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Question Number 98270 by  M±th+et+s last updated on 12/Jun/20
prove  Fg=G((m_1 m_2 )/r^2 )
$${prove} \\ $$$${Fg}={G}\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$
Answered by Rio Michael last updated on 12/Jun/20
Newton:s law of universal gravitation states that: ′′ the force  between two point mass m_1  and  m_2  seperated by a distance r  is directly proportional to the product of thier masses and inversly  proportional to the square of the distance seperating them′′  thus F ∝ m_1 m_2  .....(i)  F ∝ (1/(r^2  )) ......(ii)  ⇒ F ∝ ((m_1 m_2 )/r^2 ) ⇒ F = k ((m_1  m_2 )/r^2 )  ⇒ k = ((Fr^2 )/(m_1 m_2  )) according to the cavendish experiment,  this ratio = 6.67 × 10^(−11)  N m^2  kg^(−2)    in physics this constant = G   ⇒ k = G  thus  F = G ((m_1 m_2 )/r^2 )
$$\mathrm{Newton}:\mathrm{s}\:\mathrm{law}\:\mathrm{of}\:\mathrm{universal}\:\mathrm{gravitation}\:\mathrm{states}\:\mathrm{that}:\:''\:\mathrm{the}\:\mathrm{force} \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{point}\:\mathrm{mass}\:{m}_{\mathrm{1}} \:\mathrm{and}\:\:{m}_{\mathrm{2}} \:\mathrm{seperated}\:\mathrm{by}\:\mathrm{a}\:\mathrm{distance}\:{r} \\ $$$$\mathrm{is}\:\mathrm{directly}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{thier}\:\mathrm{masses}\:\mathrm{and}\:\mathrm{inversly} \\ $$$$\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{seperating}\:\mathrm{them}'' \\ $$$$\mathrm{thus}\:{F}\:\propto\:{m}_{\mathrm{1}} {m}_{\mathrm{2}} \:…..\left({i}\right) \\ $$$${F}\:\propto\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} \:}\:……\left({ii}\right) \\ $$$$\Rightarrow\:{F}\:\propto\:\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{{r}^{\mathrm{2}} }\:\Rightarrow\:{F}\:=\:{k}\:\frac{{m}_{\mathrm{1}} \:{m}_{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{k}\:=\:\frac{{Fr}^{\mathrm{2}} }{{m}_{\mathrm{1}} {m}_{\mathrm{2}} \:}\:\mathrm{according}\:\mathrm{to}\:\mathrm{the}\:\mathrm{cavendish}\:\mathrm{experiment}, \\ $$$$\mathrm{this}\:\mathrm{ratio}\:=\:\mathrm{6}.\mathrm{67}\:×\:\mathrm{10}^{−\mathrm{11}} \:\mathrm{N}\:\mathrm{m}^{\mathrm{2}} \:\mathrm{kg}^{−\mathrm{2}} \: \\ $$$$\mathrm{in}\:\mathrm{physics}\:\mathrm{this}\:\mathrm{constant}\:=\:{G}\: \\ $$$$\Rightarrow\:{k}\:=\:{G} \\ $$$$\mathrm{thus}\:\:{F}\:=\:{G}\:\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{{r}^{\mathrm{2}} }\: \\ $$
Commented by  M±th+et+s last updated on 12/Jun/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by smridha last updated on 13/Jun/20
this is very silly question!!don′t  mind..but it is.we derived  this formula from Newton′s  gravitational law...which is a  statement whoes mathematical  expression is that...you cannot  prove any postulate!!  but it′s true that it can be shown  from Einstein General relativity  with much more approximation.  And once more fact I want to  share that dont call Newton′s   gravitational Law is universal  law...because it does not vallid  for strong gravitational field,  but Einstein′ GTR can  explain   how gravity works...this the  regorus description about  gravity.....      G_(𝛍𝛎) = R_(𝛍ν) −(1/2)Rg_(𝛍𝛎) =((8𝛑G)/c^4 )T  this is called the Einstein′s   tensor eq^n .this is the explanation  of how gravity works...what is  gravity???
$$\boldsymbol{{this}}\:{is}\:\boldsymbol{{very}}\:\boldsymbol{{silly}}\:\boldsymbol{{question}}!!\boldsymbol{{don}}'\boldsymbol{{t}} \\ $$$$\boldsymbol{{mind}}..\boldsymbol{{but}}\:\boldsymbol{{it}}\:\boldsymbol{{is}}.\boldsymbol{{we}}\:\boldsymbol{{derived}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{formula}}\:\boldsymbol{{from}}\:\boldsymbol{{N}}{ew}\boldsymbol{{ton}}'{s} \\ $$$$\boldsymbol{{gravitational}}\:\boldsymbol{{law}}…\boldsymbol{{which}}\:\boldsymbol{{is}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{statement}}\:\boldsymbol{{whoes}}\:\boldsymbol{{mathematical}} \\ $$$$\boldsymbol{{expression}}\:\boldsymbol{{is}}\:\boldsymbol{{that}}…\boldsymbol{{you}}\:\boldsymbol{{cannot}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{any}}\:\boldsymbol{{postulate}}!! \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{it}}'{s}\:\boldsymbol{{true}}\:\boldsymbol{{that}}\:\boldsymbol{{it}}\:\boldsymbol{{can}}\:\boldsymbol{{be}}\:\boldsymbol{{shown}} \\ $$$$\boldsymbol{{from}}\:\boldsymbol{{E}}{i}\boldsymbol{{nstein}}\:\boldsymbol{{G}}{e}\boldsymbol{{neral}}\:\boldsymbol{{relativity}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{much}}\:\boldsymbol{{more}}\:\boldsymbol{{approximation}}. \\ $$$$\boldsymbol{{And}}\:\boldsymbol{{once}}\:\boldsymbol{{more}}\:\boldsymbol{{fact}}\:\boldsymbol{{I}}\:\boldsymbol{{want}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{share}}\:\boldsymbol{{that}}\:\boldsymbol{{dont}}\:\boldsymbol{{call}}\:\boldsymbol{{Newton}}'\boldsymbol{{s}} \\ $$$$\:{gra}\boldsymbol{{vitational}}\:\boldsymbol{{L}}{aw}\:\boldsymbol{{is}}\:\boldsymbol{{universal}} \\ $$$$\boldsymbol{{law}}…\boldsymbol{{because}}\:\boldsymbol{{it}}\:\boldsymbol{{does}}\:\boldsymbol{{not}}\:\boldsymbol{{vallid}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{strong}}\:\boldsymbol{{gravitational}}\:\boldsymbol{{field}}, \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{E}}{in}\boldsymbol{{stein}}'\:\boldsymbol{{G}}{T}\boldsymbol{{R}}\:\boldsymbol{{can}}\:\:\boldsymbol{{explain}}\: \\ $$$$\boldsymbol{{how}}\:\boldsymbol{{gravity}}\:\boldsymbol{{works}}…\boldsymbol{{this}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{regorus}}\:\boldsymbol{{description}}\:\boldsymbol{{about}} \\ $$$$\boldsymbol{{gravity}}….. \\ $$$$\:\:\:\:\boldsymbol{{G}}_{\boldsymbol{\mu\nu}} =\:\boldsymbol{{R}}_{\boldsymbol{\mu}\nu} −\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{Rg}}_{\boldsymbol{\mu\nu}} =\frac{\mathrm{8}\boldsymbol{\pi{G}}}{\boldsymbol{{c}}^{\mathrm{4}} }\boldsymbol{{T}} \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{called}}\:\boldsymbol{{the}}\:\boldsymbol{{Einstein}}'\boldsymbol{{s}}\: \\ $$$$\boldsymbol{{tensor}}\:\boldsymbol{{eq}}^{\boldsymbol{{n}}} .\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{explanation}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{how}}\:\boldsymbol{{gravity}}\:\boldsymbol{{works}}…\boldsymbol{{what}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{gravity}}??? \\ $$$$ \\ $$
Commented by  M±th+et+s last updated on 13/Jun/20
nice explaining sir
$${nice}\:{explaining}\:{sir} \\ $$

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