If the time period of a satellite is T, its kinetic energy is proportional to
A. \[K.E \propto {T^{\dfrac{{ - 1}}{2}}}\].
B. \[K.E \propto {T^{\dfrac{{ - 2}}{3}}}\].
C. \[K.E \propto {T^{\dfrac{3}{2}}}\].
D. None of these
Answer
673.8k+ views
Hint: In this question, we will use the concept of kepler’s law of planetary motion for circular orbits. We will use the formula of kinetic energy and time period to find out the relationship between them. Then we will get the proportionality between them.
Formula used: Centripetal force = $\dfrac{{m{v^2}}}{r}$, force of gravitation = $\dfrac{{GMm}}{{{r^2}}}$, $K.E = \dfrac{1}{2}m{v_0}^2$ and orbital speed, v \[ = \dfrac{{2\pi r}}{T}\].
Complete Step-by-Step solution:
Given that, the time period of the satellite is T.
According to kepler’s law of planetary motion,
The force of gravitation between the sun and the satellite provides the necessary centripetal force.
We know that, centripetal force = $\dfrac{{m{v^2}}}{r}.$ and force of gravitation = $\dfrac{{GMm}}{{{r^2}}}$.
$\therefore $ $\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
${v^2} = \dfrac{{GM}}{r}$ ………..(i)
But orbital speed, v = \[\dfrac{{circumference}}{{period{\text{ }}of{\text{ }}revolution}} = \dfrac{{2\pi r}}{T}\].
Putting this value of v in equation (i), we get
$\therefore $ $\dfrac{{4{\pi ^2}{r^2}}}{{{T^2}}} = \dfrac{{GM}}{r}$ , solving this for T, we get
$ \Rightarrow {T^2} = \dfrac{{4{\pi ^2}}}{{GM}}{r^3} = {K_S}{r^3}$,
Thus we can say that, ${T^2} \propto {r^3}.$
It can also be written as ${T^{\dfrac{2}{3}}} \propto r.$ ……..(ii)
We know that, kinetic energy of a satellite,
$ \Rightarrow K.E = \dfrac{1}{2}m{v_0}^2$, here we have ${v^2} = \dfrac{{GM}}{r}$, then
$ \Rightarrow K.E = \dfrac{1}{2}m\left( {\dfrac{{GM}}{r}} \right) = \dfrac{1}{2}\dfrac{{GMm}}{r}$.
This shows that the kinetic energy, $K.E \propto \dfrac{1}{r}$. …..(iii)
Now, by relating the equation (ii) and (iii), we can say that
$ \Rightarrow K.E \propto {T^{\dfrac{{ - 2}}{3}}}$.
Hence, the correct answer is option (B).
Note: Whenever we ask such a question, first we have to find the term that is common in the given quantities. Then we have to find out the relation of those quantities with that term using their formula. After that, using that term we can easily find out the relationship and proportionality between the given quantities. Thus we will get our answer.
Formula used: Centripetal force = $\dfrac{{m{v^2}}}{r}$, force of gravitation = $\dfrac{{GMm}}{{{r^2}}}$, $K.E = \dfrac{1}{2}m{v_0}^2$ and orbital speed, v \[ = \dfrac{{2\pi r}}{T}\].
Complete Step-by-Step solution:
Given that, the time period of the satellite is T.
According to kepler’s law of planetary motion,
The force of gravitation between the sun and the satellite provides the necessary centripetal force.
We know that, centripetal force = $\dfrac{{m{v^2}}}{r}.$ and force of gravitation = $\dfrac{{GMm}}{{{r^2}}}$.
$\therefore $ $\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
${v^2} = \dfrac{{GM}}{r}$ ………..(i)
But orbital speed, v = \[\dfrac{{circumference}}{{period{\text{ }}of{\text{ }}revolution}} = \dfrac{{2\pi r}}{T}\].
Putting this value of v in equation (i), we get
$\therefore $ $\dfrac{{4{\pi ^2}{r^2}}}{{{T^2}}} = \dfrac{{GM}}{r}$ , solving this for T, we get
$ \Rightarrow {T^2} = \dfrac{{4{\pi ^2}}}{{GM}}{r^3} = {K_S}{r^3}$,
Thus we can say that, ${T^2} \propto {r^3}.$
It can also be written as ${T^{\dfrac{2}{3}}} \propto r.$ ……..(ii)
We know that, kinetic energy of a satellite,
$ \Rightarrow K.E = \dfrac{1}{2}m{v_0}^2$, here we have ${v^2} = \dfrac{{GM}}{r}$, then
$ \Rightarrow K.E = \dfrac{1}{2}m\left( {\dfrac{{GM}}{r}} \right) = \dfrac{1}{2}\dfrac{{GMm}}{r}$.
This shows that the kinetic energy, $K.E \propto \dfrac{1}{r}$. …..(iii)
Now, by relating the equation (ii) and (iii), we can say that
$ \Rightarrow K.E \propto {T^{\dfrac{{ - 2}}{3}}}$.
Hence, the correct answer is option (B).
Note: Whenever we ask such a question, first we have to find the term that is common in the given quantities. Then we have to find out the relation of those quantities with that term using their formula. After that, using that term we can easily find out the relationship and proportionality between the given quantities. Thus we will get our answer.
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