8.0 Problem solving technique

  Gravitation

Most of the problem of gravitation are solved by two conservation laws.

1. Conservation of angular momentum
2. Conservation of mechanical energy about the sun

Consider a planet of mass $m$ moving around the sun in an elliptical orbit is as shown in the figure.

Conserving angular momentum about the sun in aphelion and perihelion,

$${L_A} = {L_P}$$$$m{v_A}\left[ {a\left( {1 + e} \right)} \right] = m{v_P}\left[ {a\left( {1 - e} \right)} \right]$$$$\frac{{{v_P}}}{{{v_A}}} = \frac{{1 + e}}{{1 - e}}\quad ...(i)$$
Conserving mechanical energy about the sun in aphelion and perihelion,
$${E_A} = {E_P}$$$${U_A} + {K_A} = {U_P} + {K_P}$$$$ - \frac{{GMm}}{{a(1 + e)}} + \frac{1}{2}mv_A^2 = - \frac{{GMm}}{{a(1 - e)}} + \frac{1}{2}mv_P^2$$$$\frac{1}{2}m\left( {v_P^2 - v_A^2} \right) = \frac{{GMm}}{a}\left[ {\frac{1}{{(1 - e)}} - \frac{1}{{\left( {1 + e} \right)}}} \right]$$$$\frac{1}{2}m\left( {v_P^2 - v_A^2} \right) = \frac{{GMm\left( {2e} \right)}}{{a\left( {1 - {e^2}} \right)}}$$$$\left( {v_P^2 - v_A^2} \right) = \frac{{4eGM}}{{a\left( {1 - {e^2}} \right)}}\quad ...(ii)$$
From equation $(i)$ and $(ii)$ we get, $$v_P^2 - v_P^2\frac{{{{\left( {1 - e} \right)}^2}}}{{{{\left( {1 + e} \right)}^2}}} = \frac{{4eGM}}{{a\left( {1 - {e^2}} \right)}}$$$$\frac{{v_P^2{{\left( {1 + e} \right)}^2} - v_P^2{{\left( {1 - e} \right)}^2}}}{{{{\left( {1 + e} \right)}^2}}} = \frac{{4eGM}}{{a\left( {1 + e} \right)\left( {1 - e} \right)}}$$$$\frac{{4ev_P^2}}{{\left( {1 + e} \right)}} = \frac{{4eGM}}{{a\left( {1 - e} \right)}}$$$$v_P^2 = \frac{{GM\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}$$$${v_P} = \sqrt {\frac{{GM\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}} $$ Similarly, $${v_A} = \sqrt {\frac{{GM\left( {1 - e} \right)}}{{a\left( {1 + e} \right)}}} $$
Total energy of a planet, $${E_A} = {E_P} = E$$$$E = \frac{1}{2}mv_P^2 - \frac{{GMm}}{{a(1 - e)}}$$$$E = \frac{1}{2}m\left[ {\frac{{GM\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}} \right] - \frac{{GMm}}{{a(1 - e)}}$$$$E = \frac{{GMm}}{{2a\left( {1 - e} \right)}}\left( {1 + e - 2} \right)$$$$E = - \frac{{GMm}}{{2a}}$$
Therefore we get, $${v_P} = \sqrt {\frac{{GM\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}} $$ $${v_A} = \sqrt {\frac{{GM\left( {1 - e} \right)}}{{a\left( {1 + e} \right)}}} $$$$E = - \frac{{GMm}}{{2a}}$$
We can solve many problems of gravitation from the above three equations.

It is recommended to use the concept instead of formulas.