Question

Problem 1.08. Suppose that the gravitational force varies inversely as the nth power of distance. In that situation,what will be the time period of a planet in a circular orbit of
radius R around the sun ?​

Answer 1

Answer:

$\displaystyle{T=2\pi \times\left[\dfrac{R^{n+1}}{GM}\right]^{1/2}}$

Explanation:

Let mass of sun is M , mass of planet m and radius be R .

Then centripetal force will be

F = m v² / R  ... ( i )

Since gravitational force is inversely power of radius then

F = G M m / Rⁿ   ... ( ii ) .

From ( i )  and  ( ii )  we have

m v² / R = G M m / Rⁿ

v² / R = G M / Rⁿ

$\displaystyle{v=\left[\dfrac{GM}{R^{n-1}}\right]^{1/2}}$

Now we know period of revolution :

T = 2 π R / v

$\displaystyle{T=\dfrac{2\pi R}{\left[\dfrac{GM}{R^{n-1}}\right]^{1/2}}}\\\\\\\displaystyle{T=2\pi R\times\left[\dfrac{R^{n-1}}{GM}\right]^{1/2}}\\\\\\\displaystyle{T=2\pi \times\left[\dfrac{R^{n+1}}{GM}\right]^{1/2}}$

Thus we get time period of a planet in a circular orbit of

radius R around the sun .

Answer 2

SOLUTION:-

The necessary centripetal force required for a planet to move round the sun:

=) Gravitational force exerted on it

$i.e \frac{Mv {}^{2} }{R} = \frac{GMem}{R} \\ \\ = > v = ( \frac{GM}{R{}^{n - 1} } ) {}^{ \frac{1}{2} }$

Now,

$T = \frac{2\pi \: R}{v} = 2\pi \: R \times ( \frac{ {R}^{n - 1} }{GMe} ) {}^{ \frac{1}{2} } \\ \\ = > 2\pi( \frac{ {R}^{2} \times {R}^{n - 1} }{GMe} ) {}^{ \frac{1}{2} } \\ \\ = > 2\pi( \frac{R{}^{ \frac{(n + 1)}{2} } }{(GMe) {}^{ \frac{1}{2} } } ) \\ \\ = > T =2π ×R {}^{( \frac{n + 1)}{2} }$

Hence, the time period planet in circular orbit of radius R around the sun.

Hope it helps ☺️