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InterviewSolution
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A planet moves aruond the sun in an elliptical orbit such that its kinetic energy is `K_(1)` and `K_(2)` when it is nearest to the sun and farthest from the sun respectively. The smallest distance and the largest distance between the planet and the sun are `r_(1)` and `r_(2)` respectively.A. If total energy of the planet is `E` then `(r_(2))/(r_(1))=(E-K_(1))/(K_(2)-E)`B. If the total energy of the planet is `E`, then `(r_(2))/(r_(1))=(E-K_(2))/(E-K_(1))`C. If `r_(2)=2r_(1)`, the total energy of the planet energy of the planet in terms of `K_(1)` and `K_(2)` is `(2K_(1)-K_(2))`D. If `r_(2)=2r_(1)`, the total energy of the plenet energy of the planet in terms of `K_(1)` and `K_(2)` is `(2K_(2)-K_(1))` |
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Answer» Correct Answer - D `Fpropr^(-5/2)` `rArr F=(GMm)/(r^(5//2))` `rArr (mV_(0)^(2))/r=(GMm)/(r^(5//2))` `=V_(0)^(2)=(GM)/(r^(5//2))xxr` `rArr V_(0)^(2)=(GM)/(r^(3//2))` `:. V_(0)=(sqrt(GM))/(r^(3//4))` take logrithm on both sides `logV_(0)=lgosqrt(GM)-logr^(3//4)` `log_(e)V_(0)=-3/4log_(e)r+log_(e)sqrt(GM)` `y=mx+c` slope of `log_(e)V_(0)` versus `log_(e)r` is `m=-3/4` |
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