Q.A geostationary satellite has a time period of exactly $24\ \text{h}$. Taking $M = 6 \times 10^{24}\ \text{kg}$, $R = 6400\ \text{km}$ and $G = 6.674 \times 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$, find

Given $M = 6\times 10^{24}\ \text{kg}$, $R = 6400\ \text{km}$, $G = 6.674\times 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$, so $GM \approx 4.00\times 10^{14}\ \text{m}^3/\text{s}^2$, and $T = 24\ \text{h} = 86400\ \text{s}$.

1. Orbital radius.

   $r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} = \left(\frac{(4.00\times 10^{14})(86400)^2}{4\pi^2}\right)^{1/3} \approx 4.24\times 10^7\ \text{m} = 4.24\times 10^4\ \text{km}$

2. Height above the surface.

   $h = r - R \approx 42{,}400\ \text{km} - 6400\ \text{km} \approx 3.60\times 10^4\ \text{km}$

3. Orbital speed. …

Find $r$ from the rearranged period relation, subtract $R$ for the height, then find $v_o$ either from $2\p …

• Forgetting to subtract the Earth's own radius $R$ when asked for the HEIGHT above the surface, rather than the orbital radius from the Earth's centre. …