# Can speed be a imaginary number! validity of work energy theoram in 1D

Consider mass $$m_{1}$$and $$m_{2}$$with position vector (from an inertial frame) $$\overrightarrow{x_{1}}$$ and $$\overrightarrow{x_{2}}$$ respectively and distance between them be $$x_{0}$$.

$$m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{1}}=$$$$\overrightarrow{F}$$

$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}(\overrightarrow{x_{2}}-\overrightarrow{x_{0}})=\overrightarrow{F}$$ because its assumed that $$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}$$

$$\Rightarrow-\overrightarrow{F}\frac{m_{1}}{m_{2}}+m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}$$ because $$m_{2}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{2}}=-\overrightarrow{F}$$

$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})$$

$$\Rightarrow\left(m_{1}\frac{d^{2}}{dt^{2}}x_{0}\right)\hat{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})$$ because its assumed that $$\frac{d\hat{x_{0}}}{dt}=0$$

$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}x_{0}=F(1+\frac{m_{1}}{m_{2}})$$ because $$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}\Leftrightarrow\hat{x_{0}}=\hat{F}$$, $$F$$ = Gravitational Force

$$\Rightarrow m_{1}\int_{x_{i}}^{x_{f}}\frac{d^{2}}{dt^{2}}x_{0}dx_{0}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}$$ because normally we have $$F$$ as $$F(x_{0})$$.

$$\Rightarrow\left.\frac{1}{2}m_{1}v^{2}\right|_{v=v_{i}}^{v=v_{f}}=(1+\frac{m_{1}}{m_{2}})\int_{x_{i}}^{x_{f}}Fdx_{0}$$ because $$\overrightarrow{v_{0}}=\overrightarrow{v}$$,$$\left|\overrightarrow{v_{0}}\right|^{2}=(\frac{dx_{0}}{dt})^{2}$$

in earth and an object case, let $$m_{1}=m$$, $$m_{2}=m_{e},x_{i}=x_{e}$$ and $$F=G\frac{mm_{e}}{x_{0}^{2}}$$.

$$\Rightarrow\left.\frac{1}{2}mv^{2}\right|_{v=v_{i}}^{v=v_{f}}=\left.-G\frac{m(m+m_{e})}{x_{0}}\right|_{x_{0}=x_{i}}^{x_{0}=x_{f}}$$

Escape Velocity = $$v_{i}$$ when $$v_{f}=0$$ and $$x_{f}\rightarrow\infty$$.

$$\Rightarrow-\frac{1}{2}mv_{i}^{2}=G\frac{m(m+m_{e})}{x_{e}}$$

$$\Rightarrow\frac{1}{2}mv_{i}^{2}=-G\frac{m(m+m_{e})}{x_{e}}$$

$$\Rightarrow v_{initial}=i\left(G\frac{m+m_{e}}{x_{e}}\right)^{\frac{1}{2}}$$

$$\Rightarrow$$I dont understand how much speed is this.

Its not just escape velocity, if we dont include imaginery number domian in speed then it implies that if $$v(x_{f})>v(x_{i})$$ then $$x_{f}>x_{i}$$ which is also incorrect.

so question is, can speed be a imaginary number ? How do i give imaginary escape speed to an object?

A speed can't be imaginary, you must have done something wrong in your calculations (a minus sign missing somewhere or something like this). However, I would suggest that you add some comments so we can understand what is happening in your calculations. As it is now, I don't really understand what you're trying to calculate.

i am deriving work energy theoram in 1d and proving that speed can be imaginary.
$$F,x_{1},x_{2},x_{0},v,>0$$ as these are modulous part of vector quantities.

Consider mass $$m_{1}$$and $$m_{2}$$with position vector (from an inertial frame) $$\overrightarrow{x_{1}}$$ and $$\overrightarrow{x_{2}}$$ respectively and distance between them be $$x_{0}$$.
variable definations
$$m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{1}}=$$$$\overrightarrow{F}$$
this is newton's law.
$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}(\overrightarrow{x_{2}}-\overrightarrow{x_{0}})=\overrightarrow{F}$$ because its assumed that $$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}$$

either $$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}$$
or $$\overrightarrow{x_{2}}-\overrightarrow{x_{1}}=\overrightarrow{x_{0}}$$
is true. i chose first. however it does not make any difference in
the results.

$$\Rightarrow\left(m_{1}\frac{d^{2}}{dt^{2}}x_{0}\right)\hat{x_{0}}=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})$$ because its assumed that $$\frac{d\hat{x_{0}}}{dt}=0$$
It has to be assumed that system is not rotating as it is nesscery condition for work energy theoram.

Its not just escape velocity, if we dont include imaginery number domian in speed then it implies that if $$v(x_{f})>v(x_{i})$$ then $$x_{f}>x_{i}$$ which is also incorrect.

Let $$x_{f}>x_{i}$$.

Since $$\left.\frac{1}{2}mv^{2}\right|_{v=v_{i}}^{v=v_{f}}=\left.-G\frac{m(m+m_{e})}{x_{0}}\right|_{x_{0}=x_{i}}^{x_{0}=x_{f}}$$

$$\Rightarrow v_{f}^{2}>v_{i}^{2}$$

$$\Rightarrow v_{f}>v_{i}$$

Cthugha
$$m_{1}\frac{d^{2}}{dt^{2}}\overrightarrow{x_{1}}=$$$$\overrightarrow{F}$$

$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}(\overrightarrow{x_{2}}-\overrightarrow{x_{0}})=\overrightarrow{F}$$ because its assumed that $$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}$$

I am too lazy to check whether that matters for your calculations and whether you used the different definition all the time, but if you assume $$\vec{x}_1 -\vec{x}_2=\vec{x}_0$$
then you have to replace $$\vec{x}_1$$ by $$\vec{x}_2 +\vec{x}_0$$ instead of $$\vec{x}_2 -\vec{x}_0$$.

Also, I am puzzled what force $$F=G\frac{mm_{e}}{x_{0}^{2}}$$ is supposed to be. This is a force which has the magnitude of gravity, but is repulsive instead of attractive. Gravitational force would be given by $$F=-G\frac{mm_{e}}{x_{0}^{2}}$$.

Alternatively, it would be more sensible to assume $$\hat{x_{0}}=-\hat{F}$$.
$$\vec{x}_0$$ is always the vector starting at $$\vec{x}_2$$ and pointing towards $$\vec{x}_1$$. The gravitational force on $$\vec{x}_1$$ also pointing away from $$\vec{x}_2$$ does not make much sense.

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$$\overrightarrow{x_{1}}-\overrightarrow{x_{2}}=\overrightarrow{x_{0}}$$ is wrong assumption.

the correct assumption is $$\overrightarrow{x_{2}}-\overrightarrow{x_{1}}=x_{0}\hat{F}$$

$$\Rightarrow-m_{1}\frac{d^{2}}{dt^{2}}(x_{0}\hat{F})=\overrightarrow{F}(1+\frac{m_{1}}{m_{2}})$$

$$\Rightarrow m_{1}\frac{d^{2}}{dt^{2}}x_{0}=-F(1+\frac{m_{1}}{m_{2}})$$ ..[1]

$$\Rightarrow\left.\frac{1}{2}mv^{2}\right|_{v=v_{i}}^{v=v_{f}}=\left.G\frac{m(m+m_{e})}{x_{0}}\right|_{x_{0}=x_{i}}^{x_{0}=x_{f}}$$

$$\Rightarrow v_{esacpe}=\left(G\frac{m+m_{e}}{x_{e}}\right)^{\frac{1}{2}}$$

ok everything is alright thanks everyone

Also, I am puzzled what force $$F=G\frac{mm_{e}}{x_{0}^{2}}$$ is supposed to be. This is a force which has the magnitude of gravity, but is repulsive instead of attractive. Gravitational force would be given by $$F=-G\frac{mm_{e}}{x_{0}^{2}}$$.

F is magnitude of force vector and its always positive whether its attractive or repulsive.