- A$24 \,hr$
- ✓$100 \,min$
- C$84.6 \,min$
- D$6 \,hr$
Time period of polar satellites is about $100$ minutes polar satellites are low altitude satellites. $(h=500-800 \,km )$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $-I$ | Column $-II$ |
| $(A)$ Kinetic energy | $(p)$ $ - \frac{{G{M_E}m}}{{2r}}$ |
| $(B)$ Potential energy | $(q)$ $\sqrt {\frac{{G{M_E}}}{r}} $ |
| $(C)$ Total energy | $(r)$ $ - \frac{{G{M_E}m}}{{r}}$ |
| $(D)$ Orbital energy | $(s)$ $ \frac{{G{M_E}m}}{{2r}}$ |
(where $M_E$ is the mass of the earth, $m$ is mass of the satellite and $r$ is the radius of the orbit)

(IMAGE)
($1$) The total kinetic energy of the ring is
$[A]$ $\mathrm{M} \omega_0^2 \mathrm{R}^2$ $[B]$ $\frac{1}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$ $[C]$ $\mathrm{M \omega}_0^2(\mathrm{R}-\mathrm{r})^2$ $[D]$ $\frac{3}{2} \mathrm{M} \omega_0^2(\mathrm{R}-\mathrm{r})^2$
($2$) The minimum value of $\omega_0$ below which the ring will drop down is
$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$ $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$ $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$ $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$
Givin the answer quetion ($1$) and ($2$)