- #1
coki2000
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Hi,
In gravitational force law [tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex]if i want to find the distance function which depends on time then
[tex]\frac{d^{2}x}{dt^{2}}=-G\frac{M}{x^2}[/tex]
[tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-G\frac{M}{x^2}[/tex]
[tex]vdv=-G\frac{M}{x^2}dx[/tex]
integrate the both sides
[tex]\frac{v^2}{2}=G\frac{M}{x}\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}}\Rightarrow dt=\sqrt{\frac{x}{2Gm}}dx[/tex]
if we solve that differential equation then [tex]t=\frac{2}{3}\sqrt{\frac{1}{2GM}}x^{3/2}[/tex]
this time distance function [tex]x(t)=\sqrt[3]{\frac{9}{2}GM}.t^{2/3}[/tex] Is that function right else what is the right formula? Please help me
In gravitational force law [tex]F=-G\frac{M_{1}M_{2}}{x^{2}}[/tex]if i want to find the distance function which depends on time then
[tex]\frac{d^{2}x}{dt^{2}}=-G\frac{M}{x^2}[/tex]
[tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-G\frac{M}{x^2}[/tex]
[tex]vdv=-G\frac{M}{x^2}dx[/tex]
integrate the both sides
[tex]\frac{v^2}{2}=G\frac{M}{x}\Rightarrow \frac{dx}{dt}=\sqrt{2G\frac{M}{x}}\Rightarrow dt=\sqrt{\frac{x}{2Gm}}dx[/tex]
if we solve that differential equation then [tex]t=\frac{2}{3}\sqrt{\frac{1}{2GM}}x^{3/2}[/tex]
this time distance function [tex]x(t)=\sqrt[3]{\frac{9}{2}GM}.t^{2/3}[/tex] Is that function right else what is the right formula? Please help me
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