Question

The velocity of a satellite in a parking orbit is

• A. $8Km/s$
• B. $3.1Km/s$
• C. $2.35Km/s$
• D. zero

Answer 1

The correct option is B $3.1Km/s$

The parking orbit is the orbits of geo-stationary satellite. Here time period, $T=24hr$

Using Kepler's law, $T^2=\frac{4{\pi }^2{r}^3}{GM}$ or $r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}$

Now , $v^2=\frac{GM}{r}=GM\times {\left(\frac{4{\pi }^2}{GM{T}^2}\right)}^{1/3}$

or $v={\left(\frac{2\pi GM}{T}\right)}^{1/3}={\left(\frac{2\pi \times 6.67\times {10}^{-11}\times 6\times {10}^{24}}{24\times 3600}\right)}^{1/3}=3.1\times {10}^{3}m/s=3.1km/s$