Gravitational Force - Revision history

Lduffy8: /* Difficult */

2019-06-14T19:28:35Z

Difficult

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	::<math>\vec{\mathbf{r}} = \vec{\mathbf{r_M}}-\vec{\mathbf{r_E}}</math>		::<math>\vec{\mathbf{r}} = \vec{\mathbf{r_M}}-\vec{\mathbf{r_E}}</math>
	::<math>\vec{\mathbf{r}}</math> = <<math>L,h,0></math>-<<math>L,0,0></math>		::<math>\vec{\mathbf{r}}</math> = <<math>L,h,0></math>-<<math>L,0,0></math>
	::<math>\vec{\mathbf{r}}</math> = <<math>0,h,0> m</math>		::<math>\vec{\mathbf{r}}</math> = <<math>0,h,0> </math>

	:: <math>\|\vec{\mathbf{r}}\| = \sqrt{(0^2+h^2+0^2)}</math>		:: <math>\|\vec{\mathbf{r}}\| = \sqrt{(0^2+h^2+0^2)}</math>
	:: <math>\|\vec{\mathbf{r}}\| = h </math> ~~''m''~~		:: <math>\|\vec{\mathbf{r}}\| = h </math>

	::<math>\mathbf{\hat{r}} = \frac {\vec{\mathbf{r}}}{\|\vec{\mathbf{r}}\|}</math>		::<math>\mathbf{\hat{r}} = \frac {\vec{\mathbf{r}}}{\|\vec{\mathbf{r}}\|}</math>

Lduffy8: /* Difficult */

2019-06-14T19:27:21Z

Difficult

← Older revision		Revision as of 15:27, 14 June 2019
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	::<math> \|\vec{\mathbf{F}}_{g_1}\| = G \frac{m_E m_M}{h^2} </math>		::<math> \|\vec{\mathbf{F}}_{g_1}\| = G \frac{m_E m_M}{h^2} </math>

	::<math> \vec{\mathbf{F}}_{g_1} = -\|\vec{\mathbf{F}}_{g_1}\|*{mathbf{\hat{r}}}</math>		::<math> \vec{\mathbf{F}}_{g_1} = -\|\vec{\mathbf{F}}_{g_1}\|*{\mathbf{\hat{r}}}</math>
	::<math> \vec{\mathbf{F}}_{g_1} = <0,-G \frac{m_E m_M}{h^2},0> N </math>		::<math> \vec{\mathbf{F}}_{g_1} = <0,-G \frac{m_E m_M}{h^2},0> N </math>

Lduffy8: /* See also */

2019-06-14T19:25:51Z

Lduffy8: /* Videos */

2019-06-14T19:25:37Z

Videos

← Older revision		Revision as of 15:25, 14 June 2019
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	::<math>\vec{\mathbf{F}}_{net} = <0,-G \frac{m_E m_M}{h^2},0> +<-G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0></math>		::<math>\vec{\mathbf{F}}_{net} = <0,-G \frac{m_E m_M}{h^2},0> +<-G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0></math>
	::<math>\vec{\mathbf{F}}_{net} = <-G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_E m_M}{h^2}-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0> N </math>		::<math>\vec{\mathbf{F}}_{net} = <-G \frac{m_S m_M L}{(L^2+h^2)^{3/2}},-G \frac{m_E m_M}{h^2}-G \frac{m_S m_M h}{(L^2+h^2)^{3/2}},0> N </math>

	~~== Videos ==~~
	~~:* This video here is much like the experiment done in lecture, and attempts to show how gravity acts on small objects by using a conical shape [https://www.youtube.com/watch?v=MTY1Kje0yLg Watch].~~

	~~:* [https://www.youtube.com/watch?v=p_o4aY7xkXg Here] is a short video that reinforces the concepts laid out above.~~

	==Connectedness==		==Connectedness==

Lduffy8: /* History */

2019-06-14T19:25:09Z

History

← Older revision		Revision as of 15:25, 14 June 2019
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	== History ==		== History ==
	[[File:SirIsaacNewton.jpg ‎\|150px\|thumb\|right\|Portrait of Isaac Newton]]		[[File:SirIsaacNewton.jpg ‎\|150px\|thumb\|right\|Portrait of Isaac Newton]]
	Sir Isaac Newton, the 17<sup>th</sup> century English scientist is considered one of the most important minds of the Scientific Revolution. He is popularly associated with discovering the force of gravity but scientists had been studying gravity for centuries before Newton. Based on Galileo’s and Kepler’s works on heliocentric solar system, Newton published “Principia” in 1687, in which he explained that the force that causes the planets to orbit the Sun is the same force that makes objects fall to and remain on the surface of the Earth. Newton explained that gravity is a force that attracts different masses and is inversely proportional to the square distance between the objects. However ~~Newtons~~ did not fully understand how gravitational force acted from one object to another over vast distances due to the limited tools at his disposal. Despite this flaw, the law he developed has been used for more than three centuries and is still used without issue ~~by all scientists~~. Albert Einstein rejected it in his explanation of the Theory of Relativity, saying that gravitational ~~forces~~ is ~~just~~ a ~~product~~ of the curvature of ~~space0time~~. ~~Even Einstein could not fully~~ explain ~~gravitational force~~, ~~showing how difficult~~ a ~~concept it was~~ for ~~Newton to develop~~.		Sir Isaac Newton, the 17<sup>th</sup> century English scientist is considered one of the most important minds of the Scientific Revolution. He is popularly associated with discovering the force of gravity but scientists had been studying gravity for centuries before Newton. Based on Galileo’s and Kepler’s works on heliocentric solar system, Newton published “Principia” in 1687, in which he explained that the force that causes the planets to orbit the Sun is the same force that makes objects fall to and remain on the surface of the Earth. Newton explained that gravity is a force that attracts different masses and is inversely proportional to the square distance between the objects. However Newton did not fully understand how gravitational force acted from one object to another over vast distances due to the limited tools at his disposal. Despite this flaw, the law he developed has been used for more than three centuries and is still used mostly without issue to approximate the motion of bodies acting under gravitational force. Albert Einstein rejected it in his explanation of the Theory of Relativity, saying that the gravitational force is a result of the curvature of space-time. This theory explains a number of phenomena we observe around extremely massive bodies, including the curving of light around such bodies.

			While Newton's theory of gravity falls apart under scrutiny near extremely massive bodies, and cannot explain some of the phenomena we see around these bodies, it is a good enough approximation for most situations that are encountered.

	== See also ==		== See also ==

Lduffy8: /* Examples */

2019-06-14T19:21:47Z

Examples

← Older revision		Revision as of 15:21, 14 June 2019
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	The solution of the problem involves substituting known values of G (<math>6.673 x 10^{-11}</math>N m<sup>2</sup>/kg<sup>2</sup>), m<sub>1</sub> <math>6 x 10^ {24}</math> kg, m<sub>2</sub> <math>7 x 10^ {22}</math> kg and d <math><2.8 x 10^8,0,-2.8 x 10^8></math> m m into the universal gravitation equation and solving for F<sub>grav</sub>. The solution is as follows:		The solution of the problem involves substituting known values of G (<math>6.673 x 10^{-11}</math>N m<sup>2</sup>/kg<sup>2</sup>), m<sub>1</sub> <math>6 x 10^ {24}</math> kg, m<sub>2</sub> <math>7 x 10^ {22}</math> kg and d <math><2.8 x 10^8,0,-2.8 x 10^8></math> m m into the universal gravitation equation and solving for F<sub>grav</sub>. The solution is as follows:
	[[File:Force_earth_moon.png\|~~700px~~\|thumb\|right\|Gravitational Force Between Moon and Earth]]		[[File:Force_earth_moon.png\|400px\|thumb\|right\|Gravitational Force Between Moon and Earth]]

	: '''(a)''' The position vector of Moon relative to Earth is,		: '''(a)''' The position vector of Moon relative to Earth is,

Lduffy8: /* Middling */

2019-06-14T19:21:31Z

Middling

← Older revision		Revision as of 15:21, 14 June 2019
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	The solution of the problem involves substituting known values of G (<math>6.673 x 10^{-11}</math>N m<sup>2</sup>/kg<sup>2</sup>), m<sub>1</sub> <math>6 x 10^ {24}</math> kg, m<sub>2</sub> <math>7 x 10^ {22}</math> kg and d <math><2.8 x 10^8,0,-2.8 x 10^8></math> m m into the universal gravitation equation and solving for F<sub>grav</sub>. The solution is as follows:		The solution of the problem involves substituting known values of G (<math>6.673 x 10^{-11}</math>N m<sup>2</sup>/kg<sup>2</sup>), m<sub>1</sub> <math>6 x 10^ {24}</math> kg, m<sub>2</sub> <math>7 x 10^ {22}</math> kg and d <math><2.8 x 10^8,0,-2.8 x 10^8></math> m m into the universal gravitation equation and solving for F<sub>grav</sub>. The solution is as follows:
	[[File:~~Gravitation_force_earth_moon~~.~~gif~~\|700px\|thumb\|right\|Gravitational Force Between Moon and Earth]]		[[File:Force_earth_moon.png\|700px\|thumb\|right\|Gravitational Force Between Moon and Earth]]

	: '''(a)''' The position vector of Moon relative to Earth is,		: '''(a)''' The position vector of Moon relative to Earth is,

Lduffy8: /* Simple */

2019-06-14T19:10:43Z

Simple

← Older revision		Revision as of 15:10, 14 June 2019
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	::<math> \|\mathbf{F_{g}}\| = G \frac{m_E m_M}{r^2} </math>		::<math> \|\mathbf{F_{g}}\| = G \frac{m_E m_M}{r^2} </math>
	::<math> \|\mathbf{F_{g}}\| = 6.7x10^{-11} \frac{6e24 * 70}{6.38 x 10^6} </math>		::<math> \|\mathbf{F_{g}}\| = 6.7x10^{-11} \frac{6e24 * 70}{(6.38 x 10^6)^2} </math>
	::<math> \|\mathbf{F_{g}}\| =686 N </math>		::<math> \|\mathbf{F_{g}}\| =686 N </math>

Lduffy8: /* Vector Form */

2019-06-14T19:09:51Z

Vector Form

← Older revision		Revision as of 15:09, 14 June 2019
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	::* Calculate the unit vector <math>-\mathbf{\hat{r}}</math> =<math>\vec{\mathbf{r}}</math> /<math>\|\vec{\mathbf{r}}\|</math>		::* Calculate the unit vector <math>-\mathbf{\hat{r}}</math> =<math>\vec{\mathbf{r}}</math> /<math>\|\vec{\mathbf{r}}\|</math>
	::* Multiply the Magnitude, <math>\|\vec{\mathbf{F}}_{grav}\|</math>, by the unit vector.		::* Multiply the Magnitude, <math>\|\vec{\mathbf{F}}_{grav}\|</math>, by the unit vector.


	~~Gravitational Force near the surface of the Earth is calculated by using:~~
	~~::<math>\vec{\mathbf{F}}_{grav} = mg</math>~~
	~~:where,~~
	~~::* ''F'' is the force;~~
	~~::* ''g'' is the magnitude of gravitational field near the Earth's surface;~~
	~~::* ''m'' is the mass of the object.~~
	~~This cannot be used for objects far from Earth or when determining the gravitational force from a different planet.~~

	===A Computational Model===		===A Computational Model===

Lduffy8: /* Vector Form */

2019-06-14T19:09:21Z

Vector Form

← Older revision		Revision as of 15:09, 14 June 2019
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	::r sub x, y, and z represent respectively the x, y, and z distances to the point in question.		::r sub x, y, and z represent respectively the x, y, and z distances to the point in question.

	The direction of the gravitational force on object 2 by object 1 is specified by a unit vector,<math>-\mathbf{\hat{r}}</math>, with a minus sign in front of it. The minus sign shows that the force on object 2 by 1 is in the opposite direction to <math>\mathbf{\hat{r}}</math>.		The direction of the gravitational force on object 2 by object 1 is specified by a unit vector,<math>-\mathbf{\hat{r}}</math>, with a minus sign in front of it. The minus sign shows that the force on object 2 by 1 is in the opposite direction to <math>\mathbf{\hat{r}}</math>. Because the gravitational force is an attractive force between two bodies, the magnitude of the gravitational force on one object in the duo is the same as the magnitude of the gravitational force on the other object, but in the opposite direction.
	[[File:Gravitational_field.png\|200px\|thumb\|right\|Gravitational Field around the Earth. Near the surface of the Earth an object m experience a force of mg. The g of the field decreases with increase in distance from the Earth]]		[[File:Gravitational_field.png\|200px\|thumb\|right\|Gravitational Field around the Earth. Near the surface of the Earth an object m experience a force of mg. The g of the field decreases with increase in distance from the Earth]]
	Steps for calculating Gravitational Force		Steps for calculating Gravitational Force

← Older revision		Revision as of 15:25, 14 June 2019
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	:* For more information and applications pertaining to the gravitational force visit [http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html here]		:* For more information and applications pertaining to the gravitational force visit [http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html here]

			:* This video here is much like the experiment done in lecture, and attempts to show how gravity acts on small objects by using a conical shape [https://www.youtube.com/watch?v=MTY1Kje0yLg Watch].

			:* [https://www.youtube.com/watch?v=p_o4aY7xkXg Here] is a short video that reinforces the concepts laid out above.

	===Further reading===		===Further reading===

← Older revision		Revision as of 15:10, 14 June 2019
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	::<math> \|\mathbf{F_{g}}\| = G \frac{m_E m_M}{r^2} </math>		::<math> \|\mathbf{F_{g}}\| = G \frac{m_E m_M}{r^2} </math>
	::<math> \|\mathbf{F_{g}}\| = 6.7x10^{-11} \frac{6e24 * 70}{6.38 x 10^6} </math>		::<math> \|\mathbf{F_{g}}\| = 6.7x10^{-11} \frac{6e24 * 70}{(6.38 x 10^6)^2} </math>
	::<math> \|\mathbf{F_{g}}\| =686 N </math>		::<math> \|\mathbf{F_{g}}\| =686 N </math>