Question 201: Gravitation

Question

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely
proportional to the n
th
power of R. If the period of rotation of the particle is T, then :

Accepted Answer

We know, Central force in circular motion, F = $$m{\omega ^2}R$$
According to the question,
$$F \propto {1 \over {{R^n}}}$$
$$	herefore$$ $$m{\omega ^2}R$$ $$ \propto {1 \over {{R^n}}}$$
$$ \Rightarrow m{\omega ^2}R = {k \over {{R^n}}}$$
$$ \Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}$$
$$	herefore$$ $$\omega  \propto {1 \over {{R^{{{n + 1} \over 2}}}}}$$ .......(1)
And we know, $$T = {{2\pi } \over \omega }$$
$$	herefore$$ $$T \propto {1 \over \omega }$$ ...... (2)
From (1) and (2) we can conclude that,
$$T \propto {R^{{{n + 1} \over 2}}}$$

Answer

T $$ \propto $$ R
n/2

Answer

T $$ \propto $$ R
3/2
for any n

Answer

T $$ \propto $$ R
n/2 +1

Answer

T $$ \propto $$ R
(n+1)/2

Gravitation

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n th power of R. If the period of rotation of the particle is T, then :

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