Match the column -I with column -II for a sat | vedclass

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(D) For a satellite of mass $m$ in a circular orbit of radius $r$ around the Earth of mass $M_E$:$1$. Kinetic energy $(K)$: The centripetal force is p

Vedclass · Accepted Answer

$A-s, B-r, C-p, D-q$

Vedclass · Answer

(D) For a satellite of mass $m$ in a circular orbit of radius $r$ around the Earth of mass $M_E$:$1$. Kinetic energy $(K)$: The centripetal force is provided by gravitational force,so $\frac{mv^2}{r} = \frac{GM_Em}{r^2}$,which gives $v^2 = \frac{GM_E}{r}$. Thus,$K = \frac{1}{2}mv^2 = \frac{GM_Em}{2r}$. This matches $(s)$.$2$. Potential energy $(U)$: The gravitational potential energy is given by $U = -\frac{GM_Em}{r}$. This matches $(r)$.$3$. Total energy $(E)$: $E = K + U = \frac{GM_Em}{2r} - \frac{GM_Em}{r} = -\frac{GM_Em}{2r}$. This matches $(p)$.$4$. Orbital velocity $(v)$: $v = \sqrt{\frac{GM_E}{r}}$. This matches $(q)$.Therefore,the correct matching is $A-s, B-r, C-p, D-q$.

Column $-I$	Column $-II$
$(A)$ Kinetic energy	$(p)$ $-\frac{GM_Em}{2r}$
$(B)$ Potential energy	$(q)$ $\sqrt{\frac{GM_E}{r}}$
$(C)$ Total energy	$(r)$ $-\frac{GM_Em}{r}$
$(D)$ Orbital velocity	$(s)$ $\frac{GM_Em}{2r}$

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