Q&A - Ask Doubts and Get Answers
Q

How to solve this problem- 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 :

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 :

  • Option 1)

    R^{\left ( \frac{n+1}{2} \right )}

  • Option 2)

    R^{\left ( \frac{n-1}{2} \right )}

  • Option 3)

    R^{n}

  • Option 4)

    R^{\left ( \frac{n-2}{2} \right )}

 
Answers (1)
1744 Views

As we learnt in 

For motion of a planet in circular orbit,

Centripetal force = Gravitational force

\therefore \; \; mR\omega ^{2}=\frac{GMm}{R^{n}}\; \; or\; \; \omega =\sqrt{\frac{GM}{R^{n+1}}}

\therefore \; \; \; T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R^{n+1}}{GM}}=\frac{2\pi }{\sqrt{GM}}R^{\left ( \frac{n+1}{2} \right )}

\therefore \; \; T\; is\; proportional\; to\; R^{\left ( \frac{n+1}{2} \right )}

 

 

Newton's Law of Gravitation -

F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}

F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}

F\rightarrow Force    

G\rightarrow Gravitalional constant

m_1,m_2\rightarrow  Masses

r\rightarrow  Distance between masses

- wherein

Force is along the line joining the two masses

 

m\omega ^2R = \frac{Gm_{m}}{R^{n}}

W = \sqrt{\frac{Gm}{R^{n+1}}}

\therefore T= \frac{2\pi }{w} = 2\pi \sqrt{\frac{R^{n+1}}{Gm}}

= \frac{2\pi }{\sqrt{Gm}}R

T\ is\ Proportional\ to\ R \left ( \frac{n+1}{2} \right )

 


Option 1)

R^{\left ( \frac{n+1}{2} \right )}

Correct

Option 2)

R^{\left ( \frac{n-1}{2} \right )}

Incorrect

Option 3)

R^{n}

Incorrect

Option 4)

R^{\left ( \frac{n-2}{2} \right )}

Incorrect

Exams
Articles
Questions