Total energy of earth in its circular orbit around the sun

Question :
Find out the total energy of earth in its circular orbit around the sun in terms of gravitational constant
Answer:
Let R be the total distance between the earth and the sun. If ${M_e}$ and ${M_s}$ are the mass of earth and sun respectively, the gravitational force of motion of earth and sun is given by

$F =  - \frac{{G{M_e}{M_s}}}{{{R^2}}}$

where G is the gravitational constant. Since the centripetal force balances the gravitational force of attraction, we have

${F_c} = |{F_G}|$,

where

${F_c} = \frac{{{M_e}{v^2}}}{R}$

v being the velocity with which earth is moving. Hence we have

$\frac{{{M_e}{v^2}}}{R} = \frac{{G{M_e}{M_s}}}{{{R^2}}}$

or

${M_e}{v^2} = \frac{{G{M_e}{M_s}}}{R}$

Therefore kinetic energy of earth in motion is

$T = \frac{1}{2}{M_e}{v^2} = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}$

As we know that , force in terms of potential energy is

${F_G} = - \frac{{\partial V}}{{\partial R}}$
$V = - \int {{F_G}dR} = \int {\frac{{G{M_e}{M_s}}}{{{R^2}}}dR = - } \frac{{G{M_e}{M_s}}}{R}$

Now total energy of the earth in the orbit around the sun is

$E = T + V$
$E = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R} - \frac{{G{M_e}{M_s}}}{R}$

$E = - \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}$

This is the required expression.

Brachistochrone Problem

Question

If a particle falls from rest under the influence of gravity from higher to lower point in the minimum time, what is the curve that the particle will follow?

Solution

Suppose v is the speed of the particle along the curve, then in traversing ds portion of the curve time spent would be $\frac{{ds}}{v}$ so that total time taken by the particle in moving from highest point 1 to lowest point 2 will be

$${t_{12}} = \int\limits_1^2 {\frac{{ds}}{v}}$$

Suppose vertical distance of fall upto point 2 be x, then from the principle of conservation of energy of the particle we find that

$\begin{array}{l}\frac{1}{2}m{v^2} = mgx\\or\\v = \sqrt {2gx} \\then\\{t_{12}} = \int\limits_1^2 {\frac{{\sqrt {d{x^2} + d{y^2}} }}{{\sqrt {2gx} }}} dx\\{\rm{      = }}\int\limits_1^2 {\frac{{\sqrt {1 + {{\dot y}^2}} }}{{\sqrt {2gx} }}} dx\\{\rm{      = }}\int\limits_1^2 {fdx} \end{array}$

Where,

$f=\left ( \frac{1+\dot{y}^{2}}{2gx} \right )^{\frac{1}{2}}$

For ${t_{12}}$ to be minimum equation

$\frac{d}{{dx}}\left( {\frac{{\partial f}}{{\partial \dot y}}} \right) - \frac{{\partial f}}{{\partial y}} = 0$

must be satisfied. From expression for $f$ we find that

$\begin{array}{l}\frac{{\partial f}}{{\partial y}} = 0\\\frac{{\partial f}}{{\partial \dot y}} = \frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}\\\frac{d}{{dx}}\left( {\frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}} \right) = 0\\or\\\frac{{{{\dot y}^2}}}{{2gx(1 + {{\dot y}^2})}} = c'\end{array}$

Where c’ is the constant of integration.

Since c’ is a constant we can also write

$\frac{{{{\dot y}^2}}}{{x(1 + {{\dot y}^2})}} = c$

where c is also a constant. On integrating above equation we find

$\begin{array}{l}\frac{{{{\dot y}^2}}}{c} = x(1 + {{\dot y}^2})\\or,\\{{\dot y}^2}\left( {\frac{1}{c} - x} \right) = x\\{{\dot y}^2}\left( {\frac{x}{c} - {x^2}} \right) = {x^2}\\\dot y = \frac{x}{{\sqrt {\frac{x}{c} - {x^2}} }}\end{array}$

Putting $\frac{1}{c} = 2a$ , and on integration we get

$\int dy=\int \frac{x}{\sqrt{2ax-x^{2}}}dx$

$y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}+c''$

where c'' is new constant of integration.
In case c'' is zero then y will be zero for x=0.

As such the equation

$y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}$

And this y represents an inverted cycloid with its base along y axis and cusp at the origin and is the curve that particle will follow.

Double pendulum

Question 

In case of a double pendulum find the expression for the kinetic energy of the system

Solution

We take a simple case where lengths and masses are same.see below in the figure

Here on being displaced the co-ordinates of two pendulums are

$\begin{array}{l}{x_1} = l\sin {\theta _1}\\{y_1} = l\cos {\theta _1}\end{array}$

For the first pendulum where ${\theta _1}$ is the angle through which the first pendulum have been displaced.

For second pendulum

$\begin{array}{l}{x_2} = l\sin {\theta _1} + l\sin {\theta _2}\\{y_2} = l\cos {\theta _1} + l\cos {\theta _2}\end{array}$

Where ${\theta _2}$ is the angle through which second pendulum has been displaced.

The total kinetic energy of the system is given by the expression

$T = \frac{1}{2}m(\dot x_1^2 + \dot y_1^2) + \frac{1}{2}m(\dot x_2^2 + \dot y_2^2)$

Now

And

Which is the required result