Suppose the gravitational force varies inversely as the nth power of distance, then the time period of a planet in a circular orbit of radius r around the Sun will be proportional to

Question

Suppose the gravitational force varies inversely as the nth power of distance, then the time period of a planet in a circular orbit of radius r around the Sun will be proportional to (A) r(1/2) (n + 1) (B) r(1/2) (n - 1) (C) rn (D) r(1/2) (n - 2)

Tardigrade Admin · Accepted Answer

Given F = (k/rn) Now F = (m v2/r) (centripetal force) = m ω 2 r ⇒ (k/rn) = m (. (2 π /T) .)2 r T2 propto rn + 1 T propto r(n + 1/2)

Q. Suppose the gravitational force varies inversely as the $n^{t h}$ power of distance, then the time period of a planet in a circular orbit of radius $r$ around the Sun will be proportional to

$r^{\frac{1}{2} (n + 1)}$

$r^{\frac{1}{2} (n - 1)}$

$r^{n}$

$r^{\frac{1}{2} (n - 2)}$

Given $F = \frac{k}{r ^{n}}$
Now $F = \frac{m v ^{2}}{r}$ (centripetal force) $= m ω^{2} r$
$\Rightarrow \frac{k}{r ^{n}} = m (\frac{2 π}{T})^{2} r$
$T^{2} \propto r^{n + 1}$
$T \propto r^{\frac{n + 1}{2}}$

Q. Suppose the gravitational force varies inversely as the nth power of distance, then the time period of a planet in a circular orbit of radius r around the Sun will be proportional to

r21​(n+1)

r21​(n−1)

rn

r21​(n−2)