## 7.4.3 Energy of a Satellite in circular orbit in terms of distance from Earth, r JavaScript HTML5 Applet Simulation Model

- Details
- Parent Category: 02 Newtonian Mechanics
- Category: 08 Gravity
- Created: Thursday, 20 August 2015 15:00
- Last Updated: Wednesday, 15 August 2018 17:10
- Published: Saturday, 02 April 2016 15:12
- Hits: 5256

### About

### 7.4.3 Energy of a Satellite in circular orbit in terms of distance from Earth, r

A satellite in orbit possesses kinetic energy, KE, (by virtue of its speed motion) and

gravitational potential energy, PE, (by virtue of its position within the
Earth’s gravitational field).

Hence, total energy of a satellite,

TE = PE + KE

$TE=\left(m\right)(-\genfrac{}{}{0.1ex}{}{GM}{r})+\genfrac{}{}{0.1ex}{}{1}{2}m{v}^{2}$ --- Equation (1)

Recall that for a satellite in circular orbit, its gravitational force $F=\frac{G{M}_{1}{M}_{2}}{{r}^{2}}$ acts as the centripetal
force:

ΣF = ma

$\genfrac{}{}{0.1ex}{}{GmM}{{r}^{2}}=\frac{m{v}^{2}}{r}$

simplifying and multiply by

$\frac{GmM}{2r}=\frac{1}{2}m{v}^{2}=KE$--- Equation (2)

Substituting equation (2) into (1),

Hence total energy of a satellite, TE = PE + KE

$TE=\left(m\right)(-\genfrac{}{}{0.1ex}{}{GM}{r})+\genfrac{}{}{0.1ex}{}{GmM}{2r}=-\genfrac{}{}{0.1ex}{}{GmM}{2r}$

A typical graph showing the relationship between PE, KE and TE with
respect to the distance, r, from centre of Earth, O, is as shown.

Assuming mass of object m_{object} = 1kg, Mass of Earth M_{earth}
= 6.0x10^{24} kg, G =6.67x10^{-11} N m^{2} kg^{-2}
, The potential energy PE =
$PE=-\genfrac{}{}{0.1ex}{}{GmM}{r}=\genfrac{}{}{0.1ex}{}{6.67x1{0}^{-11}\left(1\right)\left(6.0x1{0}^{24}\right)}{r}=-\genfrac{}{}{0.1ex}{}{4x1{0}^{14}}{r}$

Similarly, determine the equation/model that describes KE and TE and key them into the simulation to test your understanding

$$

$KE=\genfrac{}{}{0.1ex}{}{GmM}{2r}=\genfrac{}{}{0.1ex}{}{6.67x1{0}^{-11}\left(1\right)\left(6.0x1{0}^{24}\right)}{2r}=\genfrac{}{}{0.1ex}{}{4x1{0}^{14}}{2*r}$

$TE=-\genfrac{}{}{0.1ex}{}{GmM}{2r}=-\genfrac{}{}{0.1ex}{}{6.67x1{0}^{-11}\left(1\right)\left(6.0x1{0}^{24}\right)}{2r}=-\genfrac{}{}{0.1ex}{}{4x1{0}^{14}}{2*r}$

### 7.4.3.1 Model:

### Translations

Code | Language | Translator | Run | |
---|---|---|---|---|

### Software Requirements

Android | iOS | Windows | MacOS | |

with best with | Chrome | Chrome | Chrome | Chrome |

support full-screen? | Yes. Chrome/Opera No. Firefox/ Samsung Internet | Not yet | Yes | Yes |

cannot work on | some mobile browser that don't understand JavaScript such as..... | cannot work on Internet Explorer 9 and below |

### Credits

Todd Timberlake, this remixed version is by lookang (This email address is being protected from spambots. You need JavaScript enabled to view it.); lookang; fu-kwun hwang

### end faq

## App

https://play.google.com/store/apps/details?id=com.ionicframework.gravity081app859103

### Versions

- http://physics.weber.edu/schroeder/software/NewtonsCannon.html
- http://iwant2study.org/lookangejss/02_newtonianmechanics_7gravity/ejs/ejs_model_NewtonsMountainwee06.jar

### Other Resources

- http://physics.weber.edu/schroeder/software/NewtonsCannon.html
- https://scratch.mit.edu/projects/898903/ Launch Simulator by Sushiboy6

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