Part 2 of 3: Capstone: APPB

Part 2 of 3

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Gravity, Energy, and Velocities

We can use some simple equations and calculations we learned in class to examine the physics of an object in orbit (we will mostly stick to the satellite mentioned in Part 1).

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Gravitational Acceleration

We know that the acceleration due to gravity on Earth is $-9.81\:\frac{m}{s^2}$ . Let's take a look at Kerbal Space Program and see how it compares (assuming we are on the surface).

$F_g=mg=\frac{-GMm}{r^2}\:\Longrightarrow\:g=\frac{-GM}{r^2}$

: mass of body in orbit; : acceleration due to gravity; : universal gravitational constant; : radius of planet(from center of mass M)

$M=5.29x10^{22}\:kg$ ; $r_{equatorial}=600\:km$ ;

$g=\frac{-\left(6.67x10^{-11}\:\frac{m^3}{k\cdot s^2}\right)\left(5.29x10^{22}\:kg\right)}{\left(600,000\:m\right)}\:=\:-9.81\:\frac{m}{s^2}$

While we're at it, we can do the same for the Mun (KSP's version of the Moon, it is pronounced the same way).

$M=9.76x10^{20}\:kg$ ; $r_{equatorial}=200\:km$ ;

$g=\frac{-\left(6.67x10^{-11}\:\frac{m^3}{k\cdot s^2}\right)\left(9.76x10^{20}\:kg\right)}{\left(200,000\:m\right)}\:=\:-1.63\:\frac{m}{s^2}$

It turns out that $g_{Mun}=g_{Moon}$

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Orbital Mechanical Energy

We can find the total energy in a spacecraft using some simple equations we've already learned.

$E=U_g+K=\frac{-GMm}{r}+\frac{1}{2}mv^2$

is the orbital velocity and is given by the equation $v_{orbital}=\sqrt{\frac{\mu}{r}}$ with $\mu$ being the gravitational parameter of the parent body ( $\mu=GM$ ).

If we don't want to solve for $v_{orbital}$ we can remove it using the equation for the net force of of circular motion ( $F_{Net}=\frac{mv^2}{r}$ ). In this case we are assuming the orbit of the craft is completely circular and F_g from the parent body is the only acting force.

$F_{Net}=\frac{mv^2}{r}=\frac{GMm}{r^2}\:\:\Longrightarrow\:\:mv^2=\frac{GMm}{r}$ We can now substitute this final value into the equation.

$E=\frac{-GMm}{r}+\frac{1}{2}\frac{GMm}{r}=\frac{-GMm}{2r}$

: total energy of the orbit (J); U_g : gravitational potential energy (J); : kinetic energy (J); :universal gravitational constant ( $\frac{m^3}{kg\cdot s^2}$ ); : mass of parent body (kg); : mass of orbiting body (kg); : radius of orbit (m); $V_{orbital}$ : orbital speed (

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We can see how the equations apply to our satellite in Part 1 (assuming that the fuel mass lost is negligible).

For the 100 km orbit ( $m_{satellite}=\text{2,340 kg}$ ):

$E_{100\:km}=\frac{-\left(6.67x10^{-11}\:\frac{m^3}{kg\cdot s^2}\right)\left(5.29x10^{22\:kg}\right)\left(2,340\:kg\right)}{700,000}=-1.18x10^{10}\:J$

$E_{200\:km}=\frac{-\left(6.67x10^{-11}\:\frac{m^3}{kg\cdot s^2}\right)\left(5.29x10^{22\:kg}\right)\left(2,340\:kg\right)}{800,000}=-1.03x10^{10}\:J$

The final answers are negative and the craft is in a stable orbit. If the total energy was , the craft would be on a hyperbolic trajectory and would overpower the parent planet's gravitational pull.

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We can also find the orbital velocity of the craft at the different altitudes.

$v_{orbital}=\sqrt{\frac{\mu}{r}}$

At 100 km:

$v_{orbital}=\sqrt{\frac{\left(6.67x10^{-11}\:\frac{m^3}{kg\cdot s^2}\right)\left(5.29x10^{22\:kg}\right)}{\left(700,000\right)}}=2245.13\frac{m}{s}$

At 200 km:

$v_{orbital}=\sqrt{\frac{\left(6.67x10^{-11}\:\frac{m^3}{kg\cdot s^2}\right)\left(5.29x10^{22\:kg}\right)}{\left(800,000\right)}}=2100.13\frac{m}{s}$

Altitude (including 600 km radius) km	$v_{orbital}$ (m/s)
100	2245.13
200	2100.13
300	1980.02
400	1878.41
500	1791.00
600	1714.75
700	1647.48
800	1587.55
9000	153.72
1,000	1485.02

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[{:content=>"Gravity, Energy, and Velocities \nWe can use some simple equations and calculations we learned in class to examine the physics of an object in orbit (we will mostly stick to the satellite mentioned in Part 1).", :section_type=>"rich_text"}, {:content=>"<hr><hr>", :section_type=>"html"}, {:content=>"Gravitational Acceleration\nWe know that the acceleration due to gravity on Earth is <img class=\"equation_image\" title=\"-9.81\\:\\frac{m}{s^2}\" src=\"/equation_images/-9.81%255C%253A%255Cfrac%257Bm%257D%257Bs%255E2%257D\" alt=\"-9.81\\:\\frac{m}{s^2}\">. Let's take a look at Kerbal Space Program and see how it compares (assuming we are on the surface).\n<img class=\"equation_image\" title=\"F_g=mg=\\frac{-GMm}{r^2}\\:\\Longrightarrow\\:g=\\frac{-GM}{r^2}\" src=\"/equation_images/F_g%253Dmg%253D%255Cfrac%257B-GMm%257D%257Br%255E2%257D%255C%253A%255CLongrightarrow%255C%253Ag%253D%255Cfrac%257B-GM%257D%257Br%255E2%257D\" alt=\"F_g=mg=\\frac{-GMm}{r^2}\\:\\Longrightarrow\\:g=\\frac{-GM}{r^2}\">\n<img class=\"equation_image\" title=\"m\" src=\"/equation_images/m\" alt=\"m\">: mass of body in orbit; <img class=\"equation_image\" title=\"g\" src=\"/equation_images/g\" alt=\"g\">: acceleration due to gravity; <img class=\"equation_image\" title=\"G\" src=\"/equation_images/G\" alt=\"G\">: universal gravitational constant; <img class=\"equation_image\" title=\"r\" src=\"/equation_images/r\" alt=\"r\">: radius of planet(from center of mass M)\n<img class=\"equation_image\" title=\"M=5.29x10^{22}\\:kg\" src=\"/equation_images/M%253D5.29x10%255E%257B22%257D%255C%253Akg\" alt=\"M=5.29x10^{22}\\:kg\">; <img class=\"equation_image\" title=\"r_{equatorial}=600\\:km\" src=\"/equation_images/r_%257Bequatorial%257D%253D600%255C%253Akm\" alt=\"r_{equatorial}=600\\:km\">;\n<img class=\"equation_image\" title=\"g=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{k\\cdot s^2}\\right)\\left(5.29x10^{22}\\:kg\\right)}{\\left(600,000\\:m\\right)}\\:=\\:-9.81\\:\\frac{m}{s^2}\" src=\"/equation_images/g%253D%255Cfrac%257B-%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bk%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25285.29x10%255E%257B22%257D%255C%253Akg%255Cright%2529%257D%257B%255Cleft%2528600%252C000%255C%253Am%255Cright%2529%257D%255C%253A%253D%255C%253A-9.81%255C%253A%255Cfrac%257Bm%257D%257Bs%255E2%257D\" alt=\"g=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{k\\cdot s^2}\\right)\\left(5.29x10^{22}\\:kg\\right)}{\\left(600,000\\:m\\right)}\\:=\\:-9.81\\:\\frac{m}{s^2}\">\nWhile we're at it, we can do the same for the Mun (KSP's version of the Moon, it is pronounced the same way).\n<img class=\"equation_image\" title=\"M=9.76x10^{20}\\:kg\" src=\"/equation_images/M%253D9.76x10%255E%257B20%257D%255C%253Akg\" alt=\"M=9.76x10^{20}\\:kg\">; <img class=\"equation_image\" title=\"r_{equatorial}=200\\:km\" src=\"/equation_images/r_%257Bequatorial%257D%253D200%255C%253Akm\" alt=\"r_{equatorial}=200\\:km\">;\n<img class=\"equation_image\" title=\"g=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{k\\cdot s^2}\\right)\\left(9.76x10^{20}\\:kg\\right)}{\\left(200,000\\:m\\right)}\\:=\\:-1.63\\:\\frac{m}{s^2}\" src=\"/equation_images/g%253D%255Cfrac%257B-%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bk%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25289.76x10%255E%257B20%257D%255C%253Akg%255Cright%2529%257D%257B%255Cleft%2528200%252C000%255C%253Am%255Cright%2529%257D%255C%253A%253D%255C%253A-1.63%255C%253A%255Cfrac%257Bm%257D%257Bs%255E2%257D\" alt=\"g=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{k\\cdot s^2}\\right)\\left(9.76x10^{20}\\:kg\\right)}{\\left(200,000\\:m\\right)}\\:=\\:-1.63\\:\\frac{m}{s^2}\">\nIt turns out that <img class=\"equation_image\" title=\"g_{Mun}=g_{Moon}\" src=\"/equation_images/g_%257BMun%257D%253Dg_%257BMoon%257D\" alt=\"g_{Mun}=g_{Moon}\">", :section_type=>"rich_text"}, {:content=>"<hr><hr>", :section_type=>"html"}, {:content=>"Orbital Mechanical Energy\nWe can find the total energy in a spacecraft using some simple equations we've already learned. \n<img class=\"equation_image\" title=\"E=U_g+K=\\frac{-GMm}{r}+\\frac{1}{2}mv^2\" src=\"/equation_images/E%253DU_g%2BK%253D%255Cfrac%257B-GMm%257D%257Br%257D%2B%255Cfrac%257B1%257D%257B2%257Dmv%255E2\" alt=\"E=U_g+K=\\frac{-GMm}{r}+\\frac{1}{2}mv^2\">\n<img class=\"equation_image\" title=\"v\" src=\"/equation_images/v\" alt=\"v\"> is the orbital velocity and is given by the equation <img class=\"equation_image\" title=\"v_{orbital}=\\sqrt{\\frac{\\mu}{r}}\" src=\"/equation_images/v_%257Borbital%257D%253D%255Csqrt%257B%255Cfrac%257B%255Cmu%257D%257Br%257D%257D\" alt=\"v_{orbital}=\\sqrt{\\frac{\\mu}{r}}\"> with <img class=\"equation_image\" title=\"\\mu\" src=\"/equation_images/%255Cmu\" alt=\"\\mu\"> being the gravitational parameter of the parent body (<img class=\"equation_image\" title=\"\\mu=GM\" src=\"/equation_images/%255Cmu%253DGM\" alt=\"\\mu=GM\">).\nIf we don't want to solve for <img class=\"equation_image\" title=\"v_{orbital}\" src=\"/equation_images/v_%257Borbital%257D\" alt=\"v_{orbital}\"> we can remove it using the equation for the net force of of circular motion (<img class=\"equation_image\" title=\"F_{Net}=\\frac{mv^2}{r}\" src=\"/equation_images/F_%257BNet%257D%253D%255Cfrac%257Bmv%255E2%257D%257Br%257D\" alt=\"F_{Net}=\\frac{mv^2}{r}\">). In this case we are assuming the orbit of the craft is completely circular and <img class=\"equation_image\" title=\"F_g\" src=\"/equation_images/F_g\" alt=\"F_g\"> from the parent body is the only acting force.\n<img class=\"equation_image\" title=\"F_{Net}=\\frac{mv^2}{r}=\\frac{GMm}{r^2}\\:\\:\\Longrightarrow\\:\\:mv^2=\\frac{GMm}{r}\" src=\"/equation_images/F_%257BNet%257D%253D%255Cfrac%257Bmv%255E2%257D%257Br%257D%253D%255Cfrac%257BGMm%257D%257Br%255E2%257D%255C%253A%255C%253A%255CLongrightarrow%255C%253A%255C%253Amv%255E2%253D%255Cfrac%257BGMm%257D%257Br%257D\" alt=\"F_{Net}=\\frac{mv^2}{r}=\\frac{GMm}{r^2}\\:\\:\\Longrightarrow\\:\\:mv^2=\\frac{GMm}{r}\"> We can now substitute this final value into the <img class=\"equation_image\" title=\"K\" src=\"/equation_images/K\" alt=\"K\"> equation.\n<img class=\"equation_image\" title=\"E=\\frac{-GMm}{r}+\\frac{1}{2}\\frac{GMm}{r}=\\frac{-GMm}{2r}\" src=\"/equation_images/E%253D%255Cfrac%257B-GMm%257D%257Br%257D%2B%255Cfrac%257B1%257D%257B2%257D%255Cfrac%257BGMm%257D%257Br%257D%253D%255Cfrac%257B-GMm%257D%257B2r%257D\" alt=\"E=\\frac{-GMm}{r}+\\frac{1}{2}\\frac{GMm}{r}=\\frac{-GMm}{2r}\">\n<img class=\"equation_image\" title=\"E\" src=\"/equation_images/E\" alt=\"E\">: total energy of the orbit (J); <img class=\"equation_image\" title=\"U_g\" src=\"/equation_images/U_g\" alt=\"U_g\">: gravitational potential energy (J); <img class=\"equation_image\" title=\"K\" src=\"/equation_images/K\" alt=\"K\">: kinetic energy (J); <img class=\"equation_image\" title=\"G\" src=\"/equation_images/G\" alt=\"G\">:universal gravitational constant (<img class=\"equation_image\" title=\"\\frac{m^3}{kg\\cdot s^2}\" src=\"/equation_images/%255Cfrac%257Bm%255E3%257D%257Bkg%255Ccdot%2520s%255E2%257D\" alt=\"\\frac{m^3}{kg\\cdot s^2}\" width=\"35\" height=\"32\">); <img class=\"equation_image\" title=\"M\" src=\"/equation_images/M\" alt=\"M\">: mass of parent body (kg); <img class=\"equation_image\" title=\"m\" src=\"/equation_images/m\" alt=\"m\">: mass of orbiting body (kg); <img class=\"equation_image\" title=\"r\" src=\"/equation_images/r\" alt=\"r\">: radius of orbit (m); <img class=\"equation_image\" title=\"V_{orbital}\" src=\"/equation_images/V_%257Borbital%257D\" alt=\"V_{orbital}\">: orbital speed (", :section_type=>"rich_text"}, {:content=>"<hr><hr>", :section_type=>"html"}, {:content=>"We can see how the equations apply to our satellite in Part 1 (assuming that the fuel mass lost is negligible).\nFor the 100 km orbit (<img class=\"equation_image\" title=\"m_{satellite}=\\text{2,340 kg}\" src=\"/equation_images/m_%257Bsatellite%257D%253D%255Ctext%257B2%252C340%2520kg%257D\" alt=\"m_{satellite}=\\text{2,340 kg}\">):\n<img class=\"equation_image\" title=\"E_{100\\:km}=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)\\left(2,340\\:kg\\right)}{700,000}=-1.18x10^{10}\\:J\" src=\"/equation_images/E_%257B100%255C%253Akm%257D%253D%255Cfrac%257B-%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bkg%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25285.29x10%255E%257B22%255C%253Akg%257D%255Cright%2529%255Cleft%25282%252C340%255C%253Akg%255Cright%2529%257D%257B700%252C000%257D%253D-1.18x10%255E%257B10%257D%255C%253AJ\" alt=\"E_{100\\:km}=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)\\left(2,340\\:kg\\right)}{700,000}=-1.18x10^{10}\\:J\"> \n<img class=\"equation_image\" title=\"E_{200\\:km}=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)\\left(2,340\\:kg\\right)}{800,000}=-1.03x10^{10}\\:J\" src=\"/equation_images/E_%257B200%255C%253Akm%257D%253D%255Cfrac%257B-%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bkg%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25285.29x10%255E%257B22%255C%253Akg%257D%255Cright%2529%255Cleft%25282%252C340%255C%253Akg%255Cright%2529%257D%257B800%252C000%257D%253D-1.03x10%255E%257B10%257D%255C%253AJ\" alt=\"E_{200\\:km}=\\frac{-\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)\\left(2,340\\:kg\\right)}{800,000}=-1.03x10^{10}\\:J\">\nThe final answers are negative and the craft is in a stable orbit. If the total energy was <img class=\"equation_image\" title=\"&gt;0\" src=\"/equation_images/%253E0\" alt=\"&gt;0\">, the craft would be on a hyperbolic trajectory and would overpower the parent planet's gravitational pull.", :section_type=>"rich_text"}, {:content=>"<hr><hr>", :section_type=>"html"}, {:content=>"We can also find the orbital velocity of the craft at the different altitudes.\n<img class=\"equation_image\" title=\"v_{orbital}=\\sqrt{\\frac{\\mu}{r}}\" src=\"/equation_images/v_%257Borbital%257D%253D%255Csqrt%257B%255Cfrac%257B%255Cmu%257D%257Br%257D%257D\" alt=\"v_{orbital}=\\sqrt{\\frac{\\mu}{r}}\">\nAt 100 km:\n<img class=\"equation_image\" title=\"v_{orbital}=\\sqrt{\\frac{\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)}{\\left(700,000\\right)}}=2245.13\\frac{m}{s}\" src=\"/equation_images/v_%257Borbital%257D%253D%255Csqrt%257B%255Cfrac%257B%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bkg%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25285.29x10%255E%257B22%255C%253Akg%257D%255Cright%2529%257D%257B%255Cleft%2528700%252C000%255Cright%2529%257D%257D%253D2245.13%255Cfrac%257Bm%257D%257Bs%257D\" alt=\"v_{orbital}=\\sqrt{\\frac{\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)}{\\left(700,000\\right)}}=2245.13\\frac{m}{s}\">\nAt 200 km:\n<img class=\"equation_image\" title=\"v_{orbital}=\\sqrt{\\frac{\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)}{\\left(800,000\\right)}}=2100.13\\frac{m}{s}\" src=\"/equation_images/v_%257Borbital%257D%253D%255Csqrt%257B%255Cfrac%257B%255Cleft%25286.67x10%255E%257B-11%257D%255C%253A%255Cfrac%257Bm%255E3%257D%257Bkg%255Ccdot%2520s%255E2%257D%255Cright%2529%255Cleft%25285.29x10%255E%257B22%255C%253Akg%257D%255Cright%2529%257D%257B%255Cleft%2528800%252C000%255Cright%2529%257D%257D%253D2100.13%255Cfrac%257Bm%257D%257Bs%257D\" alt=\"v_{orbital}=\\sqrt{\\frac{\\left(6.67x10^{-11}\\:\\frac{m^3}{kg\\cdot s^2}\\right)\\left(5.29x10^{22\\:kg}\\right)}{\\left(800,000\\right)}}=2100.13\\frac{m}{s}\">\n \n<table border=\"1\"><tbody>\n<tr>\n<td>\nAltitude (including\n600 km radius) km\n</td>\n<td>\n<img class=\"equation_image\" title=\"v_{orbital}\" src=\"/equation_images/v_%257Borbital%257D\" alt=\"v_{orbital}\"> (m/s)</td>\n</tr>\n<tr>\n<td>100</td>\n<td>2245.13</td>\n</tr>\n<tr>\n<td>200</td>\n<td>2100.13</td>\n</tr>\n<tr>\n<td>300</td>\n<td>1980.02</td>\n</tr>\n<tr>\n<td>400</td>\n<td>1878.41</td>\n</tr>\n<tr>\n<td>500</td>\n<td>1791.00</td>\n</tr>\n<tr>\n<td>600</td>\n<td>1714.75</td>\n</tr>\n<tr>\n<td>700</td>\n<td>1647.48</td>\n</tr>\n<tr>\n<td>800</td>\n<td>1587.55</td>\n</tr>\n<tr>\n<td>9000</td>\n<td>153.72</td>\n</tr>\n<tr>\n<td>1,000</td>\n<td>1485.02</td>\n</tr>\n</tbody></table>", :section_type=>"rich_text"}]

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