Imagine a light planet revolving around a very massive star in a circular orbit of radius .r. with a period of revolution T On what power of .r. will the square of time period depend on the gravitational force of attraction between the planet and the star is proportional to r^(-5//2)

Imagine a light planet revolving around a very massive star in a circu

1.	Imagine a light planet revolving around a very massive star in a circular orbit of radius .r. with a period of revolution T On what power of .r. will the square of time period depend on the gravitational force of attraction between the planet and the star is proportional to r^(-5//2)
Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> gravitational force <a href="https://interviewquestions.tuteehub.com/tag/provides-1171126" style="font-weight:bold;" target="_blank" title="Click to know more about PROVIDES">PROVIDES</a> <a href="https://interviewquestions.tuteehub.com/tag/necessary-1112675" style="font-weight:bold;" target="_blank" title="Click to know more about NECESSARY">NECESSARY</a> centripetal force `(mV^<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)/<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>=(K)/(r^(5//2))impliesV^(2)=K/(mr^(3//2))` <br/> So that `T = (2pir)/V=2pirsqrt(("mr"^(3//2))/K)` <br/> So `T^(2) prop r^(7//2)`</body></html>