Question

A planet of mass m revolves around the Sun in an elliptical orbit. Find the total mechanical energy using conservation laws.

Answer 1

Here, we have to use two concepts namely conservation of angular momentum and conservation of total energy. Both of these laws cannot be violated. They are known as Universal laws of conservation. We know that the energy cannot be created nor be destroyed and also the angular momentum of a system remains conserved if there is no external torque applied to the system. Energy is composed of two types one is kinetic energy and another is potential energy. Similarly, the angular momentum of one system must be equal to the angular momentum of another system. Form a relation between the angular momentum and energy of the system and find out the total mechanical energy of the system.

Formula used:

The formula for finding out the total energy is $E= K.E+P.E$

$E = \dfrac{1}{2}(m{v^2}) - \dfrac{{G{M_s}M}}{{{r^2}}}$

Where

$E$= Energy (Total)

${M_s}$= Mass of the sun

$M$= Mass of the planet

$m$= mass used in the K.E equation

$v$= velocity

$r$= distance between two planets

$G$= Gravitational Constant

Complete step by step answer:

Step 1: Writing the equation for total mechanical energy at B.

E = Kinetic Energy + Potential Energy

E= K.E+P.E

The Semi-Major axis of the ellipse is

$a= \dfrac{{{r_1} + {r_2}}}{2}$

Put the values of KE and PE in the total energy equation.

$\Rightarrow E = \dfrac{1}{2}(m{v^2}) - \dfrac{{G{M_s}M}}{{{r^2}}}$

Step 2: Applying conservation of angular momentum

$M{V_1}{r_1} = M{V_2}{r_2}$

Solving for${V_1}$

$\Rightarrow {V_1} = \dfrac{{{V_2}{r_2}}}{{{r_1}}}$

Applying conservation of mechanical energy

$\Rightarrow \dfrac{1}{2}(m{v_1}^2) - \dfrac{{G{M_s}M}}{{{r^2}}} = \dfrac{1}{2}(m{v_2}^2) - \dfrac{{G{M_s}M}}{{{r^2}}}$

Put the value of ${v_1}$in the above equation

$\Rightarrow \dfrac{1}{2} \times m \times {(\dfrac{{{v_2}{r_2}}}{{{r_1}}})^2} - \dfrac{{G{M_s}M}}{{{r_1}^2}} = \dfrac{1}{2}(m{v_2}^2) - \dfrac{{G{M_s}M}}{{{r_2}^2}}$

Simplify the above equation

$\Rightarrow v_2^2(\dfrac{{r_2^2}}{{r_1^2}} - 1) = G{M_s}(\dfrac{1}{{{r_1}}} - \dfrac{1}{{{r_2}}})$

Simplify more,

$\Rightarrow \dfrac{{v_2^2}}{2}\dfrac{{({r_2} + {r_1})({r_2} - {r_1})}}{{r_1^2}} = G{M_s}(\dfrac{{{r_2} - {r_1}}}{{{r_1}{r_2}}})$

$\Rightarrow v_2^2 = 2\dfrac{{G{M_s}}}{{({r_2} + {r_1})}}\dfrac{{{r_1}}}{{{r_2}}}$

Put the above value in the total energy equation

$\Rightarrow E = \dfrac{1}{2}m(\dfrac{{2G{M_s}}}{{({r_1} + {r_2})}} \times \dfrac{{{r_1}}}{{{r_2}}}) - \dfrac{{G{M_s}m}}{{{r_2}}}$

Solve the above equation

$\Rightarrow E = \dfrac{1}{2}m{M_s} \times 2 \times G \times \dfrac{1}{{{r_1} + {r_2}}} \times \dfrac{{{r_1}}}{{{r_2}}} - \dfrac{{G{M_s}m}}{{{r_2}}}$

$\Rightarrow E = \dfrac{{G{M_s}m}}{{{r_2}}}(\dfrac{{{r_1}}}{{{r_1} + {r_2}}} - 1)$

Simplifying the above equation

$\Rightarrow E = -\dfrac{{G{M_s}m}}{{{r_1} + {r_2}}}$

Put the value of semi major axis in the above equation

$\Rightarrow E = -\dfrac{{G{M_s}m}}{{2a}}$

The total mechanical energy is $E =- \dfrac{{G{M_s}m}}{{2a}}$.

Note: Here, in this question find out the semi-major axis and then find the total energy. Establish the relation between angular momentum and energy of the system and then solve, there is no need for any mathematical calculation as it is a question in which the only derivation is required.