Practicing Success
An artificial satellite is describing an equatorial orbit at 3600 km above the earth’s surface. Calculate its period of revolution ? |
8.71 hrs 9.71 hrs 10.71 hrs 11.71 hrs |
8.71 hrs |
The time period of satellite is given by $T^2=\frac{4 \pi^2}{\mathrm{GM}}(\mathrm{R}+\mathrm{h})^3=\frac{4 \pi^2}{\mathrm{~g}} \frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{R}^2}$ $\mathrm{~T}=\frac{2 \pi}{\mathrm{R}} \sqrt{\frac{(\mathrm{R}+\mathrm{h})^3}{\mathrm{~g}}}=\frac{2 \pi}{6400 \times 10^3} \sqrt{\frac{10^{21}}{9.8}}\left[\text { as } \mathrm{R}+\mathrm{h}=(6400+3600) \times 10^3 \mathrm{~m}=10^7 \mathrm{~m}\right]$ = 31360.78 sec = 8.71 hrs |