Question 445: Gravitation

Question

If gravitational attraction between two points masses be given by $F=G \frac{m_1 m_2}{r^n}$, then the period of a satellite in a circular orbit will be proportional to

Accepted Answer

The problem involves determining how the period of a satellite, in a circular orbit, relates to the radius $ r $ when the gravitational force between two point masses is given by:
$ F = G \frac{m_1 m_2}{r^n} $
To solve this, let's analyze the forces and relationships involved:
Centripetal Force and Gravitational Force:
The gravitational force $ F $ provides the necessary centripetal force for the satellite's circular motion:
$    \frac{m v^2}{r} = \frac{G m_1 m_2}{r^n}    $
Thus, solving for velocity $ v $, we find:
$    v \propto r^{\left(\frac{1-n}{2}\right)}    $
Relating Velocity to the Period $ T $:
The period $ T $ of the satellite is the time to complete one orbit, given by the circumference of the orbit divided by velocity:
$    T = \frac{2 \pi r}{v}    $
Therefore, substituting the expression for $ v $ from above:
$    T \propto \frac{r}{r^{\left(\frac{1-n}{2}\right)}}    $
Simplifying the Proportionality:
Simplifying the expression for $ T $, we get:
$    T \propto r^{\frac{1+n}{2}}    $
In conclusion, the period of the satellite in a circular orbit is proportional to $ r^{\frac{n+1}{2}} $.

Answer

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

Answer

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

Answer

$r^{\frac{n}{2}}$

Answer

independent of $n$

Gravitation

If gravitational attraction between two points masses be given by F=G \frac{m_1 m_2}{r^n}, then the period of a satellite in a circular orbit will be proportional to

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