Question

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

• A. $ R^{\left(\frac{n+1}{2}\right)} $
• B. $ R^{\left(\frac{n-1}{2}\right)} $
• C. $ R^n $
• D. $ R^{\left(\frac{n-2}{2}\right)} $

Answer 1

The Correct Option is A

The necessary centripetal force required for a planet to move round the sun
$=$ gravitational force exerted on it
i.e., $\frac{mv^{2}}{R} = \frac{GM_{e} m}{R^{n}}$
$ v = \left(\frac{GM_{e}}{R^{n-1}}\right)^{1/2}$
Now, $T = \frac{2\pi R}{v} = 2\pi R \times \left(\frac{R^{n-1}}{GM_{e}}\right)^{1/2}$
$ \Rightarrow 2\pi \left(\frac{R^{2} \times R^{n-1}}{GM_{e}}\right)^{1/2} $
$T = 2\pi\left(\frac{R^{\left(n+1\right)/2}}{\left(GM_{e}\right)^{1/2}}\right) $
$\therefore T \propto R^{\left(n+1\right)/2}$

Answer 2

The gravitational force typically varies inversely as the square of the distance (\( n = 2 \)) according to Newton's law of universal gravitation. However, if the gravitational force varies inversely as the \( n \)-th power of the distance, the relationship between the time period \( T \) of a planet in a circular orbit of radius \( R \) and the radius \( R \) itself will change.

Let's denote the force as:
\[ F \propto \frac{1}{R^n} \]

 Step-by-Step Solution:

1. Expression for the Gravitational Force:
  The gravitational force between two masses \( M \) (mass of the Sun) and \( m \) (mass of the planet) at a distance \( R \) can be written as:
  \[ F = \frac{kMm}{R^n} \]
  where \( k \) is a constant of proportionality.

2. Centripetal Force and Orbital Motion:
  For a planet in a circular orbit, the gravitational force provides the necessary centripetal force to keep the planet in orbit. Therefore,
  \[ \frac{kMm}{R^n} = \frac{mv^2}{R} \]
  where \( v \) is the orbital speed of the planet.

3. Orbital Speed:
  Solving for \( v \):
  \[ \frac{kM}{R^n} = \frac{v^2}{R} \]
  \[ v^2 = \frac{kM}{R^{n-1}} \]
  \[ v = \sqrt{\frac{kM}{R^{n-1}}} \]

4. Orbital Period:
  The time period \( T \) of the planet is the time it takes to complete one full orbit, which is the circumference of the orbit divided by the orbital speed:
  \[ T = \frac{2\pi R}{v} \]
  Substituting \( v \) from above:
  \[ T = \frac{2\pi R}{\sqrt{\frac{kM}{R^{n-1}}}} \]
  \[ T = \frac{2\pi R}{\sqrt{kM} \cdot R^{\frac{1-n}{2}}} \]
  \[ T = \frac{2\pi R^{\frac{n+1}{2}}}{\sqrt{kM}} \]

5. Proportionality:
  Ignoring the constants \( 2\pi \) and \( \sqrt{kM} \), the time period \( T \) is proportional to:
  \[ T \propto R^{\frac{n+1}{2}} \]

 Conclusion:
The time period \( T \) of a planet in a circular orbit of radius \( R \) around the sun, when the gravitational force varies inversely as the \( n \)-th power of the distance, is proportional to \( R^{\frac{n+1}{2}} \).
Hence the correct Answer is Option(A)(1): \( R^{\frac{n+1}{2}} \)