JEE Main 2026 — 21 January, Morning Shift — Question 41

Energy of a satellite in a circular orbit is given as $\mathrm{E}=\frac{-\mathrm{GM}_{\mathrm{E}} \mathrm{m}}{2 \mathrm{r}} ; \mathrm{r}=$ radius of circular orbit Required energy to be supplied $=\Delta \mathrm{E}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}$

$$\begin{aligned} \Delta \mathrm{E} & =\left(\frac{-\mathrm{GM}_{\mathrm{E}} \mathrm{~m}}{2\left(3 \mathrm{R}_{\mathrm{E}}\right)}\right)-\left(\frac{-\mathrm{GM}_{\mathrm{E}} \mathrm{~m}}{2\left(1.5 \mathrm{R}_{\mathrm{E}}\right)}\right) & =\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{~m}}{6 \mathrm{R}_{\mathrm{E}}} \end{aligned}$$

Now, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^{2}} \Rightarrow \frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}=\mathrm{gR}_{\mathrm{E}}$

$$\begin{aligned} \therefore \quad \Delta \mathrm{E} & =\frac{1}{6} \mathrm{gmR}_{\mathrm{E}} & =\frac{1}{6} \times 10 \times 100 \times 6 \times 10^{6} & =1000 \times 10^{6} \end{aligned}$$

$\alpha=1000$