(“Scaling up” worksheet adapted from Yishun JC which is the lead school)
2013 Easy Java Simulation (EJS) Gravitation Worksheet 1
Geostationary Orbit and Kepler’s 3rd Law
……………………………………………….
Name:
Aims: (1)
(2)
Apparatus:
Class:………………….
To understand the characteristics of Geostationary Satellite;
To understand and appreciate Kepler’s 3rd law through the Solar System.
Computer must be installed with Java runtime and the two java applets, titled
“Geostationary Orbit Model” & “Kepler System Model”, which can be downloaded
from
https://dl.dropbox.com/u/44365627/lookangEJSworkspace/export/ejs_EarthAndSatelite.jar
and
https://dl.dropbox.com/u/44365627/lookangEJSworkspace/export/ejs_KeplerSystem3rdLaw03.jar
This will serve as your working document. Write your responses here while you work on the Easy
Java Simulation (EJS) that you have downloaded. By viewing and ‘playing’ around with the variables,
you would be able to see the changes as it happens on the screen. After making keen observations
and inferences from the EJS, transfer your answers to the following google drive.
https://docs.google.com/a/moe.edu.sg/spreadsheet/viewform?usp=drive_web&formkey=dEZFY0twR0FYM2J
UWnIxbnhlN1FPUmc6MQ#gid=0
The suggested answers will be given to you later via LMS.
To have easier access to the hyperlinks in these documents, please download the softcopies in LMS.
________________________________________________________________
(1) Understanding the characteristics of a Geostationary Satellite:
A geostationary satellite is a satellite that rotates in a certain orbit such that it is
always positioned vertically above a fixed point on the Earth’s surface.
View the Youtube clip below so as to familiarise yourself on how to navigate Geostationary
EJS.
http://youtu.be/JzfFxJirJ54
Carry out the following steps to deduce the characteristics of
a geostationary satellite.
1) Open the Easy-Java-Simulation (EJS) Open Source
file titled “Geostationary Orbit Model”. This model
allows you to visualise the motion of Earth rotating
about its own axis and a satellite rotating around it.
An ‘Intro Page’ will pop up. The geostationary satellite is 35 700 km from the Earth’s surface.
Close the Intro Page.
1
2) You can zoom in and out of the view on the screen by holding down the ‘shift’ key, while you
‘left-click & hold’ and move your mouse forward and backward.
3) You may also re-orientate your point of view on the model when you ‘left-click & hold’ and
move your mouse forward and backward (without holding down the ‘shift’ key) until you get a
clear 3D perspective as shown below.
4) You may check the Equatorial Plane as shown:
plane of Earth.
to display the equatorial
5) You may check the axes of the Earth and the Satellite as shown:
display both axis of rotation for the Earth and the Satellite.
to
6) You can change the mode of display by checking the different buttons as shown:
7) However please note that before you change the mode of display or check any buttons, you
need to pause the simulation by clicking
button! To reset the simulation, you may click
button.
Procedure
Observation and Answers
Mode 1 shows a geostationary satellite
directly above Singapore.
Play the simulation by clicking
button
and observe the relative positions of the
satellite and Singapore.
What do you notice about the relative
positions of the satellite and Singapore?
Watch ‘t’ changes. Observe the time for one
complete orbit.
The satellite always stays above Singapore as
it rotates around the Earth.
What is the period of rotation of the Earth
about its own axis?
24 hr
What is the period of rotation of the
geostationary satellite?
24 hr
Pause the simulation and check the Free
Body button
to display the
forces acting on both Earth and satellite.
What type of forces are these?
What do you notice about the magnitude of
the forces acting on Earth and on satellite
(indicated by the length of arrow)? Give a
reason for your answer.
Gravitational forces.
The forces have the same magnitude but acting
in opposite directions.
Because of Newton’s 3rd law.
The gravitational force by Earth on the satellite.
What force is keeping the satellite in the
circular orbit?
2
Observation and Answers
Procedure
Mode 2 shows another geostationary
satellite moving above a point in Africa.
Mode 3 shows yet another geostationary
satellite moving above a point in America.
What do you notice about the relative
positions of the satellite and Africa /
America?
The respective satellites always stays above
Africa or America as they rotate around the
Earth.
Determine the periods of rotation of these
geostationary satellites.
24 hr.
Hence why do you think the satellite is
called a “geostationary satellite”, when the
satellite is obviously not stationary?
The satellite always stays above a fixed point
as it rotates around Earth.
It appears to be ‘stationary’ only from the
perspective of the fixed point near or on the
Earth’s equator.
Comment on the angular velocity of the
geostationary satellite with respect to the
angular velocity of the Earth’s rotation.
The angular velocity of the satellite is the same
as that of Earth.
Procedure
Observation and Answers
Mode 4
Check
the
Geostationary
button
to display a sample
geostationary satellite (in red) and another
satellite. Click
button and observe
their motions.
Both satellites have periods equals to 24hr.
What do you notice about the radius of the
satellite’s orbit, compared to the radius of
the geostationary orbit?
Both radii are the same.
State the direction of rotation of the (red)
geostationary satellite.
The direction of the red (geostationary) satellite
is from ………… (West/East)
to ………….(West/East).
State the direction of rotation of the other
satellite.
The direction of the other satellite is
from …………(West/East)
to ………….(West/East).
Does this other satellite stay directly above
a fixed point on Earth?
Explain whether the other satellite is
considered to be geostationary.
No.
This satellite is moving from East to West,
instead of West to East. It is moving in the
opposite direction of the Earth’s rotation and so
is never ‘stationary’ relative to a point on the
Earth’s equator.
3
Mode 5 has a satellite with radius of orbit
around 3 times the radius of the Earth.
This is less than the radius of the sample
geostationary (red) satellite orbit.
Click
button. Observe the motion of
the satellite.
The radius of the satellite’s orbit is shorter than
that of the geostationary orbit.
Determine the period of rotation of the
satellite.
7.2 hr
Does this other satellite stay directly above
a fixed point on Earth?
Explain whether the other satellite is
considered to be geostationary.
No.
The angular speed is greater than that of
Earth’s;
or the period of rotation is shorter than that of
Earth’s.
so it is never ‘stationary’ relative to a point on
the Equator.
Mode 5.1 has a satellite with radius of
orbit around 10 times the radius of the
Earth. This is more than the radius of the
sample geostationary (red) satellite orbit.
Click
button. Observe the motion of
the satellite.
The radius of the satellite’s orbit is greater
than that of the geostationary orbit.
Determine the period of rotation of the
satellite.
48 hr
Does this other satellite stay directly above
a fixed point on Earth?
Explain whether the other satellite is
considered to be geostationary.
No.
The angular speed is more than that of Earth’s;
or the period of rotation is more than that of
Earth’s (more than 24 hr).
So it is never ‘stationary’ relative to a point on
the Equator.
Mode 6
Select this mode and click
button.
Observe the motion of this satellite.
Determine the period of rotation of the
satellite
24 hr
Comment on the axis of rotation for Earth
(green arrow) and the axis of rotation of
the satellite (violet arrow).
Their axes of rotation are not in the same
direction.
Does this other satellite stay directly above
a fixed point on Earth?
Explain whether the other satellite is
considered to be geostationary.
No.
It did not rotate in the equatorial plane.
Its axis of rotation is different from that of
Earth’s.
4
Mode 6.1
Select this mode and click
This is a satellite in polar orbit.
button.
Observe the motion of this satellite and
explain why this is called the polar orbit.
Its plane of rotation is perpendicular to the
equator plane, such that the satellite passes
directly above both the poles.
Does this other satellite stay directly above
a fixed point on Earth?
Explain whether the other satellite is
considered to be geostationary.
No.
It did not rotate in the equatorial plane.
Its axis of rotation is different from that of
Earth’s.
Mode 7
Select this mode and make sure that you
check the Free Body button
.
Click
button and observe the motion
of this satellite carefully.
Does this satellite stay above a fixed point
on Earth?
No.
Is this kind of orbit possible?
Give a reason for your answer.
(Hint: Observe the direction of the force
acting on the satellite.)
This kind of orbit is not possible.
The direction of the centripetal force for the
orbit is not towards the Earth’s centre. Or the
gravitational force acting on the satellite is not
pointing towards the centre of orbit.
To orbit about the Earth, the centripetal force
which is provided by the gravitation force must
point towards the centre of the Earth.
The plane of rotation of the orbit must intersect
the Earth’s centre.
Having completed the above exercises, can you make some conclusions about the conditions for
geostationary orbit?
Summarise the 3 conditions that must be satisfied for a satellite to be considered a
geostationary satellite.
(1) ………………………………………………………………………………………..………………
(2) …………………………………………………………………………………………………..……
(3) ………………………………………………………………………………………………………..
(1) It must travel in the same direction as the rotation of the Earth ie from West to East .
(2) Must be placed vertically above the equator or be in the equatorial plane.
(3) The satellite’s orbital period must be equal to Earth’s rotational period (24 hrs).
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(2) Understanding and appreciating Kepler’s 3rd law through Solar System:
Through much trial and error scrutinising the planetary data, Kepler was able to find a connection
between the period T of orbits of the planets in the Solar System and its mean radii R of orbits about
the Sun which he set out in his 3rd Law about planetary system.
Now put yourself in the shoes of Kepler and try to find out the connection yourself. You have the EJS
below at your disposal. Scroll down the Intro Page to familiarise yourself with the information.
https://dl.dropbox.com/u/44365627/lookangEJSworkspace/export/ejs_KeplerSystem3rdLaw03.jar
Carry out the following steps to tabulate period of revolution T and radius r of the orbit, for the 5
planets, namely Mercury, Venus, Mars, Jupiter and Saturn:
1) Go to the URL stated above or open the Easy-Java-Simulation (EJS) Open Source file titled
“Kepler System Model”. This model allows you to observe the orbit of different planets (one
orbit at a time), around the Sun.
2) Check the buttons for (i) labels – to display the descriptive texts for planets; (ii) Earth trail – to
display the orbital path of Earth; (iii) Planet trail – to display the orbital path of planet, as shown:
.
3) Select the planet you wish to observe using the ‘Select Planet’ button, as shown below. Start
with planet Mercury (the planet closest to Sun, with the shortest radius of orbit).
4) Orientate your point of view by ‘left-click & hold’ on the model and move your mouse to view
the Solar System in different 3D perspectives.
5) The radius of orbit for Earth rE is about 1 au. (in astronomical unit, 1 au = 1.496 × 1011 m), The
radius of orbit (in au) for the planet r and duration of orbit (in years), are displayed as shown
below.
radius of orbit for planet, r (in au)
Duration of orbit (in yr)
radius of orbit for Earth, rE (in au)
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6) Check the boxes for rE and r to display the radii of orbits. The time step can be adjusted by
using the slider
to control the speed of simulation (how fast the
simulation runs, not the angular speed of planet!). To watch the simulation slowly at step by
step run, click the
button one click at a time.
7) Record the average radius of orbit r for planet Mercury and convert the value from ‘au’ to
‘m’.
8) To determine the period of orbit T for planet Mercury, ensure that the model is reset by clicking
the
clicking
button such that the duration of orbit starts from 0.00 year. Play the simulation by
button and click
button when planet Mercury has rotated exactly one
round. (You may also click the
button one click at a time to ensure that planet Mercury
complete exactly one round) Record the duration of orbit (for one round) as the period of orbit
T (in years) for planet Mercury. Change T in years to seconds. 1 year = 365x24x3600s.
9) Repeat step 2 to 8 for the other four planets, namely Venus, Mars, Jupiter and Saturn.
10) Tabulate the results as follows. Notice from the EJS, the radius of orbit of a planet is not
constant because the orbit is elliptical and not circular. So here we have taken the average
radius for r . To save time, the values for the average radius r for each planetary orbit are
given.
Planet
T/year
T/s
r / au
r/m
lg (r/m)
lg (T/s)
Mercury
0.24
7.57E+06
0.388
5.80E+10
10.764
6.879
Venus
0.723
1.08E+11
11.034
Mars
1.524
2.28E+11
11.358
Jupiter
5.209
7.79E+11
11.892
Saturn
9.527
1.43E+12
12.154
1 astronomical unit (average radius of Earth’s Orbit round the Sun)= 1 au = 1.496 × 10 11 m
1 year = 365×24×3600s.
Note : 5.80E+10 m = 5.80 × 1010 m
Assume the period of revolution T is related to the radius r of the orbit by the equation
T = A r n where A and n are constants.
a)
Draw a suitable graph to deduce the values of ‘n’.
b)
Kepler’s 3rd law states that “the square of the revolutionary periods (T) is directly
proportional to the cube of their semi-major axes (r).”
Check if your value of n found in (a) is consistent with Kepler’s 3rd Law (n = 1.5).
You can use Excel to help you plot your graph to get the relation and the value of n.
--- End of worksheet ---
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Solution from Excel
Planet
T/year
T/s
r / au
r/m
lg (r/m)
lg (T/s)
Mercury
0.24
7.57E+06
0.388
5.80E+10
10.764
6.879
Venus
0.61
1.92E+07
0.723
1.08E+11
11.034
7.284
Mars
1.87
5.90E+07
1.524
2.28E+11
11.358
7.771
Jupiter
11.8
3.72E+08
5.209
7.79E+11
11.892
8.571
Saturn
29.5
9.30E+08
9.527
1.43E+12
12.154
8.969
lg (T/s) vs lg (r/m)
9.500
9.000
y = 1.5024x - 9.2927
R² = 1
Saturn
Jupiter
8.500
8.000
Mars
7.500
7.000
Venue
Mercury
6.500
10.500
11.000
11.500
Since n = 1.5 (ie 3/2) , Kepler’s 3rd law is verified.
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12.000
12.500