EJSS Two Body Circular Orbits Model

EJSS Two Body Circular Orbits Model by Wolfgang Christian, customized by lookang for A level Physics.

http://weelookang.blogspot.com/2014/05/ejss-two-body-circular-orbits-model.html
https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_TwoBodyOrbitswee/TwoBodyOrbitswee_Simulation.xhtml
source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_TwoBodyOrbitswee.zip
authors: Wolfgang Christian, this version by lookang
author of EJS 5: Paco.

Two Body Orbit Theory by Wolfgang Christian

Because two objects (e.g., a star and a planet) with masses

m1 and

m2 interacting gravitationally orbit a common center of mass (0,0), we can determine their relative positions of the objects if their mass ratio

mr=m1m2

is known.

Placing the center of mass (barycenter) at the (0,0) center of the coordinate system, the ratio or the distance from the center is inversely proportional to the object mass ratio

m1m2=r2r1

where

r=r1+r2 is the separation.

In other words, if

m1 is twice as massive as

m2, then the two objects will be on opposite sides of the origin and

m1 will be half as far from the origin as

m2. This ratio relation is also observed when using a lever or seesaw to balance two masses.

when

m1=1,m2=1 implies

r2r1=1

when

m1=1,m2=0.5 implies

r2r1=2

when

m1=1,m2=0.25 implies

r2r1=4

when

m1=1,m2=0.1 implies

r2r1=10

The elliptical motion of two gravitationally attracting objects about their center of mass is a classical mechanics problem and was first solved by Newton. This two-body problem can be re-formulated as the motion of a single object (if you observe carefully, it is white color of the model) with a mass

μ=m1m2m1+m2

that is attracted toward the origin and obeys Newton's second law:

F=μa=Gm1m2r2

in x and y axis,

Fx=μax=−xGm1m2r3

Fy=μay=−yGm1m2r3

Substituting for

μ gives the following differential equation for the separation vector

a=G(m1+m2)r2

in x and y axis,

ax=−xG(m1+m2)r3

ay=−yG(m1+m2)r3

Ordinary Differential Equations:

dxdt=vx

dydt=vy

dvxdt=ax=−xG(m1+m2)r3

dvydt=ay=−yG(m1+m2)r3

where

totalMass=redMass+blueMass=m1+m2

GM=G=4π2

rcubed=r3

The choice of units determines the numerical value of the gravitational constant G. In this simulation, we compute the orbit using time in Earth years and distance in Earth radii (astronomical units). Using these units, the value of

G=4π2

rather than the MKS value of

G=6.67300×10−11m3kgs2.

Motion of Mass 1 and 2

To get the motion of the 2 bodies, says call them body 1 and body 2 where

xblue=x2=xmr1+mr

yblue=y2=ymr1+mr

xred=x1=−x11+mr

yred=y1=−y11+mr

vblue=vx1=vmr1+mr

vblue=vy1=vmr1+mr

vred=vx2=−v11+mr

vred=vy2=−v11+mr

ablue=ax1=amr1+mr

ablue=ay1=amr1+mr

ared=ax2=−a11+mr

ared=ay2=−a11+mr

where

mr=m1m2=mredmblue

the velocities can be determined using the following

v1=v2x1+v2y1−−−−−−−√=v2x+v2y−−−−−−√mr1+mr

v2=v2x2+v2y2−−−−−−−√=v2x+v2y−−−−−−√11+mr

similarly for the acceleration,

a1=a2x1+a2y1−−−−−−−√=a2x+a2y−−−−−−√mr1+mr

a2=a2x2+a2y2−−−−−−−√=a2x+a2y−−−−−−√11+mr

similarly the forces,

F1=m1a1

F2=m2a2

EJSS Two Body Circular Orbits Model by Wolfgang Christian

The Two-Body Orbits JavaScript Model shows the motion of two objects (e.g., binary star or moon-planet system) interacting via Newton's law of universal gravitation. It is designed to teach physics, Earth science, and environmental science topics by showing the spatial path of objects around their common center of mass (barycenter). Default units are chosen for Earth orbit about our Sun so that the distance unit is one astronomical unit and the time unit is one year.

Credits:

The Two-Body Orbits JS Model was developed by Wolfgang Christian using the Easy Java Simulations version 5 (EJS 5) modeling tool. Although EJS is a Java program, EJS 5 can create stand alone JavaScript programs that run in almost any PC or tablet browser.

Information about EJS is available at: <http://www.um.es/fem/Ejs/> and in the OSP ComPADRE collection <http://www.compadre.org/OSP/>.

Problem 1:

Example 1 (N84/P2/Q7) Binary Stars
Two stars of equal mass M move with constant speed v in a circular orbit of radius R about their common centre of mass. What is the net force on each star?

(N84/P2/Q7) Binary Stars

GM24R2

Mv22R

zero

D

2Mv2R

GM2R2

Simulation Equivalent:

Example Simulation Equivalent (N84/P2/Q7) Binary Stars
Two stars of equal mass M move with constant speed v in a circular orbit of radius R about their common centre of mass. What is the net force on each star?
http://weelookang.blogspot.com/2014/05/ejss-two-body-circular-orbits-model.html
https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_TwoBodyOrbitswee/TwoBodyOrbitswee_Simulation.xhtml
source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_TwoBodyOrbitswee.zip
authors: Wolfgang Christian, this version by lookang
author of EJS 5: Paco.

Working

Using Newton's Gravitational Law in Circular Motion

F=GM1M2(R1+R2)2=GMM4R2

Answer is A.

Problem 2:

Example 2 (N09/I/16) Binary Stars
Two stars of mass M and 2M, a distance 3x apart rotate in circles about their common centre of mass O.
The gravitational force acting on the stars can be written as

kGM2x2

(N09/I/16) Binary Stars Two stars of mass M and 2M, a distance 3x apart rotate in circles about their common centre of mass O. The gravitational force acting on the stars can be written as

kGM2x2
What is the value of k ?

What is the value of k ?

A 0.22

B 0.50

C 0.67

D 2.0

Simulation Equivalent:

(N09/I/16) Binary Stars Two stars of mass M and 2M, a distance 3x apart rotate in circles about their common centre of mass O.The gravitational force acting on the stars can be written as

kGM2x2
What is the value of k ?
http://weelookang.blogspot.com/2014/05/ejss-two-body-circular-orbits-model.html
https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_TwoBodyOrbitswee/TwoBodyOrbitswee_Simulation.xhtml
source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_TwoBodyOrbitswee.zip
authors: Wolfgang Christian, this version by lookang
author of EJS 5: Paco.

Working

Using Newton's Gravitational Law in Circular Motion

F=GM1M2(R1+R2)2=G2MM(3x)2=29GMM(x)2

Answer is A.

"M1=5.9726,M2=0.073242,Approximate_Earth_Moon"

"M1=5.9726,M2=0.073242,Approximate_Earth_Moon" as suggested by soo kok
http://weelookang.blogspot.com/2014/05/ejss-two-body-circular-orbits-model.html
https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_model_TwoBodyOrbitswee/TwoBodyOrbitswee_Simulation.xhtml
source: https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejss_src_TwoBodyOrbitswee.zip
authors: Wolfgang Christian, this version by lookang
author of EJS 5: Paco.

using the M1 = 5.9726 , M2 =0.073242, the approximate earth moon system can be simulated but the values for the time is not correct. only the relative values are accurate!

Changes made by lookang include

added the invisible mass for which the binary system's motion are derived from.
added the velocities v1x, v1y and v2x v2y
added velocities checkbox
added units labels clearly showing units a.u. and year
added combo drop-down list for ease of use
added numerical values for sense making such as F1, F2, a1, a2, R1 and R2 v1 and v2
made forces a bit longer to see more clearly