JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Velocity of a Planet in Terms of Eccentricity

Applying the law of conservation of angular momentum at perigee and apogee

\[m{{v}_{p}}{{r}_{p}}=m{{v}_{a}}{{r}_{a}}\]

\[\Rightarrow \] \[\frac{{{v}_{p}}}{{{v}_{a}}}=\frac{{{r}_{a}}}{{{r}_{p}}}=\frac{a+c}{a-c}=\frac{1+e}{1-e}\]    

[As \[{{r}_{p}}=a-c,\,\,\,\,\,\,\,{{r}_{a}}=a+c\] and eccentricity \[e=\frac{c}{a}\]]

Applying the conservation of mechanical energy at perigee and apogee

\[\frac{1}{2}m{{v}_{p}}^{2}-\frac{GMm}{{{r}_{p}}}=\frac{1}{2}m{{v}_{a}}^{2}-\frac{GMm}{{{r}_{a}}}\]

\[\Rightarrow \] \[{{v}_{p}}^{2}-{{v}_{a}}^{2}=2GM\,\left[ \frac{1}{{{r}_{p}}}-\frac{1}{{{r}_{a}}} \right]\]

\[\Rightarrow \] \[{{v}_{a}}^{2}\,\left[ \frac{{{r}_{a}}^{2}-{{r}_{p}}^{2}}{{{r}_{p}}^{2}} \right]\,=\,\,2\,GM\,\left[ \frac{{{r}_{a}}-{{r}_{p}}}{{{r}_{a}}{{r}_{p}}} \right]\]        [As \[{{v}_{p}}=\frac{{{v}_{a}}{{r}_{a}}}{{{r}_{p}}}\]]

\[\Rightarrow \] \[{{v}_{a}}^{2}=\frac{2\,GM}{{{r}_{a}}+{{r}_{p}}}\,\,\left[ \frac{{{r}_{p}}}{{{r}_{a}}} \right]\]Þ \[{{v}_{a}}^{2}=\frac{2\,GM}{a}\,\left( \frac{a-c}{a+c} \right)\,=\,\frac{GM}{a}\,\left( \frac{1-e}{1+e} \right)\]

Thus the speeds of planet at apogee and perigee are

\[{{v}_{a}}=\sqrt{\frac{GM}{a}\left( \frac{1-e}{1+e} \right)}\],                      

\[{{v}_{p}}=\sqrt{\frac{GM}{a}\,\left( \frac{1+e}{1-e} \right)}\]

Note :

• The gravitational force is a central force so torque on planet relative to sun is always zero, hence angular momentum of a planet or satellite is always constant irrespective of shape of orbit.