitation 9
Using both the Work-Energy and Momentum Principles
Solution
Purpose: The purpose of this recitation is to allow you to gain experience working with
both the Work-Energy Principle and the Momentum principle
Readings: Chapter 7.1-7.6
Learning Objectives:
6.8.1 Calculate the potential energy for a system of two or more particles due to
gravitational or electrical interactions
6.8.2 Define the change in potential energy of a system in terms of the work done by
internal forces
6.8.3 Write down the multi-particle energy principle
6.9.1 Relate force to potential energy
6.9.2 Write down the formula for gravitational potential energy
7.6.1 List the three fundamental forms of energy in a multi-particle system
Challenge Problem:
Consider a satellite with a mass of 200 kg in a low Earth circular orbit just above the
Earth’s atmosphere, which is about 50 km thick. Putting this satellite into low Earth orbit
is one step in the process of putting it into the final orbit it must be in if it is to function as
a communication satellite that relays messages between two specific ground stations. To
do this, the satellite’s final circular orbit must run above the equator and be
geosynchronous, i.e., have an orbital period of 24 hours, so that the satellite is always
visible from both ground stations.
• Determine the minimum amount of energy that would be required to move the
satellite from low Earth to geosynchronous orbit.
Gather Information:
Goal:
The goal of this problem is to determine the minimum amount of energy to move a
satellite in a circular orbit 50 km above the surface of the Earth to one that is
geosynchronous obit.
Indentify your system:
Satellite and The Earth
Identify the objects that interacting with your system:
The Sun and The moon (massive objects located near the Earth)
Fundamental Principles:
Work-Energy Principle, Momentum principle
MEarth = 6 x 1024 kg, REarth = 6.4 x 106 m, G = 6.7 x 10-11 Nm2/kg2, MSun = 2 x 1030 kg
Diagram:
Assumptions/ Approximations:
The only force acting on the system is the gravitational force due to the Earth. The Sun
and Moon are either too far away or not massive enough to affect our system. There is no
space junk or dust that interacts with the satellite.
Outline Solution:
1. Use the momentum principle to determine the speed of the satellite in the near
Earth orbit.
2. Use the momentum principle to determine the speed of the satellite in the
geosynchronous Earth orbit.
3. Use the work-energy principle to determine the amount of work needed to change
the orbit.
Analyze Solution:
1. Calculate the speed of satellite in the near Earth orbit
If we apply the momentum principle to the system consisting of the satellite
alone, we obtain an equation that allows us to determine the satellite’s speed.
Since its orbit is circular, the radius of curvature of the satellite’s orbit is equal to
its distance from the center of the Earth and the force exerted on the satellite by
the Earth and the time rate of change of the satellite’s momentum are both
perpendicular to the satellite’s orbit (momentum and velocity). So, applying the
momentum principle to the satellite in Low Earth (LE) orbit implies that


2
v LE
dp  | v LE |
Mm 
=| pLE |
≈m
= G 2 = Fon satellite by Earth ,
dt
R
R
R
24
where M = 6 x 10 kg is the mass of the Earth, R ( = 6.4 x 103 km+ 50 km) is the
distance of the satellite from the Earth’s center, and where m is mass of the
satellite. Solving, we conclude that
2
vLE
Mm
GM
m
= G 2 ⇒ vLE =
≈ 7.9 km/s .
R
R
R
2. Calculate the speed of satellite in the near Earth orbit
Proceeding as with the near-Earth orbit applying the momentum principle to the
satellite in a circular, geosynchronous orbit, yields a relationship between the
satellite’s speed and the radius of its orbit analogous to the one we found before,
GM
.
vgeo =
Rgeo
Since we know the period of the orbit is 24 hours, the satellite’s speed and its
2π Rgeo
orbital radius also satisfy the equation vgeo =
, where Tgeo is the 24 hour
Tgeo
orbital period. Substituting this into the previous equation and solving for R geo
and, then, solving for v geo yields
2π Rgeo
86400 s
=
2π Rgeo
GM
(6.7 × 10−11 Nm2 /kg2 )(6 × 1024 kg)
≈
⇒ Rgeo ≈ 4.2 × 107 m
Rgeo
Rgeo
2π (4.2 × 107 m)
≈ 3.1 km/s .
Tgeo
86400 s
Determine work that must be done on the satellite-Earth system necessary to
move the satellite from one orbit to another.
We can compute the energy of the satellite-Earth system in its initial low Earth
orbit state and its final geosynchronous orbit state. The sum of the change of the
system’s kinetic and potential energy between its initial and final orbits must be
provided by some object, external to the satellite-Earth system. We need not
concern ourselves as to how this occurs. Thus,
# 1
1
1&
2
2
Wexternal = ΔEsys = ΔK satellite + ΔU satellite = m(vgeo
− v LE
) − GMm %%
− (( ,
2
$ Rgeo R '
where we have made the easily justified assumption that the kinetic energy of the
Earth does not change significantly. Plugging in the values of the things we
know, we find that
W = 5.2 × 109 J .
v geo =
≈
Learn:
This problem forces you to utilize both the momentum principle and the energy principle
to solve. Once we have the different velocities from the momentum principle, you must
calculate the work done by some object, external to the satellite-Earth system. We are
not concerned with what this object is, only that it has done work on our system.