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If a planet was suddenly stopped in its orbit supposed to be circular, show that it would fall onto the sun in a time equal to (sqrt2 //8)times the period of planet.s revolution.

Answer» <html><body><p></p>Solution :If M is mass of the <a href="https://interviewquestions.tuteehub.com/tag/sun-1234558" style="font-weight:bold;" target="_blank" title="Click to know more about SUN">SUN</a> and r is radius of the planet.s orbit, its orbital <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> is `V_0 = sqrt((GM)/(r ))`<br/>Orbital time period `T = (2pi r)/(v_0) " or " T^2 = (4pi^2 r^3)/(GM)`<br/> After stopping the planet, if it has velocity <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>, when it is at a distance x from the sun, from conservation of <a href="https://interviewquestions.tuteehub.com/tag/mechanical-2780" style="font-weight:bold;" target="_blank" title="Click to know more about MECHANICAL">MECHANICAL</a> energy,<br/>` 1/2 mv^2 - (GMm)/(x) = - (GMm)/(r ) <a href="https://interviewquestions.tuteehub.com/tag/ie-496825" style="font-weight:bold;" target="_blank" title="Click to know more about IE">IE</a>, - ((dx)/(dt))^2 = (2GM)/(r ) [ (r - x)/(x)]`<br/>`(-dx)/(dt) = sqrt((2GM)/(r )) sqrt((r-x)/(x)) ,int_0^t dt = - sqrt((r)/(2GM)) int_r^0 ( sqrt((x)/(r - x)) )dx ` After solving , we get `t = (sqrt2)/(8) T`</body></html>


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