Match List-I with List-II:List-IList-II(A) Ki | vedclass

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(A) For a planet in a circular orbit of radius $a$ around the Sun of mass $M$:$1$. Kinetic Energy $(KE)$: The orbital velocity is $v = \sqrt{\frac{GM}

Vedclass · Accepted Answer

$(A) - II, (B) - I, (C) - IV, (D) - III$

Vedclass · Answer

(A) For a planet in a circular orbit of radius $a$ around the Sun of mass $M$:$1$. Kinetic Energy $(KE)$: The orbital velocity is $v = \sqrt{\frac{GM}{a}}$. Thus,$KE = \frac{1}{2}mv^2 = \frac{GMm}{2a}$. This matches $(2)$.$2$. Gravitational Potential Energy $(PE)$: The potential energy of the system is $PE = -\frac{GMm}{a}$. This matches $(1)$.$3$. Total Mechanical Energy $(TE)$: $TE = KE + PE = \frac{GMm}{2a} - \frac{GMm}{a} = -\frac{GMm}{2a}$. This matches $(4)$.$4$. Escape energy at the surface of the planet (mass $m_p$,radius $r$): The energy required to escape is equal to the magnitude of the gravitational potential energy at the surface,which is $\frac{GM_p m}{r}$. For a unit mass object $(m=1)$,this is $\frac{GM_p}{r}$. Assuming the question refers to the planet's own gravitational field,this matches $(3)$.Therefore,the correct matching is $(A) - II, (B) - I, (C) - IV, (D) - III$.

List-$I$	List-$II$
$(A)$ Kinetic energy of planet	$(1)$ $-\frac{GMm}{a}$
$(B)$ Gravitational potential energy of Sun-planet system	$(2)$ $\frac{GMm}{2a}$
$(C)$ Total mechanical energy of planet	$(3)$ $\frac{GM}{r}$
$(D)$ Escape energy at the surface of planet for unit mass object	$(4)$ $-\frac{GMm}{2a}$

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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