- #1
ManishR
- 88
- 0
Consider mass [tex]m_{1}[/tex]and [tex]m_{2}[/tex]with position vector (from an inertial frame) [tex]\overrightarrow{x_{1}}[/tex] and [tex]\overrightarrow{x_{2}}[/tex] respectively and distance between them be [tex]x_{0}[/tex].
[tex]m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{1}}=[/tex][tex]\overrightarrow{F}[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}(\overrightarrow{x_{2}}-\overrightarrow{x_{0}})=\overrightarrow{F}[/tex] because its assumed that [tex]\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}[/tex]
[tex]\Rightarrow-\overrightarrow{F}\frac{m_{1}}{m_{2}}+m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}[/tex] because [tex]m_{2}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{2}}=-\overrightarrow{F}[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})[/tex]
[tex]\Rightarrow\left(m_{1}\frac{d^{2}}{dt^{2}}x_{0}\right)\hat{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})[/tex] because its assumed that [tex]\frac{d\hat{x_{0}}}{dt}=0[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}x_{0}=F(1+\frac{m_{1}}{m_{2}})[/tex] because [tex]\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}\Leftrightarrow\hat{x_{0}}=\hat{F}[/tex], [tex]F[/tex] = Gravitational Force
[tex]\Rightarrow m_{1}\int_{x_{i}}^{x_{f}}\frac{d^{2}}{dt^{2}}x_{0}dx_{0}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}[/tex] because normally we have [tex]F[/tex] as [tex]F(x_{0})[/tex].
[tex]\Rightarrow\left.\frac{1}{2}m_{1}v^{2}\right|_{v=v_{i}}^{v=v_{f}}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}[/tex] because [tex]\overrightarrow{v_{0}}=\overrightarrow{v}[/tex],[tex]\left|\overrightarrow{v_{0}}\right|^{2}=(\frac{dx_{0}}{dt})^{2}[/tex]
in Earth and an object case, let [tex]m_{1}=m[/tex], [tex]m_{2}=m_{e},x_{i}=x_{e}[/tex] and [tex]F=G\frac{mm_{e}}{x_{0}^{2}}[/tex].
[tex]\Rightarrow\left.\frac{1}{2}mv^{2}\right|_{v=v_{i}}^{v=v_{f}}=\left.-G\frac{m(m+m_{e})}{x_{0}}\right|_{x_{0}=x_{i}}^{x_{0}=x_{f}}[/tex]
Escape Velocity = [tex]v_{i}[/tex] when [tex]v_{f}=0[/tex] and [tex]x_{f}\rightarrow\infty[/tex].
[tex]\Rightarrow-\frac{1}{2}mv_{i}^{2}=G\frac{m(m+m_{e})}{x_{e}}[/tex]
[tex]\Rightarrow\frac{1}{2}mv_{i}^{2}=-G\frac{m(m+m_{e})}{x_{e}}[/tex]
[tex]\Rightarrow v_{initial}=i\left(G\frac{m+m_{e}}{x_{e}}\right)^{\frac{1}{2}}[/tex]
[tex]\Rightarrow[/tex]I don't understand how much speed is this.
Its not just escape velocity, if we don't include imaginery number domian in speed then it implies that if [tex]v(x_{f})>v(x_{i})[/tex] then [tex]x_{f}>x_{i}[/tex] which is also incorrect.
so question is, can speed be a imaginary number ? How do i give imaginary escape speed to an object?
[tex]m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{1}}=[/tex][tex]\overrightarrow{F}[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}(\overrightarrow{x_{2}}-\overrightarrow{x_{0}})=\overrightarrow{F}[/tex] because its assumed that [tex]\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}[/tex]
[tex]\Rightarrow-\overrightarrow{F}\frac{m_{1}}{m_{2}}+m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}[/tex] because [tex]m_{2}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{2}}=-\overrightarrow{F}[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})[/tex]
[tex]\Rightarrow\left(m_{1}\frac{d^{2}}{dt^{2}}x_{0}\right)\hat{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})[/tex] because its assumed that [tex]\frac{d\hat{x_{0}}}{dt}=0[/tex]
[tex]\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}x_{0}=F(1+\frac{m_{1}}{m_{2}})[/tex] because [tex]\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}\Leftrightarrow\hat{x_{0}}=\hat{F}[/tex], [tex]F[/tex] = Gravitational Force
[tex]\Rightarrow m_{1}\int_{x_{i}}^{x_{f}}\frac{d^{2}}{dt^{2}}x_{0}dx_{0}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}[/tex] because normally we have [tex]F[/tex] as [tex]F(x_{0})[/tex].
[tex]\Rightarrow\left.\frac{1}{2}m_{1}v^{2}\right|_{v=v_{i}}^{v=v_{f}}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}[/tex] because [tex]\overrightarrow{v_{0}}=\overrightarrow{v}[/tex],[tex]\left|\overrightarrow{v_{0}}\right|^{2}=(\frac{dx_{0}}{dt})^{2}[/tex]
in Earth and an object case, let [tex]m_{1}=m[/tex], [tex]m_{2}=m_{e},x_{i}=x_{e}[/tex] and [tex]F=G\frac{mm_{e}}{x_{0}^{2}}[/tex].
[tex]\Rightarrow\left.\frac{1}{2}mv^{2}\right|_{v=v_{i}}^{v=v_{f}}=\left.-G\frac{m(m+m_{e})}{x_{0}}\right|_{x_{0}=x_{i}}^{x_{0}=x_{f}}[/tex]
Escape Velocity = [tex]v_{i}[/tex] when [tex]v_{f}=0[/tex] and [tex]x_{f}\rightarrow\infty[/tex].
[tex]\Rightarrow-\frac{1}{2}mv_{i}^{2}=G\frac{m(m+m_{e})}{x_{e}}[/tex]
[tex]\Rightarrow\frac{1}{2}mv_{i}^{2}=-G\frac{m(m+m_{e})}{x_{e}}[/tex]
[tex]\Rightarrow v_{initial}=i\left(G\frac{m+m_{e}}{x_{e}}\right)^{\frac{1}{2}}[/tex]
[tex]\Rightarrow[/tex]I don't understand how much speed is this.
Its not just escape velocity, if we don't include imaginery number domian in speed then it implies that if [tex]v(x_{f})>v(x_{i})[/tex] then [tex]x_{f}>x_{i}[/tex] which is also incorrect.
so question is, can speed be a imaginary number ? How do i give imaginary escape speed to an object?