Orbital Dynamics - Revision history

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2015-06-16T01:04:09Z

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2015-06-16T00:52:05Z

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Farred: correct format

2015-06-15T18:17:41Z

correct format

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2015-06-15T18:00:17Z

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2015-06-14T15:01:51Z

correct spelling

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2015-06-14T14:56:40Z

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2015-06-14T14:40:54Z

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2015-06-06T04:06:23Z

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Farred: I could not get the proper formula to work with math coding but the proper formula is more important than math coding

2012-05-05T21:39:29Z

I could not get the proper formula to work with math coding but the proper formula is more important than math coding

76.24.108.150 at 12:24, 5 May 2012

2012-05-05T12:24:37Z

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 === Example ===
-If we want to give a small rocket ship an impulsive acceleration with a mass driver on the moon so that it enters hohmann transfer to 30 km altitude and then circularize the orbit at altitude, we start by getting data.  The radius of the moon is 1737.100 km.  For the purpose of the example we keep three more significant digits than the source of the data justifies to help keep track of calculations in detecting error and to avoid loss of significance in calculation.  The superfluous digits will be stripped away for the final answer.  The surface gravity is 1.622000 m/sec<sup>2</sup>.
+If we want to give a small rocket ship an impulsive acceleration with a [[Mass_Drivers|mass driver]] on the moon so that it enters hohmann transfer to 30 km altitude and then circularize the orbit at altitude, we start by getting data.  The radius of the moon is 1737.100 km.  For the purpose of the example we keep three more significant digits than the source of the data justifies to help keep track of calculations in detecting error and to avoid loss of significance in calculation.  The superfluous digits will be stripped away for the final answer.  The surface gravity is 1.622000 m/sec<sup>2</sup>.
 r<sub>s</sub> = 1737.100 km

@@ Line 129: / Line 129: @@
 V<sub>o+30</sub> - V<sub>ta</sub> = 7.138 m/sec = circularization impulse
-−
+==== Final Answer ====
 [[Category:Missions]]

← Older revision		Revision as of 18:17, 15 June 2015
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	V<sub>o+30</sub> = (g<sub>+30</sub> * r<sub>+30</sub>)<sup>0.5</sup>		V<sub>o+30</sub> = (g<sub>+30</sub> * r<sub>+30</sub>)<sup>0.5</sup>
		+
		+
	V<sub>o+30</sub> = (1.567394 m/sec <sup>2</sup> * 1.767100 E6 m)<sup>0.5</sup> = 1664.254 m/sec		V<sub>o+30</sub> = (1.567394 m/sec <sup>2</sup> * 1.767100 E6 m)<sup>0.5</sup> = 1664.254 m/sec

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 In the case in which the tabulated data available is surface gravity and radius, as is the case for some amateurs dealing with moons and planets, then surface orbital velocity can be found as the square root of the product of the radius and the surface gravity.
-V<sub>os</sub> = (g<sub>s</sub> * r) ^ 0.5
+V<sub>os</sub> = (g<sub>s</sub> * r<sub>s</sub>)<sup>0.5</sup>
@@ Line 44: / Line 44: @@
 g<sub>a</sub> = g<sub>s</sub> * (r<sub>s</sub><sup>2</sup> / r<sub>a</sub><sup>2</sup>)
-V<sub>oa</sub> = (g<sub>a</sub> * r<sub>a</sub>) ^ 0.5
+V<sub>o+30</sub> - V<sub>ta</sub> = 7.138 m/sec = circularization impulse

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 == Simple Celestial Mechanics for Simple Problems ==
-In the case in which the tabulated data available is surface gravity and radius, as is the case for some amatures dealing with moons and planets, then surface orbital velocity can be found as the square root of the product of the radius and the surface gravity.
+In the case in which the tabulated data available is surface gravity and radius, as is the case for some amateurs dealing with moons and planets, then surface orbital velocity can be found as the square root of the product of the radius and the surface gravity.
 V<sub>os</sub> = (g<sub>s</sub> * r) ^ 0.5

← Older revision		Revision as of 14:56, 14 June 2015
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	V<sub>os</sub> = (g<sub>s</sub> * r) ^ 0.5		V<sub>os</sub> = (g<sub>s</sub> * r) ^ 0.5
		+
		+
		+	To find circular orbital velocity at some altitude, first find the gravity at that altitude. Then the orbital velocity at that altitude is the square root of the product of gravity at altitude the radius of the orbit.
		+
		+	g<sub>a</sub> = g<sub>s</sub> * (r<sub>s</sub><sup>2</sup> / r<sub>a</sub><sup>2</sup>)
		+
		+	V<sub>oa</sub> = (g<sub>a</sub> * r<sub>a</sub>) ^ 0.5

	{{Physics Stub}}		{{Physics Stub}}

← Older revision		Revision as of 14:40, 14 June 2015
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	These assumptions lead to an elliptical orbit. In fact, both bodies will orbit around a point that is located along the line connecting the center of mass of the two bodies. The location on this line is the center of mass <math>m_2/(m_1+m_2)</math>. For a man-made satellite orbiting the Earth, the center of mass of the combined system is essentially at the center of mass of the earth. For the [[Moon]] orbiting the Earth, the center of mass of the combined system is displaced 2949 miles from the center of mass of the Earth. It is 1014 miles below the surface of the Earth.		These assumptions lead to an elliptical orbit. In fact, both bodies will orbit around a point that is located along the line connecting the center of mass of the two bodies. The location on this line is the center of mass <math>m_2/(m_1+m_2)</math>. For a man-made satellite orbiting the Earth, the center of mass of the combined system is essentially at the center of mass of the earth. For the [[Moon]] orbiting the Earth, the center of mass of the combined system is displaced 2949 miles from the center of mass of the Earth. It is 1014 miles below the surface of the Earth.
		+

		+	== Simple Celestial Mechanics for Simple Problems ==
		+	In the case in which the tabulated data available is surface gravity and radius, as is the case for some amatures dealing with moons and planets, then surface orbital velocity can be found as the square root of the product of the radius and the surface gravity.
		+
		+	V<sub>os</sub> = (g<sub>s</sub> * r) ^ 0.5

	{{Physics Stub}}		{{Physics Stub}}

← Older revision		Revision as of 04:06, 6 June 2015
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	#No other forces act on the system.		#No other forces act on the system.

−	These assumptions lead to an elliptical orbit. In fact, both bodies will orbit around a point that is located along the line connecting the center of mass of the two bodies. The location on this line is the center of mass <math>m_2/(m_1+m_2)</math>. For a man-made satellite orbiting the Earth, the center of mass of the combined system is essentially at the center of mass of the earth. For the [[Moon]] orbiting the Earth, the center of mass of the combined system is ~~significantly~~ displaced from the center of mass of the Earth.	+	These assumptions lead to an elliptical orbit. In fact, both bodies will orbit around a point that is located along the line connecting the center of mass of the two bodies. The location on this line is the center of mass <math>m_2/(m_1+m_2)</math>. For a man-made satellite orbiting the Earth, the center of mass of the combined system is essentially at the center of mass of the earth. For the [[Moon]] orbiting the Earth, the center of mass of the combined system is displaced 2949 miles from the center of mass of the Earth. It is 1014 miles below the surface of the Earth.

@@ Line 13: / Line 13: @@
 The [[exd:centripetal force|centripetal force]] of an object that is held in rotation around a fixed point is expressed as
-<math>F = \frac{m_{2}V}{r}</math>
+F = (m<sub>2</sub> * V<sup>2</sup>) / r
-−
 where V is the [[exd:linear velocity|linear velocity]] of the object, which in our case is the satellite.  Notice that the [[exd:velocity|velocity]] (and later the period) are independent of the [[exd:mass|mass]] of the satellite.  Combining these two equations together and solving for the velocity of the satellite gives an orbital velocity of a circular orbit as