Question

The period of revolution (t) of a planet around the sun depends upon radius (r) of the orbit, mass (m) of the sun and gravitational constant (G). Prove that t² ∝ r³

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Answer 1

$\large \bf \clubs \:  To \: Prove :-$

The square of Time Period of a Planet revolving around the Sun is directly proportional to cube of Radius of orbit .

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$\large \bf \clubs \:  Proof:-$

Here,

The Gravitational Force between Planet and the Sun provide enough Centripetal force for the Revolution of the Planet .

Hence Equating Gravitational Force and centripetal Force :

$\sf \dfrac{GMm}{ {r}^{2} } = mr { \omega}^{2} \\ \\ \sf\frac{GM}{ {r}^{3} } = { \bigg(\frac{2 \pi}{t} \bigg)}^{2} \: \: \: ( \because \: \omega = { \frac{2\pi}{t} )}\\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf  {t}^{2} = \frac{4 {\pi}^{2} \: {r}^{3} }{GM} \propto {r}^{3} } }}}$

Hence,

$\purple{ \Large :\longmapsto  \underline {\boxed{{\bf  {t}^{2} \propto {r}^{3} } }}}$

$\red{ \underline{ \underline{\LARGE\bf \bigstar \: Hence \: Proved \: \bigstar}}}$

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$\large \bf \clubs \:  Notations \: Used:-$

• t = Time Period of Revolution

• m = mass of Planet

• M = mass of Sun

• r = radius of Orbit

• ω = Angular Frequency

• G = Gravitational Constant

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Answer 2

Answer:

Dimensional Analysis:

Given :

The period of revolution (t) of a planet around the sun depends upon radius (r) of the orbit, mass (m) of the sun and gravitational constant (G), mathematically :

$\implies \sf [T] \propto [r]^a [m]^b [G]^c$

Adding k as dimensionaless constant :

$\implies \sf [T] = k [r]^a [m]^b [G]^c \: \: \: \: \:... (i)$

• DIMENSIONS:

$\dag \bf \: Dimensions \: of \: period\:of\: revolution \: (t) = [M^0L^0T^1]$

$\dag \bf \: Dimensions \: of \: radius \: (r) = [M^0L^1T^0]$

$\dag \bf \: Dimensions \: of \: mass \: (m) = [M^1L^0T^0]$

$\dag \bf \: Dimensions \: of \: gravitational \:constant \: (G) = [M^{ - 1} L^3T^{ - 2} ]$

$\implies \sf [M^0L^0T^1] =[M^0L^1T^0] ^a [M^1L^0T^0]^b [M^{ - 1} L^3T^{ - 2} ]^c$

$\implies \sf [M^0L^0T^1] =[M]^{b - c} [L]^{a + 3c} [T]^{ - 2c}$

• On comparing the powers we have:

$\implies b - c = 0$

$\implies b = c \: \: \: ....(ii)$

$\implies a + 3c = 0 \: \: \: ....(iii)$

$\implies - 2c = 1$

$\implies \bf c = \dfrac{ -1 }{2} \: \: \: ....(iv)$

Substituting value of c = -1/2 from equation (iv) to equation (ii) :

$\implies b = c$

$\implies \bf b = \dfrac{ - 1}{2}$

Also substitue the value of c from equation (iv) to equation (ii)

$\implies a + 3c = 0$

$\implies a + 3 \times \dfrac{ - 1}{2} = 0$

$\implies a - \dfrac{ 3}{2} = 0$

$\implies \bf a = \dfrac{ 3}{2}$

Now, Substituting the value of a, b and c in the equation (I):

$\implies \sf [T] = k [r]^{ \frac{3}{2} }[m]^{ \frac{ - 1}{2} } [G]^{ \frac{ - 1}{2} }$

$\implies \sf [T]^{2} = k [r]^{3 }[m]^{ - 1 } [G]^{ - 1 }$

$\implies \sf [T]^{2} = k [r]^{3 }$

$\implies \sf [T]^{2} \propto [r]^{3 }$

$\implies \underline{ \boxed{ \orange{\bf T^{2} \propto r^{3 }}}}$

Hence, Proved!