Answer
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Hint ${V_p} = a\left( {1 + e} \right)$ and ${V_a} = a\left( {1 - e} \right)$ [where, $a$ is the semi-major axis of the ellipse and $e$ is the eccentricity]. Divide both the velocities and cross multiply both sides. Take the terms on the same sides.
Formula used: Let, $a$ be the semi-major axis of the ellipse and $e$ is the eccentricity of the elliptical orbit.
So, ${V_p} = a\left( {1 + e} \right)$ and ${V_a} = a\left( {1 - e} \right)$
Complete step by step solution:
Eccentricity of an ellipse is defined as the ratio of the distance between the center of the ellipse and each focus to the length of semi-major axis.
It is given that ${V_p}$ and ${V_a}$ are the velocities of the planet at the perigee and apogee respectively.
Now, let $a$ be the semi-major axis of the ellipse and $e$ is the eccentricity of the elliptical orbit.
Then, $\dfrac{{{V_p}}}{{{V_a}}} = \dfrac{{a\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}$
or, ${V_p}\left( {1 - e} \right) = {V_a}\left( {1 + e} \right)$ [cross multiplying both the sides]
or, ${V_p} - e{V_p} = {V_a} + e{V_a}$
or, $e\left( {{V_p} + {V_a}} \right) = \left( {{V_p} - {V_a}} \right)$ [taking the like terms on the same sides]
or, $e = \dfrac{{{V_p} - {V_a}}}{{{V_p} + {V_a}}}$
Option D is correct answer
Additional information We should also focus on three of Kepler’s law:
Kepler’s 1st law: Every planet moves in an elliptical orbit with the sun at one of its foci.
Kepler’s 2nd law: The line joining the sun and a planet sweeps out equal areas in equal intervals of time. It means the areal velocity of a planet is constant.
Kepler’s 3rd law: The square of the time period of revolution of a planet is directly proportional to the cube of the length of the semi-major axis of its elliptical orbit.
These orbits of planets are known as Keplerian orbits while their motions are known as Keplerian motions.
Note: It should be always remembered that the eccentricities of the orbits of the planets in the solar system maintain all Kepler’s laws. We can say that the orbits are almost circular. As the circle is a special case of an ellipse (although for circles $e = 0$ ), Kepler’s laws are equally applicable for the circular orbits too. So, we can say all Kepler’s laws are applicable and valid for both the circular and elliptical orbits, which means for orbits having $0 \leqslant e < 1$ .
Formula used: Let, $a$ be the semi-major axis of the ellipse and $e$ is the eccentricity of the elliptical orbit.
So, ${V_p} = a\left( {1 + e} \right)$ and ${V_a} = a\left( {1 - e} \right)$
Complete step by step solution:
Eccentricity of an ellipse is defined as the ratio of the distance between the center of the ellipse and each focus to the length of semi-major axis.
It is given that ${V_p}$ and ${V_a}$ are the velocities of the planet at the perigee and apogee respectively.
Now, let $a$ be the semi-major axis of the ellipse and $e$ is the eccentricity of the elliptical orbit.
Then, $\dfrac{{{V_p}}}{{{V_a}}} = \dfrac{{a\left( {1 + e} \right)}}{{a\left( {1 - e} \right)}}$
or, ${V_p}\left( {1 - e} \right) = {V_a}\left( {1 + e} \right)$ [cross multiplying both the sides]
or, ${V_p} - e{V_p} = {V_a} + e{V_a}$
or, $e\left( {{V_p} + {V_a}} \right) = \left( {{V_p} - {V_a}} \right)$ [taking the like terms on the same sides]
or, $e = \dfrac{{{V_p} - {V_a}}}{{{V_p} + {V_a}}}$
Option D is correct answer
Additional information We should also focus on three of Kepler’s law:
Kepler’s 1st law: Every planet moves in an elliptical orbit with the sun at one of its foci.
Kepler’s 2nd law: The line joining the sun and a planet sweeps out equal areas in equal intervals of time. It means the areal velocity of a planet is constant.
Kepler’s 3rd law: The square of the time period of revolution of a planet is directly proportional to the cube of the length of the semi-major axis of its elliptical orbit.
These orbits of planets are known as Keplerian orbits while their motions are known as Keplerian motions.
Note: It should be always remembered that the eccentricities of the orbits of the planets in the solar system maintain all Kepler’s laws. We can say that the orbits are almost circular. As the circle is a special case of an ellipse (although for circles $e = 0$ ), Kepler’s laws are equally applicable for the circular orbits too. So, we can say all Kepler’s laws are applicable and valid for both the circular and elliptical orbits, which means for orbits having $0 \leqslant e < 1$ .
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