
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
580.8k+ 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.
Recently Updated Pages
The height of a solid metal cylinder is 20cm Its r-class-10-maths-ICSE

If a train crossed a pole at a speed of 60kmhr in 30 class 10 physics CBSE

Name the Writs that the High Courts are empowered to class 10 social science CBSE

A tower is 5sqrt 3 meter high Find the angle of el-class-10-maths-CBSE

Immediate cause of variations of A Mutations B Environmental class 10 biology CBSE

A rectangular container whose base is a square of side class 10 maths CBSE

Trending doubts
Who composed the song Vande Mataram A RabindraNath class 10 social science CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

The revolutionary who died after 63 days of the hunger class 10 social science CBSE

The slogan of Bande Mataram was first adopted during class 10 social science CBSE

Why is Sardar Vallabhbhai Patel called the Iron man class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

