QNEET-2024-Ph-61: NEET 2024 Physics - Class 11 NEET-2024-Ph-61 Solution: The minimum energy required to launch a satellite of mass m…

**Core concept & governing formulas** - Gravitational potential energy of a mass $m$ at a distance $r$ from the centre of Earth: $U(r)=-\dfrac{GMm}{r}$. - Kinetic energy of a satellite in a circular orbit of radius $r$: $K=\dfrac12 mv^{2}$, with orbital speed $v$ obtained from the centripetal condition $\dfrac{mv^{2}}{r}= \dfrac{GMm}{r^{2}} \;\Rightarrow\; v^{2}= \dfrac{GM}{r}$. - Mechanical energy is conserved: $U_i+K_i = U_f+K_f$, where the subscript $i$ denotes the initial state (on the Earth’s surface) and $f$ the final state (circular orbit at altitude $2R$). --- **Step 1: Initial mechanical energy on the Earth’s surface** The satellite is initially at the Earth’s surface ($r=R$). Its potential energy is

If the satellite is launched from rest relative to the Earth’s surface, its initial kinetic energy is zero, $K_i=0$. Thus --- **Step 2: Final mechanical energy in the required orbit** The required circular orbit is at an altitude $2R$ above the surface, so the orbital radius is *Potential energy at $r_f$* *Orbital speed at $r_f$* *Total final mechanical energy* --- **Step 3: Minimum work (energy) required to reach the orbit** The minimum external energy that must be supplied equals the increase in mechanical energy: --- **Final conclusion & correct option** The minimum energy required is which corresponds to option **A**.