Question

If a satellite is revolving around the planter of mass $M$ in an elliptical orbit of the semi major axis $a$. Show that the orbital speed of the satellite when it is a distance $r$ from the focus will be given by $\left( {{v^2} = GM\left[ {\dfrac{2}{r} - \dfrac{1}{a}} \right]} \right).$

Answer 1

The orbital speed of the satellite is given by the law of the conversation of the energy. The total energy of the satellite in the elliptical orbit with the semi major axis is equal to the sum of the potential energy of the satellite and kinetic energy of the satellite.

Complete step by step solution:
Given that,
The mass of the planet is $M$.
The semi major axis of the ellipse is $a$.
The distance from the focus is $r$.
The total energy of the any satellite in elliptical orbit with semi major axis $a$ is given by,
$E = - \dfrac{{GMm}}{{2a}}$
Where, $E$ is the total energy of the satellite, $G$ is the gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the satellite and $a$ is the semi major axis.
Now, by the law of conservation of the energy, total energy is equal to the sum of kinetic energy and the potential energy.
Potential energy is given by,
$P.E = - \dfrac{{GMm}}{r}$
Where, $P.E$ is the potential energy, $G$ is the gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the satellite and $r$ is the distance from the focus.
Kinetic energy is given by,
$K.E = \dfrac{1}{2}m{v^2}$
Where, $K.E$ is the kinetic energy of the satellite, $m$ is the mass of the satellite and $v$ is the velocity of the satellite.
By law of conservation of energy,
$\Rightarrow - \dfrac{{GMm}}{{2a}} = - \dfrac{{GMm}}{r} + \dfrac{1}{2}m{v^2}$
By taking the potential energy from the RHS to LHS, then
$\Rightarrow - \dfrac{{GMm}}{{2a}} + \dfrac{{GMm}}{r} = \dfrac{1}{2}m{v^2}$
By taking the term $GMm$ as a common in the LHS, then
$\Rightarrow GMm\left( { - \dfrac{1}{{2a}} + \dfrac{1}{r}} \right) = \dfrac{1}{2}m{v^2}$
By cancelling the term $m$ on both sides, then
$\Rightarrow GM\left( { - \dfrac{1}{{2a}} + \dfrac{1}{r}} \right) = \dfrac{1}{2}{v^2}$
Now by keeping the velocity in one side and the other terms in other side, then
$\Rightarrow GM \times 2\left( { - \dfrac{1}{{2a}} + \dfrac{1}{r}} \right) = {v^2}$
By multiplying the term $2$ inside the bracket, then
$\Rightarrow GM\left( { - \dfrac{2}{{2a}} + \dfrac{2}{r}} \right) = {v^2}$
On further simplification, then
$\Rightarrow GM\left( { - \dfrac{1}{a} + \dfrac{2}{r}} \right) = {v^2}$
By rearranging the terms, then
$\Rightarrow GM\left( {\dfrac{2}{r} - \dfrac{1}{a}} \right) = {v^2}$
Hence, that the orbital speed of the satellite when it is a distance $r$ from the focus will be given by $\therefore \left( {{v^2} = GM\left[ {\dfrac{2}{r} - \dfrac{1}{a}} \right]} \right)$.

Note: The velocity of the satellite depends on the gravitational constant of the planet, mass of the planet, distance from the focus and the semi major axis of the elliptical path. From the final answer, it is clear that the velocity of the satellite is inversely proportional to the distance from the focus and the semi major axis of the elliptical path. As the distance from the focus and the semi major axis of the elliptical path decreases, the velocity of the satellite increases.