\[(a)\;\frac{4 \pi^2}{GM} \quad (b)\;\frac{GM}{4 \pi ^2} \quad (c)\;4\pi GM \quad (d)\;0 \]

Slope of grap between $T^2 \;and \;r^3$

$\quad= \large\frac{T^2}{r^3}$

$\quad= \large\frac{\bigg( \Large\frac{2 \pi r}{v_0}\bigg)^2}{r^3}$

$\large\frac{mv_0^2}{r}=\frac{GMm}{r^2}$

$\therefore v_0^2 =\large\frac{GM}{r}$

$\quad = \large\frac{(2 \pi r)^2}{r^3} \times \large\frac{1}{GM}$$r$

$\quad= \large\frac{4 \pi ^2}{GM}$

Hence a is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...