
A planet is revolving around the sun in an elliptical orbit, its closest distance from sun is \[{r_{\min }}\], and farthest distance from sun is \[{r_{\max }}\]. If the orbital angular velocity of planet when it is nearest to sun is \[\omega \], then orbital angular velocity at the point when it is at the farthest distance of sun is
A) \[\omega \sqrt {\dfrac{{{r_{\min }}}}{{{r_{\max }}}}} \]
B) \[\omega \sqrt {\dfrac{{{r_{\max }}}}{{{r_{\min }}}}} \]
C) \[\omega {\left( {\dfrac{{{r_{\max }}}}{{{r_{\min }}}}} \right)^2}\]
D) \[\omega {\left( {\dfrac{{{r_{\min }}}}{{{r_{\max }}}}} \right)^2}\]
Answer
150.6k+ views
Hint: As there is no external torque so the angular momentum is conserved, that means \[L = \omega {r^2}\]is constant on all the path of elliptical orbit where is \[\omega \] angular velocity of the way and \[r\] is distance of that point from sun.
Complete step by step answer:
As we are given in question with minimum distance of sun as, \[{r_{\min }}\]
and we are also given with maximum distance as, \[{r_{\max }}\]
and with angular velocity at min. distance as, \[\omega \]
and to find angular velocity at max. distance as, \[{\omega _1}\]
so we know the angular momentum is conserved so we have
\[L = {L_1}\]
\[\omega {{r^2}_{\min }} = {\omega _1}{{r^2}_{\max }}\]
\[{\omega _1} = \dfrac{{\omega {{r^2}_{\min }}}}{{{{r^2}_{\max }}}}\]
So we get angular velocity at maximum distance as \[\dfrac{{\omega {{r^2}_{\min }}}}{{{{r^2}_{\max }}}}\].
So, The correct option is D.
Additional information:
An important role is played by Johannes Kepler, the physicist who gave us the three laws of planetary motion. The three laws are:
The law of ellipses:
The path of the planets about the sun is elliptical in shape, with the centre of the sun being located at one focus.
The law of equal areas:
An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time.
The Law of Harmonies:
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.
It was his second law, the law of equal areas which was further manipulated to find out that the angular momentum remains conserved during the whole planetary motion of a planet around the Sun.
Note: We know \[L = mvr\] and we can substitute \[v = \omega r\] so that we can make the formula \[L = m\omega {r^2}\]and \[m\] is constant and external torque is absent so we can say that angular momentum is constant and can calculate the angular velocity at any point on that path if we know the distance of that point from sun.
Complete step by step answer:
As we are given in question with minimum distance of sun as, \[{r_{\min }}\]
and we are also given with maximum distance as, \[{r_{\max }}\]
and with angular velocity at min. distance as, \[\omega \]
and to find angular velocity at max. distance as, \[{\omega _1}\]
so we know the angular momentum is conserved so we have
\[L = {L_1}\]
\[\omega {{r^2}_{\min }} = {\omega _1}{{r^2}_{\max }}\]
\[{\omega _1} = \dfrac{{\omega {{r^2}_{\min }}}}{{{{r^2}_{\max }}}}\]
So we get angular velocity at maximum distance as \[\dfrac{{\omega {{r^2}_{\min }}}}{{{{r^2}_{\max }}}}\].
So, The correct option is D.
Additional information:
An important role is played by Johannes Kepler, the physicist who gave us the three laws of planetary motion. The three laws are:
The law of ellipses:
The path of the planets about the sun is elliptical in shape, with the centre of the sun being located at one focus.
The law of equal areas:
An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time.
The Law of Harmonies:
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.
It was his second law, the law of equal areas which was further manipulated to find out that the angular momentum remains conserved during the whole planetary motion of a planet around the Sun.
Note: We know \[L = mvr\] and we can substitute \[v = \omega r\] so that we can make the formula \[L = m\omega {r^2}\]and \[m\] is constant and external torque is absent so we can say that angular momentum is constant and can calculate the angular velocity at any point on that path if we know the distance of that point from sun.
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