A planet moves around the sun in nearly circular orbit. It's the period of revolution ‘ $ T $ ’ upon i) radius ‘ $ r $ ’ of the orbit ii) Mass ‘ $ M $ ’ of the sun and iii) gravitational constant ‘ $ G $ ’. Show dimensionally that $ {T^2} \propto {r^3} $ .
Last updated date: 21st Mar 2023
•
Total views: 183k
•
Views today: 4.61k
Answer
183k+ views
Hint: Law of Gravitation – Every object in the universe attract other object along the line of center for two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.
Complete step by step answer:
According to the law of Gravitation,
$ F = \dfrac{{GMm}}{{{r^2}}} $ where $ F $ is the gravitational force, $ M $ is the mass of the sun, $ m $ is the mass of the planet, $ r $ is the distance between the sun and the planet.
It is stated that the planet moves round the sun in a nearly circular orbit.
Let ,
$ T = K{r^x}{M^y}{G^z} $ …(i) where $ K $ is a dimensionless constant. The dimensions of various quantities are-
$ \left[ T \right] = T $ , $ \left[ r \right] = L $ , $ \left[ M \right] = M $
$ \left[ G \right] = \dfrac{{F{r^2}}}{{Mm}} $ $ = \dfrac{{ML{T^{ - 2}}{L^2}}}{{MM}} $
$ = {M^{ - 1}}{L^3}{T^{ - 2}} $
By substituting these dimensionally in equation (i), we get
$ {M^0}{L^0}{T^1} = {M^{y - z}}{L^{x + 3z}}{T^{ - 2z}} $
Equating the dimensions,
$ y - z = 0 $ , $ x + 3z = 0 $ , $ - 2z = 1 $
On solving,
$ x = \dfrac{3}{2} $ , $ y = - \dfrac{1}{2} $ , $ z = - \dfrac{1}{2} $
$ T = K{r^{\dfrac{3}{2}}}{M^{ - \dfrac{1}{2}}}{G^{ - \dfrac{1}{2}}} $
Squaring both sides,
$ {T^2} = \dfrac{{{K^2}{R^3}}}{{MG}} $
$ \Rightarrow {T^2} \propto {r^3} $ .
Note:
Planets revolve in elliptical orbits around the sun. It is closest to the sun at perihelion and farthest from the sun at aphelion.
Applying Kepler’s Third Law – We assume that the planets revolve in circular orbits around the sun.
$ {F_C} = {F_G} $ where $ {F_C} $ is the centripetal force and $ {F_G} $ is the gravitational force. The centripetal force arises due to the gravitational force.
$ \dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}} $
$ v = \sqrt {\dfrac{{GM}}{r}} $ where $ v $ is the velocity with which the planet moves in an orbit around the sun.
$ T = \dfrac{{2\pi r}}{v} $ where $ T $ is the time period.
$ T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }} $
$ T = \dfrac{{2\pi {r^{\dfrac{3}{2}}}}}{{{G^{\dfrac{1}{2}}}{M^{\dfrac{1}{2}}}}} $
$ T \propto {r^{\dfrac{3}{2}}} $
Squaring both sides,
$ {T^2} \propto {r^3} $
Complete step by step answer:
According to the law of Gravitation,
$ F = \dfrac{{GMm}}{{{r^2}}} $ where $ F $ is the gravitational force, $ M $ is the mass of the sun, $ m $ is the mass of the planet, $ r $ is the distance between the sun and the planet.
It is stated that the planet moves round the sun in a nearly circular orbit.
Let ,
$ T = K{r^x}{M^y}{G^z} $ …(i) where $ K $ is a dimensionless constant. The dimensions of various quantities are-
$ \left[ T \right] = T $ , $ \left[ r \right] = L $ , $ \left[ M \right] = M $
$ \left[ G \right] = \dfrac{{F{r^2}}}{{Mm}} $ $ = \dfrac{{ML{T^{ - 2}}{L^2}}}{{MM}} $
$ = {M^{ - 1}}{L^3}{T^{ - 2}} $
By substituting these dimensionally in equation (i), we get
$ {M^0}{L^0}{T^1} = {M^{y - z}}{L^{x + 3z}}{T^{ - 2z}} $
Equating the dimensions,
$ y - z = 0 $ , $ x + 3z = 0 $ , $ - 2z = 1 $
On solving,
$ x = \dfrac{3}{2} $ , $ y = - \dfrac{1}{2} $ , $ z = - \dfrac{1}{2} $
$ T = K{r^{\dfrac{3}{2}}}{M^{ - \dfrac{1}{2}}}{G^{ - \dfrac{1}{2}}} $
Squaring both sides,
$ {T^2} = \dfrac{{{K^2}{R^3}}}{{MG}} $
$ \Rightarrow {T^2} \propto {r^3} $ .
Note:
Planets revolve in elliptical orbits around the sun. It is closest to the sun at perihelion and farthest from the sun at aphelion.
Applying Kepler’s Third Law – We assume that the planets revolve in circular orbits around the sun.
$ {F_C} = {F_G} $ where $ {F_C} $ is the centripetal force and $ {F_G} $ is the gravitational force. The centripetal force arises due to the gravitational force.
$ \dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}} $
$ v = \sqrt {\dfrac{{GM}}{r}} $ where $ v $ is the velocity with which the planet moves in an orbit around the sun.
$ T = \dfrac{{2\pi r}}{v} $ where $ T $ is the time period.
$ T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{GM}}{r}} }} $
$ T = \dfrac{{2\pi {r^{\dfrac{3}{2}}}}}{{{G^{\dfrac{1}{2}}}{M^{\dfrac{1}{2}}}}} $
$ T \propto {r^{\dfrac{3}{2}}} $
Squaring both sides,
$ {T^2} \propto {r^3} $
Recently Updated Pages
Which element possesses the biggest atomic radii A class 11 chemistry JEE_Main

The highly efficient method of obtaining beryllium class 11 chemistry JEE_Main

Which of the following sulphates has the highest solubility class 11 chemistry JEE_Main

Amongst the metal Be Mg Ca and Sr of group 2 of the class 11 chemistry JEE_Main

Which of the following metals is present in the greencolored class 11 chemistry JEE_Main

To prevent magnesium from oxidation in the electrolytic class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

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

List out three methods of soil conservation

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE
