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If a planet were suddenly stopped in its circular orbit, how much time it would take to fall onto the Sun. Assume the planets time-period of revolution as T. |
Answer» Solution : For the circular orbit, `T^(2)PROPR^(3)`. When the planet is falling onto the SUN, we can think of its path as a completely flattened ellipse, WHOSE semi-major axis `=(r)/(2)`. `rArr(T')^(2)prop((r)/(2))^(3)` Now,`((T')/(T))^(2)=((r//s)/(r))^(3)`, `(T')^(2)=(T^2)/(8)`, `T'=(T)/(2sqrt2)` The required TIME is `(T')/(2)` `rArrt=(T')/(2)=(T)/(4sqrt2)` (or) `t=(Tsqrt2)/(8)` |
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