Suppose the gravitational force varies inversely as the $n t h$ power of distance. Then the time period of a planet in circular orbits of radius ' $R$ ' around the sun will be proportional to

R^{n}

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R^{⎛ ⎝ \frac{n - 1}{2} ⎞ ⎠}

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R^{⎛ ⎝ \frac{n + 1}{2} ⎞ ⎠}

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R^{⎛ ⎝ \frac{n - 2}{2} ⎞ ⎠}

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Solution

The correct option is C $R^{⎛ ⎝ \frac{n + 1}{2} ⎞ ⎠}$
$F = K R^{- n} = M R ω^{2} \Rightarrow ω^{2} = K R^{- (n + 1)}$
or $ω = K^{'} R^{\frac{- (n + 1)}{2}}$ [where $K^{'} = K^{1 / 2}$ , a constant]
$\frac{2 π}{T} \propto R^{\frac{- (n + 1)}{2}}$
$∴ T \propto R^{\frac{(n + 1)}{2}}$

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Similar questions

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to:

Q. A particle is moving with a uniform speed in a circular orbit of radius

R

in a central force inversely proportional to the

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power of

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. If the period of rotation of the particle is

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Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbits of radius 'R' around the sun will be proportional to

The correct option is C R⎛⎝n+12⎞⎠F=KR−n=MRω2⇒ω2=KR−(n+1)or ω=K′R−(n+1)2 [where K′=K1/2, a constant]2πT∝R−(n+1)2∴T∝R(n+1)2

Suppose the gravitational force varies inversely as the $n t h$ power of distance. Then the time period of a planet in circular orbits of radius ' $R$ ' around the sun will be proportional to

The correct option is C $R^{⎛ ⎝ \frac{n + 1}{2} ⎞ ⎠}$
$F = K R^{- n} = M R ω^{2} \Rightarrow ω^{2} = K R^{- (n + 1)}$
or $ω = K^{'} R^{\frac{- (n + 1)}{2}}$ [where $K^{'} = K^{1 / 2}$ , a constant]
$\frac{2 π}{T} \propto R^{\frac{- (n + 1)}{2}}$
$∴ T \propto R^{\frac{(n + 1)}{2}}$